2D Mensuration: Paths, Costs and Combined Figures

Competitive Mathematics · SSC CGL

5 Part42 worked examples164 practice questions

The questions of this chapter look frightening in SSC papers — paths, grazing, shaded flowers — but all rest on one idea: a complicated figure is made of simple figures (rectangles, squares, triangles, circles) added or subtracted. The first job in every solution is to see which simple figures are there, and which are added and which subtracted.

◭ Diagram

Example: the shaded part = square − circle. The circle goes out and the four corner pieces remain

∑ Formula

A_{shaded} = \sum A_{added} - \sum A_{removed}

One rule for every complicated figure

$A_{shaded}$: the part whose area is asked (square units)
$A_{added}$: areas of the simple figures that are added
$A_{removed}$: areas of the simple figures that are removed

Tool	When	Example
Subtract	a figure with a hole or an inner part	path, ring, square minus circle
Add	a figure made of separate pieces	grazing (¾ + two ¼), track (rectangles + ring)
Slide	pieces can be moved to make the figure simple	crossing roads, the petal

RememberIn every question: (1) draw the figure and write all measures; (2) decide which part is asked; (3) write that part as a sum and difference of simple figures; (4) do the last step (cost / percent / ratio). All measures in one unit, and

π = \frac{22}{7}

only when the radius is a multiple of 7.

The order of this chapter: paths (round rectangles and circles), costs (fences, carpets and tiles, walls, grazing, tracks), combined figures (one inside another, packing coins, shaded regions) and changes (percent and reshaping a wire). The basic formulas for triangles and quadrilaterals, and for circles and polygons, are in the previous two chapters of this topic (parts 1 and 2).

Paths and borders

Outside · inside · crossing · circular

A path of even width round or inside a rectangle or circle, and two paths crossing in the middle.

Path outside or inside a rectangle

Outer − inner, and

2 w (l + b) \pm 4 w^{2}

The area, cost and width of an even strip round a ground, garden, frame or pool.

When an even path (of width $w$ ) runs round a rectangular ground, there is one way to find its area: area of the big figure − area of the small figure. The only question is which one is big and which small.

Path outside: the ground is the small figure. The path is added at both ends, so length and breadth both grow by $2 w$ : the big figure is $(l + 2 w) \times (b + 2 w)$ .
Path inside: the ground is the big figure and the middle part left over is the small one. Both measures shrink by $2 w$ : the small figure is $(l - 2 w) \times (b - 2 w)$ .

◭ Diagram

Outside: two long strips (

l \times w

, blue), two side strips (

b \times w

, green) and four corner squares (

w^{2}

, orange). Inside: two strips

l \times w

and two shorter strips

(b - 2 w) \times w

The picture gives the formula by itself. The outside path has four strips and four corner squares: $2 lw + 2 b w + 4 w^{2}$ . In the inside path the strips meet at the corners, so the four corner squares, which would be counted twice, are taken off: $2 lw + 2 b w - 4 w^{2}$ .

∑ Formula

A_{out} = 2 w (l + b) + 4 w^{2} A_{in} = 2 w (l + b) - 4 w^{2}

Area of the path

$A_{out}, A_{in}$: area of the outside / inside path (square units)
$l, b$: length and breadth of the ground (in units)
$w$: width of the path, in the same unit

A neat shortcut: the dotted line running exactly through the middle of the path forms a rectangle. Area of the path $= w \times$ (length of that middle line). Check: for the outside path the middle rectangle has sides $l + w$ and $b + w$ , perimeter $2 (l + b + 2 w)$ , and $w \times 2 (l + b + 2 w) = 2 w (l + b) + 4 w^{2}$ ✓. Another useful fact: outer perimeter − inner perimeter $= 8 w$ (two extra $w$ at each of the four corners).

Spot itSpot it: “path / border / frame / thickness of a wall” + “all round” → big − small. If it says “outside” increase the measures, if “inside” decrease them. Trap: in an inside path the given measure belongs to the outer rectangle.

Draw the figure; decide whether the path is outside or inside the ground.
Outside: write $l + 2 w, b + 2 w$ ; inside: $l - 2 w, b - 2 w$ (the $2 w$ is for both ends).
Find both rectangle areas and subtract (or use $2 w (l + b) \pm 4 w^{2}$ and check).
If the cost is asked, area × rate. If the width is asked, outer side − inner side $= 2 w$ .

Example 1 · Basic

A temple courtyard 30 m × 20 m has a 2 m wide border all round the outside. Find the area of the border and the cost of laying stone on it at ₹45 per sq m.

Solution, step by step

1The border is outside, so the whole figure (courtyard + border) grows by 2 m on every side: length $30 + 2 \times 2 = 34$ m and breadth $20 + 2 \times 2 = 24$ m.
2Outer area $= 34 \times 24 = 816$ sq m; courtyard $= 30 \times 20 = 600$ sq m.
3Border $= 816 - 600 = 216$ sq m. Check: $2 w (l + b) + 4 w^{2} = 2 \times 2 \times 50 + 4 \times 4 = 200 + 16 = 216$ ✓.
4Cost $= 216 \times 45 = 9720$ rupees.

Answer 216 sq m; ₹9,720

Trap: using only

2 w (l + b) = 200

misses the four corner squares (16).

Example 2 · Standard

In a hall 18 m × 12 m a carpet border 1.5 m wide is laid along the walls on the inside. Find the area of the border and the area of the bare floor in the middle.

Solution, step by step

1The border is inside, so the bare middle loses 1.5 m at both ends: $18 - 2 \times 1.5 = 15$ m and $12 - 2 \times 1.5 = 9$ m.
2Bare floor in the middle $= 15 \times 9 = 135$ sq m.
3The whole hall $= 18 \times 12 = 216$ sq m, so the border $= 216 - 135 = 81$ sq m.
4Formula check: $2 w (l + b) - 4 w^{2} = 2 \times 1.5 \times 30 - 4 \times 2.25 = 90 - 9 = 81$ ✓.

Answer border 81 sq m; bare floor 135 sq m

Middle-line shortcut: for an inside path the middle rectangle is

(18 - 1.5) \times (12 - 1.5) = 16.5 \times 10.5

, perimeter

54

, and

1.5 \times 54 = 81

✓.

Example 3 · Advanced

A square field has side 24 m. An even path runs round it on the outside, and the field and the path together cover 841 sq m. Find the width of the path and the cost of cementing the path at ₹120 per sq m.

Solution, step by step

1The whole figure is also a square (an even strip round a square), so its side $= 841 = 29$ m.
2This side exceeds the field's side by the path width twice (both ends): $29 - 24 = 5 = 2 w$ ⟹ $w = 2.5$ m.
3Area of the path $= 841 - 2 4^{2} = 841 - 576 = 265$ sq m.
4Cost $= 265 \times 120 = 31, 800$ rupees.

Answer $w = 2.5$ m; cost ₹31,800

No quadratic is needed to find the width — take the root of the total area and halve the difference of the sides.

BasicTry it yourself first

A rectangular garden measures 40 m × 25 m. A path 3 m wide runs all round it on the outside. Find the area of the path.

BasicTry it yourself first

A rectangular swimming pool measures 16 m × 10 m. A path 1 m wide runs along all four edges on the inside. Find the area of the path.

StandardTry it yourself first

A rectangular playground measures 32 m × 24 m. A 2.5 m wide border all round the outside is to be tiled at ₹140 per sq m. Find the total cost.

StandardTry it yourself first

A rectangular pond measures 15.5 m × 10 m. A path 1.5 m wide runs all round it on the outside. Find the area of the path.

StandardTry it yourself first

A square ground has side 36 m. A path of equal width runs round it on the outside. The ground and the path together cover 2,025 sq m. Find the width of the path.

AdvancedTry it yourself first

A rectangular garden measures 36 m × 28 m. An even path runs round it on the outside, and the outer boundary of the path is 40 m longer than the garden's perimeter. Find the area of the path.

AdvancedTry it yourself first

A rectangular school ground measures 25 m × 18 m. A path 2.5 m wide runs along all four edges on the inside. Find the area of the path.

AdvancedTry it yourself first

A square ground has side 50 m. A path of equal width runs round it on the outside. The ground and the path together cover 3,844 sq m. Find the width of the path.

Two paths crossing in the middle

w (l + b - w)

and the lawn left over

Two paths (parallel to the length and the breadth) cutting through the middle: area of the paths, the part left over, cost and width.

When two paths run through a park or field — one parallel to the length and one parallel to the breadth — a shared crossing (a small rectangle) forms where they meet. If you add the two path areas separately this crossing is counted twice, so it has to be subtracted once.

◭ Diagram

Left: the paths (blue) and the shared crossing (orange); four green pieces are left. Right: push the paths to the edges and the green part becomes one rectangle

(l - w_{2}) (b - w_{1})

∑ Formula

A_{roads} = l w_{1} + b w_{2} - w_{1} w_{2} A_{lawn} = (l - w_{2}) (b - w_{1})

For equal widths

w

A_{roads} = w (l + b - w)

$l, b$: length and breadth of the ground (m)
$w_{1}$: width of the path running parallel to the length (m); it takes $w_{1}$ out of the breadth $b$
$w_{2}$: width of the path running parallel to the breadth (m); it takes $w_{2}$ out of the length $l$
$A_{roads}, A_{lawn}$: area of the paths and of the part left over (the lawn), in sq m

The second formula is very handy: whether the paths run through the middle or along the edges, the part left over is the same — slide the left-over pieces together and they form one rectangle $(l - w_{2}) \times (b - w_{1})$ (the right-hand picture). So if the left-over part is asked, there is no need to find the area of the paths.

Spot itSpot it: “two paths cross each other / crossing / through the middle”. Traps: (1) adding the crossing twice; (2) taking the width of the length-wise path off

l

— it comes off

b

Decide which path is parallel to the length (width $w_{1}$ ) and which to the breadth (width $w_{2}$ ).
Area of the paths: $l w_{1} + b w_{2} - w_{1} w_{2}$ (or $w (l + b - w)$ for equal widths).
Left-over part: directly $(l - w_{2}) (b - w_{1})$ , or total $l b$ minus the paths.
Cost = area × rate; if the width is unknown, the smaller root of $w^{2} - (l + b) w + A = 0$ .

Example 4 · Basic

Through a park of 60 m × 40 m runs one path parallel to each side, both 3 m wide. Find the area of the paths and the cost of spreading gravel on them at ₹25 per sq m.

Solution, step by step

1Path along the length: $60 \times 3 = 180$ sq m; path along the breadth: $40 \times 3 = 120$ sq m.
2The shared crossing $3 \times 3 = 9$ sq m is counted in both; subtract it once: $180 + 120 - 9 = 291$ sq m.
3By the formula: $w (l + b - w) = 3 \times (100 - 3) = 291$ ✓.
4Cost $= 291 \times 25 = 7275$ rupees.

Answer 291 sq m; ₹7,275

Trap: writing

180 + 120 = 300

counts the crossing twice.

Example 5 · Standard

Two 2 m wide paths (one parallel to each side) cross in the middle of a ground 50 m × 36 m. Find the area of the paths and the area of the grassy part that is left.

Solution, step by step

1Paths: $w (l + b - w) = 2 \times (50 + 36 - 2) = 2 \times 84 = 168$ sq m.
2The whole ground $= 50 \times 36 = 1800$ sq m, so the grassy part $= 1800 - 168 = 1632$ sq m.
3Another way: push the paths to the edges and the grass $= (50 - 2) \times (36 - 2) = 48 \times 34 = 1632$ ✓ — the same answer, in one line.

Answer paths 168 sq m; grass 1632 sq m

When only the left-over part is asked,

(l - w) (b - w)

is the fastest.

Example 6 · Advanced

Through a 90 m × 60 m field runs a path 5 m wide parallel to the length and a path 4 m wide parallel to the breadth. Find (a) the area of the paths, (b) the cost of turfing the rest at ₹8 per sq m.

Solution, step by step

1Here $w_{1} = 5$ (parallel to the length, taking 5 out of the breadth 60) and $w_{2} = 4$ (parallel to the breadth, taking 4 out of the length 90).
2Paths $= l w_{1} + b w_{2} - w_{1} w_{2} = 90 \times 5 + 60 \times 4 - 5 \times 4 = 450 + 240 - 20 = 670$ sq m.
3Left-over part $= (l - w_{2}) (b - w_{1}) = (90 - 4) \times (60 - 5) = 86 \times 55 = 4730$ sq m. Check: $5400 - 670 = 4730$ ✓.
4Cost of turf $= 4730 \times 8 = 37, 840$ rupees.

Answer (a) 670 sq m (b) ₹37,840

Trap: writing

(l - w_{1}) (b - w_{2})

(swapping the widths) — that would give

85 \times 56 = 4760

, which is wrong.

BasicTry it yourself first

A rectangular garden measures 45 m × 30 m. Two paths cross each other through the middle — one parallel to the length and one parallel to the breadth — each 2 m wide. Find the total area of the paths.

BasicTry it yourself first

Through the middle of an 80 m × 50 m field run a road 4 m wide parallel to the length and a road 3 m wide parallel to the breadth. Find the total area of the roads.

StandardTry it yourself first

Two paths run through the middle of a 70 m × 40 m park, 3 m wide parallel to the length and 2 m wide parallel to the breadth. The rest is to be turfed at ₹12 per sq m. Find the cost.

StandardTry it yourself first

A rectangular pond measures 62 m × 38 m. Two paths cross each other through the middle — one parallel to the length and one parallel to the breadth — each 3.5 m wide. Find the total area of the paths.

StandardTry it yourself first

Two roads of equal width (one parallel to each side) cross in the middle of a 50 m × 30 m ground and cover 231 sq m in all. Find the width of the road.

StandardTry it yourself first

Two roads, each 2 m wide (one parallel to each side), cross in the middle of a 50 m × 40 m ground. What percent of the ground do the roads use?

AdvancedTry it yourself first

Two paths run through the middle of a 120 m × 80 m park, 6 m wide parallel to the length and 5 m wide parallel to the breadth. The rest is to be turfed at ₹9 per sq m. Find the cost.

AdvancedTry it yourself first

Two roads of equal width (one parallel to each side) cross in the middle of a 60 m × 48 m ground and cover 416 sq m in all. Find the width of the road.

Path round or inside a circular ground

π w (2 r + w)

and the difference of circumferences

An even ring round a circular park, well, fountain or pond: area, cost and width.

An even path round a circular ground is a ring between two concentric circles (same centre). The same idea as with rectangles applies: big circle − small circle. Now radii take the place of length and breadth, and the width $w$ of the path is the difference of the two radii.

Outside path (ground of radius $r$ ): outer radius $R = r + w$ , inner $r$ .
Inside path (ground of radius $R$ ): outer $R$ , inner $R - w$ .

◭ Diagram

Ring = big circle − small circle. The green line is the inner radius, the orange line the width

w

; the dashed rose circle is the middle line of the path

∑ Formula

A = π (R^{2} - r^{2}) = π (R + r) (R - r) = π w (2 r + w)

Area of the ring; for an outside path

$A$: area of the path (ring), in square units
$R, r$: outer and inner radius (in units)
$w$: width of the path $= R - r$
$π$: $\frac{22}{7}$ (radius a multiple of 7) or $3.14$

The form $π (R + r) (R - r)$ is fastest: $R - r = w$ and $R + r = 2 r + w$ (or $2 R - w$ for an inside path). Shortcut: area of the path $= w \times$ (circumference of the middle circle), since the middle circle has radius $\frac{R + r}{2}$ and $w \times 2 π \frac{R + r}{2} = π w (R + r)$ ✓. And the difference of the two circumferences $2 π R - 2 π r = 2 π w$ — very useful to find the width.

Spot itSpot it: “circular park / well / pond / fountain” + “path / thickness of wall all round”. Traps: (1) using the diameter as the radius (halve a well's diameter first); (2) in an inside path the given radius is the outer one; (3) writing

π w^{2}

(that is not the ring).

Write the outer and inner radii (halve a diameter; mind outside/inside).
Apply $A = π (R + r) (R - r)$ .
If the width is asked, use $2 π w = C_{outer} - C_{inner}$ or $R - r$ ; cost = area × rate.

Example 7 · Basic

A circular park of radius 35 m has a path 7 m wide on the outside. Find the area of the path and the cost of paving it at ₹40 per sq m. (

π = \frac{22}{7}

)

Solution, step by step

1The path is outside: inner radius $r = 35$ m, outer radius $R = 35 + 7 = 42$ m.
2Area $= π (R + r) (R - r) = \frac{22}{7} \times 77 \times 7$ . Here $\frac{22}{7} \times 7 = 22$ , so $22 \times 77 = 1694$ sq m.
3Cost $= 1694 \times 40 = 67, 760$ rupees.

Answer 1694 sq m; ₹67,760

Pick the factor that cancels the 7 first — the arithmetic gets easy.

Example 8 · Standard

A circular well has an inner diameter of 7 m and a masonry parapet 0.7 m thick round it. Find the area of the top of the parapet. (

π = \frac{22}{7}

)

Solution, step by step

1The diameter is given, so first the inner radius $r = \frac{7}{2} = 3.5$ m. Outer radius $R = 3.5 + 0.7 = 4.2$ m.
2 $R + r = 7.7$ and $R - r = 0.7$ (that is the thickness).
3Area $= \frac{22}{7} \times 7.7 \times 0.7 = \frac{22}{7} \times 5.39 = 16.94$ sq m.

Answer 16.94 sq m

Trap: taking 7 m as the radius gives

R = 7.7

r = 7

and the answer

\frac{22}{7} \times 14.7 \times 0.7 = 32.34

— wrong.

Example 9 · Advanced

The outer circumference of a path round a circular pond is 154 m and the inner circumference is 132 m. Find the width and the area of the path. (

π = \frac{22}{7}

)

Solution, step by step

1Circumference $= 2 π \times$ radius, so the difference $2 π (R - r) = 2 π w = 154 - 132 = 22$ ⟹ $w = \frac{22}{2 \times \frac{22}{7}} = \frac{22 \times 7}{44} = 3.5$ m.
2Radii: $r = \frac{132 \times 7}{44} = 21$ m, $R = 21 + 3.5 = 24.5$ m (check: $2 \times \frac{22}{7} \times 24.5 = 154$ ✓).
3Area $= π (R + r) (R - r) = \frac{22}{7} \times 45.5 \times 3.5 = \frac{22}{7} \times 159.25 = 500.5$ sq m.

Answer $w = 3.5$ m; area 500.5 sq m

Shortcut: area

= w \times

circumference of the middle circle

= 3.5 \times \frac{154 + 132}{2} = 3.5 \times 143 = 500.5

✓ — without finding the radii.

BasicTry it yourself first

A circular park has radius 28 m. A path 3.5 m wide runs round it on the outside. Find the area of the path. (

π = \frac{22}{7}

)

BasicTry it yourself first

A circular pond has radius 35 m. Inside, along its edge, runs a footpath 3.5 m wide. Find the area of the footpath. (

π = \frac{22}{7}

)

StandardTry it yourself first

A circular fountain has radius 21 m. Find the cost of laying stone at ₹60 per sq m on a 2.8 m wide border all round the outside. (

π = \frac{22}{7}

)

StandardTry it yourself first

A circular well has an inner diameter of 14 m. A masonry wall 1.4 m thick surrounds it. Find the area of the top surface of the wall. (

π = \frac{22}{7}

)

StandardTry it yourself first

The outer circumference of a path round a circular park is 264 m and the inner circumference is 220 m. Find the width of the path. (

π = \frac{22}{7}

)

StandardTry it yourself first

A circular town square has radius 56 m. A path 7 m wide runs round it on the outside. Find the area of the path. (

π = \frac{22}{7}

)

AdvancedTry it yourself first

A circular fountain basin has radius 17.5 m. Inside, along its edge, runs a footpath 3.5 m wide. Find the area of the footpath. (

π = \frac{22}{7}

)

AdvancedTry it yourself first

The outer circumference of a path round a circular park is 330 m and the inner circumference is 308 m. Find the width of the path. (

π = \frac{22}{7}

)

AdvancedTry it yourself first

A circular well has an inner diameter of 14 m. A masonry wall 0.7 m thick surrounds it. Find the area of the top surface of the wall. (

π = \frac{22}{7}

)

Cost problems

Fence · floor · walls · grazing · track

Find an area or a perimeter, then multiply by the rate: fences, carpets and tiles, wall painting, a cow on a rope and running tracks.

Fencing and posts

Perimeter × rate; posts = gaps (+1)

Cost of fencing, the gate gap, several strands of wire, fencing three sides and counting posts.

Fencing is always a question of perimeter, not area: the fence runs along the boundary. The cost formula is direct — length to fence × rate. Difficulty comes in three places: (1) a gate or open stretch where no fence is put; (2) several strands of wire, which multiply the length; (3) counting posts, where an open line and a closed loop follow different rules.

◭ Diagram

On a closed loop the first post is also the end of the last gap, so posts = gaps (8). On an open line a post is needed at both ends, so posts = gaps + 1 (4 gaps, 5 posts)

∑ Formula

cost = (P - g) \times s \times c N_{loop} = \frac{P}{d}, N_{line} = \frac{L}{d} + 1

Cost of fencing and number of posts

$P$: perimeter of the plot (m): rectangle $2 (l + b)$ , square $4 a$ , circle $π d$
$g$: width of the gate / open stretch (m)
$s$: number of strands of wire
$c$: rate of the wire (₹ per metre)
$d$: distance between two neighbouring posts (m)
$L$: total length of an open fence (m)
$N$: number of posts

Fencing three sides (the fourth lies along a river, wall or road): length $=$ one long side $+$ two short sides, i.e. $l + 2 b$ or $2 l + b$ — see in the question which side is open. This is an open line, so posts $= \frac{L}{d} + 1$ . A gate: take the gate width off a closed loop and it becomes an open line, and the rule for posts changes — this is the favourite SSC trap.

Spot itSpot it: “fence / wire / plant every … m”. First ask: is the loop closed or open? Closed ⟹ posts = gaps. Open ⟹ gaps + 1. Traps: working the cost from the area, and forgetting to subtract the gate.

Decide the length of fence (full perimeter, or three sides) and subtract the gate.
Multiply by the number of strands, then by the rate.
Posts: closed loop $\frac{P}{d}$ ; open line $\frac{L}{d} + 1$ . The gaps must come out whole (check).

Example 10 · Basic

A field 85 m × 45 m is to be fenced with 3 strands of barbed wire, leaving a 5 m wide gate open. If the wire costs ₹12 per m, what is the total cost?

Solution, step by step

1The fence runs along the boundary, so the perimeter is needed: $2 (85 + 45) = 260$ m.
2No wire across the gate: $260 - 5 = 255$ m.
33 strands: total wire $= 255 \times 3 = 765$ m.
4Cost $= 765 \times 12 = 9180$ rupees.

Answer ₹9,180

Order: subtract the gate first and then multiply by the strands — the other way also works, as long as the gate is subtracted.

Example 11 · Standard

One long side of a field 120 m × 80 m lies along a river, so only the other three sides are fenced. A post stands every 5 m (including both ends) and 2 strands of wire at ₹14 per m run between them. If a post costs ₹90, find the total cost.

Solution, step by step

1Length of fence $= 120 + 2 \times 80 = 280$ m — an open line (no fence on the fourth side).
2Gaps $= \frac{280}{5} = 56$ ; on an open line posts $= 56 + 1 = 57$ . Cost of posts $= 57 \times 90 = 5130$ rupees.
3Wire: $280 \times 2 = 560$ m, cost $= 560 \times 14 = 7840$ rupees.
4Total cost $= 5130 + 7840 = 12, 970$ rupees.

Answer ₹12,970

Trap: taking 56 posts gives

56 \times 90 = 5040

— one post short; do not forget the +1 on an open line.

Example 12 · Advanced

Posts are to be put every 2.2 m along the edge of a circular garden of diameter 70 m and 3 strands of wire at ₹15 per m run between them, but a 6.6 m wide entrance is left open (with a post at each end of the entrance). Find the number of posts and the cost of the wire. (

π = \frac{22}{7}

)

Solution, step by step

1Circumference $= π d = \frac{22}{7} \times 70 = 220$ m. The full circle is closed: gaps $= \frac{220}{2.2} = 100$ .
2The entrance $6.6$ m $= \frac{6.6}{2.2} = 3$ gaps wide, so the fenced gaps $= 100 - 3 = 97$ .
3The loop is no longer closed, it is an open line (a post at both ends): posts $= 97 + 1 = 98$ .
4Length of wire $= (220 - 6.6) \times 3 = 213.4 \times 3 = 640.2$ m; cost $= 640.2 \times 15 = 9603$ rupees.

Answer 98 posts; cost of wire ₹9,603

A closed loop would have 100 posts; once the entrance is left the fence becomes an open line and the rule turns to “gaps + 1”. So the answer is

97 + 1 = 98

, not 100.

BasicTry it yourself first

A rectangular plot measures 70 m × 40 m. It is to be fenced with 2 strands of barbed wire, leaving a 6 m wide gate open. If the wire costs ₹16 per m, find the total cost.

BasicTry it yourself first

A square plot has an area of 1,225 sq m. Find the cost of fencing it all round at ₹18 per m.

StandardTry it yourself first

A rectangular field has its sides in the proportion 5 : 4, and covers 1,620 sq m. What will it cost to fence it at ₹15 per m?

StandardTry it yourself first

One long side of a 96 m × 60 m field lies along a river, so only the other three sides are fenced. A post is put every 6 m, including both ends. How many posts are needed?

StandardTry it yourself first

A rectangular garden 80 m × 48 m is to be fenced all round with a post every 8 m (corners included). How many posts are needed?

StandardTry it yourself first

A plant is to be set every 1.8 m along the edge of a circular park of diameter 63 m. How many plants are needed? (

π = \frac{22}{7}

)

AdvancedTry it yourself first

A farm measures 112 m × 78 m. It is to be fenced with 4 strands of barbed wire, leaving a 7.5 m wide gate open. If the wire costs ₹13 per m, find the total cost.

AdvancedTry it yourself first

A rectangular field has its sides in the proportion 7 : 3, and covers 336 sq m. What will it cost to fence it at ₹22 per m?

Carpets, tiles and floors

Area ÷ width, area ÷ tile

Length of carpet, number of tiles, border tiles, the largest tile (HCF), wastage and comparing two options.

To cover a floor the basic fact is: area to cover = area of the floor. Carpet comes in a roll of fixed width, so length = area ÷ width (picture the carpet laid in strips). Tiles are counted, so tiles = area of floor ÷ area of one tile. In both the commonest mistakes are about units (mixing cm and m).

◭ Diagram

Each strip is as wide as the carpet and as long as the room; strips × length = total length of carpet

◭ Diagram

Border tiles (orange) = all tiles − the middle

(p - 2) (q - 2)

tiles. Shortcut:

2 (p + q) - 4

∑ Formula

length of carpet = \frac{l b}{w} tiles = \frac{l b}{t ^{2}} border tiles = 2 (p + q) - 4

Carpet, tiles, border tiles

$l, b$: length and breadth of the room (in one unit)
$w$: width of the carpet (in the same unit)
$t$: side of a square tile (in the same unit)
$p, q$: number of rows and tiles in each row (counts)

The largest square tile that fits the floor without cutting: its side must go into both the length and the breadth a whole number of times, so the side $=$ HCF(length, breadth) (in cm). If wastage (cutting tiles/slabs) is $x %$ , order $x %$ more than needed: need $\times (1 + \frac{x}{100})$ .

Spot itSpot it: “carpet … cm wide, ₹ per metre” → find the length. “Square tiles” → count. “Largest tile / without cutting” → HCF. “Border tiles” →

2 (p + q) - 4

. Trap: the carpet rate is per metre of length, not per sq m.

Write all measures in one unit (m or cm).
Find the area of the floor.
Carpet: area ÷ width = length. Tile: area ÷ area of a tile = count. With wastage multiply by $(1 + \frac{x}{100})$ .
Count × rate = cost; with two options compare the two total costs.

Example 13 · Basic

A carpet 75 cm wide, priced at ₹85 per metre of length, is to cover the floor of a hall 12 m × 7.5 m. How many metres of carpet are needed and what is the total cost?

Solution, step by step

1Area of the floor $= 12 \times 7.5 = 90$ sq m.
2The carpet width is in cm; change it to metres: $75$ cm $= 0.75$ m.
3Length $= \frac{90}{0.75} = 120$ m (each metre of the 0.75 m wide carpet covers 0.75 sq m).
4Cost $= 120 \times 85 = 10, 200$ rupees.

Answer 120 m; ₹10,200

Trap: taking

90 \times 85 = 7650

— the rate is on the length, not on the area.

Example 14 · Standard

The floor of a courtyard 4.8 m × 3.2 m is laid with square tiles of side 40 cm. (a) How many tiles? (b) If the border tiles are red (₹22 each) and the rest white (₹15 each), what is the total cost?

Solution, step by step

1Courtyard in cm: $480 \times 320$ . Along the length $\frac{480}{40} = 12$ tiles, along the breadth $\frac{320}{40} = 8$ tiles; total $= 12 \times 8 = 96$ .
2Middle tiles (edge removed) $= (12 - 2) \times (8 - 2) = 10 \times 6 = 60$ , so border tiles $= 96 - 60 = 36$ (shortcut: $2 (12 + 8) - 4 = 36$ ✓).
3Cost $= 36 \times 22 + 60 \times 15 = 792 + 900 = 1692$ rupees.

Answer (a) 96 tiles (b) ₹1,692

The count also comes directly from rows × columns; dividing areas is easier when the tile fits exactly.

Example 15 · Advanced

For a room 6.4 m × 4.5 m there are two options: (a) square tiles of side 30 cm at ₹15 per tile; (b) slabs at ₹155 per sq m, of which 5% extra has to be ordered for cutting waste. Which is cheaper and by how much?

Solution, step by step

1Floor $= 6.4 \times 4.5 = 28.8$ sq m $= 288, 000$ sq cm.
2(a) One tile $= 30 \times 30 = 900$ sq cm; tiles $= \frac{288000}{900} = 320$ ; cost $= 320 \times 15 = 4800$ rupees.
3(b) To be ordered $28.8 \times 1.05 = 30.24$ sq m; cost $= 30.24 \times 155 = 4687.2$ rupees.
4Difference $= 4800 - 4687.2 = 112.8$ rupees; the slabs are cheaper.

Answer slabs are cheaper by ₹112.8

Never compare without the full cost of each — the unit rates are not directly comparable (tile per piece, slab per sq m).

BasicTry it yourself first

A classroom measures 9 m × 6 m. It is to be covered with carpet 90 cm wide, sold at ₹70 per metre of length. Find the total cost.

BasicTry it yourself first

A carpet 150 m long and 80 cm wide exactly covers the floor of a rectangular room. The room is 12 m long. Find its breadth.

StandardTry it yourself first

A kitchen measures 4.5 m × 3.6 m. It is to be paved with square tiles of side 30 cm. How many tiles are needed?

StandardTry it yourself first

Square tiles are laid in a courtyard in 18 rows with 11 tiles in each row. The tiles along the outer edge are red. How many red tiles are there?

StandardTry it yourself first

A floor of 825 cm × 675 cm is to be covered, without cutting, with the largest possible square tiles. How many tiles are needed?

StandardTry it yourself first

Marble slabs are to be laid in a 12 m × 8.5 m hall. Cutting wastes 8% extra material, so that much more has to be ordered. At ₹120 per sq m find the total cost.

AdvancedTry it yourself first

For a room 7.2 m × 5 m there are two options: (a) square tiles of side 40 cm at ₹22 per tile; (b) slabs at ₹190 per sq m, to be ordered with 4% extra for waste. Which is cheaper and by how much?

AdvancedTry it yourself first

A floor of 1,050 cm × 770 cm is to be covered, without cutting, with the largest possible square tiles. How many tiles are needed?

Painting and papering walls

2 (l + b) h

, then subtract the openings

Area of four walls, doors and windows, the ceiling, several coats, length of wallpaper and room measures from the floor.

Open the four walls out like one big strip: its length is the perimeter of the room $2 (l + b)$ and its height is $h$ . So the area of the four walls is $2 (l + b) \times h$ . The floor and ceiling are not in it — each is a separate $l \times b$ . Doors and windows get no paint or paper, so subtract their area.

◭ Diagram

The four walls open into one strip (length

2 (l + b)

, height

h

); the door and window (rose) are to be subtracted

∑ Formula

A_{walls} = 2 (l + b) h - A_{openings} wallpaper length = \frac{A _{walls}}{paper width}

Area of the walls and length of the paper

$l, b, h$: length, breadth and height of the room (m)
$A_{openings}$: total area of doors and windows (sq m)
$A_{walls}$: area to be painted / papered (sq m)

If the ceiling is also painted, add $l \times b$ . For several coats multiply the area by their number. Wallpaper comes in a roll, so like carpet the length $= \frac{area}{width}$ and do not forget to change the width from cm to metres. When the floor area and a ratio are given, find $l$ and $b$ first.

Spot itSpot it: “four walls / whitewash / paint / wallpaper”. Traps: (1) using the total surface

2 (l b + bh + h l)

(a 3D formula that includes the floor); (2) forgetting to subtract the openings; (3) applying the rate to the floor area instead of the walls.

Find the area of the four walls with $2 (l + b) h$ .
Subtract the doors and windows; add $l b$ if the ceiling is included.
Multiply for several coats; for paper divide by the width (in metres) to get the length.
Multiply by the rate.

Example 16 · Basic

A classroom is 9 m × 6 m × 4 m. It has one door of 2 m × 1 m and two windows of 1.5 m × 1.2 m. Find the cost of painting its four walls at ₹18 per sq m.

Solution, step by step

1Four walls $= 2 (l + b) h = 2 (9 + 6) \times 4 = 120$ sq m.
2Openings: door $2 \times 1 = 2$ and two windows $2 \times (1.5 \times 1.2) = 3.6$ , total $5.6$ sq m.
3Area to paint $= 120 - 5.6 = 114.4$ sq m.
4Cost $= 114.4 \times 18 = 2059.2$ rupees.

Answer ₹2,059.20

Only the surface that gets painted counts — the window glass is not painted.

Example 17 · Standard

The four walls and the ceiling of a hall 12 m × 8 m × 5 m are whitewashed twice. It has two doors of 2.5 m × 2 m and four windows of 2 m × 1.5 m. What is the total cost at ₹6 per sq m (per coat)?

Solution, step by step

1Four walls $= 2 (12 + 8) \times 5 = 200$ sq m; ceiling $= 12 \times 8 = 96$ sq m.
2Openings $= 2 \times (2.5 \times 2) + 4 \times (2 \times 1.5) = 10 + 12 = 22$ sq m (they come off the walls only, not the ceiling).
3Area for one coat $= 200 - 22 + 96 = 274$ sq m.
4Two coats: $274 \times 2 = 548$ sq m; cost $= 548 \times 6 = 3288$ rupees.

Answer ₹3,288

Trap: the rate is “per coat” — two coats means double the area, the rate stays the same.

Example 18 · Advanced

The length of a room is twice its breadth and the floor area is 98 sq m. The room is 3.5 m high. Find the cost of papering its four walls (leaving out a door of 2 sq m) with wallpaper 50 cm wide at ₹22 per m.

Solution, step by step

1Let the breadth be $b$ m, so the length is $2 b$ : $2 b \times b = 98$ ⟹ $b^{2} = 49$ ⟹ $b = 7$ m, $l = 14$ m.
2Four walls $= 2 (14 + 7) \times 3.5 = 147$ sq m; after the door $147 - 2 = 145$ sq m.
3Paper width $50$ cm $= 0.5$ m, so the length $= \frac{145}{0.5} = 290$ m.
4Cost $= 290 \times 22 = 6380$ rupees.

Answer ₹6,380

The floor area only served to find

l

and

b

; the paper does not go on it.

BasicTry it yourself first

A hall measures 8 m × 6 m × 3.5 m and has 1 opening of 1.5 m × 2 m and 2 openings of 1.2 m × 1 m. Find the cost of painting its four inside walls once at ₹22 per sq m (per coat).

StandardTry it yourself first

A dining hall measures 15 m × 10 m × 4.5 m and has 2 openings of 2 m × 3 m. Find the cost of painting its four inside walls and the ceiling 2 coats at ₹8 per sq m (per coat).

StandardTry it yourself first

The four walls of a 7.5 m × 6 m × 3.2 m room (leaving out 4.6 sq m of openings) are to be covered with wallpaper 50 cm wide, at ₹18 per m of length. Find the cost.

StandardTry it yourself first

In a room the length : breadth is 5 : 4; the ceiling is 3.6 m high and the floor measures 80 sq m. What does whitewashing the four walls cost at ₹12 per sq m?

StandardTry it yourself first

Painting the four walls of an 8 m × 5.5 m room at ₹14 per sq m cost ₹1,209.6 in all (no openings). Find the height of the room.

AdvancedTry it yourself first

A library room measures 18 m × 12 m × 6 m and has 2 openings of 3 m × 2.5 m and 5 openings of 2.5 m × 1.5 m. Find the cost of painting its four inside walls and the ceiling 3 coats at ₹9 per sq m (per coat).

AdvancedTry it yourself first

The four walls of a 14 m × 9 m × 4 m room (leaving out 12.4 sq m of openings) are to be covered with wallpaper 80 cm wide, at ₹26 per m of length. Find the cost.

AdvancedTry it yourself first

In a room the length : breadth is 7 : 3; the ceiling is 3.2 m high and the floor measures 84 sq m. What does whitewashing the four walls cost at ₹17.5 per sq m?

A cow on a rope: the grazing area

A sector by the angle available

The area a tethered animal can graze: ¼, ½, ¾ circle, wrapping round when the rope is long, lengthening the rope.

A tethered animal that pulls the rope tight and walks round can graze a sector (part of a circle) whose radius is the length of the rope. The whole question now rests on one thing: through what angle can the rope swing? In the open round a peg $36 0^{\circ}$ ; on a straight wall $18 0^{\circ}$ ; at an inside corner of a rectangular field $9 0^{\circ}$ ; at an outside corner of a house $36 0^{\circ} - 9 0^{\circ} = 27 0^{\circ}$ .

◭ Diagram

The rope (rose) sweeps out the region. Inside a field at a corner:

9 0^{\circ}

, a quarter circle. Outside a house at a corner:

27 0^{\circ}

, three quarters of a circle

∑ Formula

A = \frac{θ}{360} π r^{2}

Grazing area

$A$: area that can be grazed (square units)
$θ$: the angle (in degrees) through which the rope can swing
$r$: length of the rope (in units)
$π$: $\frac{22}{7}$ when $r$ is a multiple of 7

Situation	Angle $θ$	Area
open ground, a peg	$36 0^{\circ}$	$π r^{2}$
on a straight wall	$18 0^{\circ}$	$\frac{1}{2} π r^{2}$
inside corner of a rectangle/square	$9 0^{\circ}$	$\frac{1}{4} π r^{2}$
corner of an equilateral triangular field	$6 0^{\circ}$	$\frac{1}{6} π r^{2}$
corner of a regular hexagon	$12 0^{\circ}$	$\frac{1}{3} π r^{2}$
outside corner of a house	$27 0^{\circ}$	$\frac{3}{4} π r^{2}$
outside, rope $R$ longer than the side $a$	¾ + two ¼	$\frac{3}{4} π R^{2} + \frac{1}{2} π (R - a)^{2}$

Letters in the table: $r$ = rope length, $R$ = rope when it is longer than the side of the house, $a$ = side of the (square) house. Understand the last row: at an outside corner the rope first sweeps $\frac{3}{4}$ of a circle; $a$ m of the rope run along the wall to the next corner, and the remaining $R - a$ m turn there and sweep a new quarter circle of $9 0^{\circ}$ . This happens at both neighbouring corners, so two quarter circles, i.e. $\frac{1}{2} π (R - a)^{2}$ . (Condition: $R - a \leq a$ , otherwise the two quarter circles start to meet behind the house.)

◭ Diagram

Rope

R

longer than the side

a

: a big ¾ circle (blue, radius

R

) and two quarter circles of radius

R - a

at the two neighbouring corners (green and orange)

Spot itSpot it: “cow/goat/horse tied by a rope”, “can graze”. First ask: is the corner inside or outside, and is the rope shorter or longer than the side? Traps: (1) taking the rope as a diameter; (2) taking an outside corner as ¼ too; (3) counting the common part twice when two ropes meet.

Draw the figure: where is the peg and on which side are the wall/field?
Decide the angle $θ$ (inside: the corner angle; outside: $36 0^{\circ} -$ the corner angle; straight wall: $18 0^{\circ}$ ).
If the rope is longer than the side, add the quarter circles at the turns (radius = leftover rope).
Apply $\frac{θ}{360} π r^{2}$ ; for “area left”, subtract from the whole.

Example 19 · Basic

A cow is tied at a corner of a 20 m × 15 m grass field by a 7 m rope. How much area can it graze and how much of the field is left? (

π = \frac{22}{7}

)

Solution, step by step

1The corner is inside the field and the 7 m rope is shorter than both sides (15 and 20), so the rope swings $9 0^{\circ}$ — a quarter circle.
2Grazed area $= \frac{1}{4} \times \frac{22}{7} \times 7^{2} = \frac{1}{4} \times 22 \times 7 = 38.5$ sq m.
3The whole field is $20 \times 15 = 300$ sq m, so the left-over $= 300 - 38.5 = 261.5$ sq m.

Answer 38.5 sq m; 261.5 sq m left

Check the condition: if the rope were longer than a side the quarter circle would spill out of the field.

Example 20 · Standard

A goat is tied by a 14 m rope at a corner of the outer wall of a square water tank of side 7 m. How much area can it graze? (

π = \frac{22}{7}

)

Solution, step by step

1The corner is outside, so the rope first sweeps $27 0^{\circ}$ — $\frac{3}{4}$ of a circle of radius 14 m: $\frac{3}{4} \times \frac{22}{7} \times 196 = \frac{3}{4} \times 616 = 462$ sq m.
2The rope (14 m) is longer than the side of the tank (7 m). 7 m of rope reach the next corner along the wall, and with the remaining $14 - 7 = 7$ m the goat swings round that corner, grazing a $9 0^{\circ}$ quarter circle of radius 7 m.
3There are two such corners (both neighbours): $2 \times \frac{1}{4} \times \frac{22}{7} \times 49 = \frac{1}{2} \times 154 = 77$ sq m.
4Total $= 462 + 77 = 539$ sq m.

Answer 539 sq m

Trap: taking only the

\frac{3}{4}

circle (462) misses the rest; whenever the rope is longer than the side, think of the “turns”.

Example 21 · Advanced

A cow and a goat are tied by 14 m ropes at two adjacent corners of a square field of side 28 m. How much area do they graze in all and how much of the field is left? (

π = \frac{22}{7}

)

Solution, step by step

1Both are at inside corners, so each area is a quarter circle: $\frac{1}{4} \times \frac{22}{7} \times 1 4^{2} = \frac{1}{4} \times 616 = 154$ sq m.
2Do they overlap? The adjacent corners are 28 m apart and the two ropes add up to $14 + 14 = 28$ m — the quarter circles just touch at one point, with no common part.
3Total grazed $= 2 \times 154 = 308$ sq m. The field $= 2 8^{2} = 784$ sq m, left $= 784 - 308 = 476$ sq m.

Answer 308 sq m; 476 sq m left

When the two ropes add up to more than the distance between the corners a common part forms — then simple adding would be wrong.

BasicTry it yourself first

A cow is tied by a 14 m rope at a corner of a rectangular field (every side of the field is longer than the rope). How much area can it graze inside the field? (

π = \frac{22}{7}

)

BasicTry it yourself first

A goat is tied by a 21 m rope at a corner outside a rectangular house (every wall is longer than the rope). How much area can it graze? (

π = \frac{22}{7}

)

StandardTry it yourself first

A horse is tied by a 21 m rope at a vertex of an equilateral triangular field (every side of the field is longer than the rope). How much area can it graze inside the field? (

π = \frac{22}{7}

)

StandardTry it yourself first

A goat is tied by a 21 m rope at a vertex of a regular hexagonal field (every side of the field is longer than the rope). How much area can it graze inside the field? (

π = \frac{22}{7}

)

StandardTry it yourself first

A goat is tied by a 14 m rope to a peg in the middle of a straight wall, and the field lies on one side of the wall (the wall is longer than the rope on both sides). Find the area it can graze. (

π = \frac{22}{7}

)

AdvancedTry it yourself first

A goat is tied by a 21 m rope at a corner of the outer wall of a square water tank of side 14 m. How much area can it graze? (

π = \frac{22}{7}

)

StandardTry it yourself first

A cow is tied at a corner of a 40 m × 30 m grass field by a 14 m rope. How much of the field's area is left ungrazed? (

π = \frac{22}{7}

)

AdvancedTry it yourself first

A goat is tied by a 35 m rope at a corner of the outer wall of a square water tank of side 21 m. How much area can it graze? (

π = \frac{22}{7}

)

AdvancedTry it yourself first

A calf is tied at a corner of a square pasture (side 38 m) by a 14 m rope. By how much does the grazing area grow if the rope is lengthened to 28 m? (

π = \frac{22}{7}

)

Running tracks and shortcuts across a ground

2 s + 2 π r

· lane gap

2 π w

· diagonal saving

Lap, lanes, area and start lines of a track made of two straights and two semicircles; the saving across a rectangular ground by the diagonal.

The running track of a stadium is made of two parallel straights (length $s$ ) and two semicircular bends (inner radius $r$ ). The two semicircles together make one full circle, so one lap $= 2 s + 2 π r$ — the straights twice, and the bends as one full circle.

◭ Diagram

The rose dot runs one lap. The blue line is the straight

s

, green the inner radius

r

, orange the lane width

w

. The blue band is the track

∑ Formula

L = 2 s + 2 π r Δ L = 2 π w per lane A_{track} = 2 s w + π (R^{2} - r^{2})

Lap, gap per lane, area of the track

$L$: length of one lap on the inner edge (m)
$s$: length of one straight (m)
$r, R$: inner and outer radius of the bends (m), $R = r + w$
$w$: width of the track (or of one lane) (m)
$Δ L$: increase of the lap on moving to the next lane (m)

Moving a lane outwards leaves the straights the same; only the radius of the bends grows by $w$ , so the lap grows by $2 π w$ . This is why the start line is staggered in a race: a runner $n$ lanes outside starts $2 π w n$ m ahead so that every lap is equal. The area of the track = two rectangular straights ( $s \times w$ each) + one ring $π (R^{2} - r^{2})$ for the bends.

The diagonal shortcut across a rectangular ground is a direct idea of the same kind: along two sides $a + b$ m, across the diagonal $a^{2} + b^{2}$ m. Saving $= a + b - a^{2} + b^{2}$ ; time saved $= \frac{saving}{speed}$ . Recognise multiples of the Pythagorean triples $(3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25)$ — the path length comes out at once.

◭ Diagram

Blue route:

a + b

. The rose diagonal

d = a^{2} + b^{2}

is always shorter; both runners start together and the rose one arrives first

Spot itSpot it: “running track / stadium / lane / lap / start line” and “saving from one corner to the opposite corner”. Traps: (1) counting the semicircles as two full circles (the two together make one circle); (2) adding the straights in the lane gap; (3) using

2 w

for

w

Draw the track: write $s, r, w$ ; $R = r + w$ .
Lap $2 s + 2 π r$ ; lane gap $2 π w$ (per lane); start distance = $2 π w \times$ number of lanes.
Area: $2 s w + π (R^{2} - r^{2})$ .
Across a ground: the diagonal by Pythagoras, then $a + b - d$ .

Example 22 · Basic

The running track of a college has two straights of 100 m and two semicircles of inner radius 21 m at their ends. How long is one lap along the inner edge? (

π = \frac{22}{7}

)

Solution, step by step

1One lap has two straights: $2 \times 100 = 200$ m.
2The two semicircles together make one full circle: $2 π r = 2 \times \frac{22}{7} \times 21 = 132$ m.
3Lap $= 200 + 132 = 332$ m.

Answer 332 m

Check: a 400 m race needs a little more than one lap; the outer lanes get longer in the same way.

Example 23 · Standard

The two straights of a running track are 90 m long, the bends are semicircles of inner radius 35 m, and the track is 3.5 m wide throughout. (a) Find the area of the track. (b) By how much should the runner in lane 3 start ahead of the runner in lane 1 (each lane 3.5 m wide)? (

π = \frac{22}{7}

)

Solution, step by step

1Outer radius $R = 35 + 3.5 = 38.5$ m. The two straight rectangular parts: $2 \times (90 \times 3.5) = 630$ sq m.
2The two bends together make one ring: $π (R^{2} - r^{2}) = \frac{22}{7} (38. 5^{2} - 3 5^{2}) = \frac{22}{7} (1482.25 - 1225) = \frac{22}{7} \times 257.25 = 808.5$ sq m.
3(a) Track $= 630 + 808.5 = 1438.5$ sq m.
4(b) Lane 3 is two lanes outside lane 1, so the bend radius is $2 \times 3.5 = 7$ m larger. The lap differs by $2 π \times 7 = 44$ m, so start the lane-3 runner 44 m ahead.

Answer (a) 1438.5 sq m (b) 44 m

The straights are equal in every lane, so the 90 m plays no part in (b).

Example 24 · Advanced

A boy goes from one corner of a rectangular ground 72 m × 54 m to the opposite corner at 1.5 m/s. How much distance and time does he save by going straight along the diagonal instead of along the sides?

Solution, step by step

1Along the sides: $72 + 54 = 126$ m.
2Diagonal (right triangle): $7 2^{2} + 5 4^{2}$ . Both sides share 18: $72 = 18 \times 4$ , $54 = 18 \times 3$ , i.e. $18$ times the triple $(3, 4, 5)$ , so the diagonal $= 18 \times 5 = 90$ m.
3Saving $= 126 - 90 = 36$ m.
4Time: along the sides $\frac{126}{1.5} = 84$ seconds; across the diagonal $\frac{90}{1.5} = 60$ seconds; time saved $= 24$ seconds (or directly $\frac{36}{1.5} = 24$ ).

Answer 36 m; 24 seconds

Spotting the Pythagorean triple removed the square root: look for a common multiple (here 18).

BasicTry it yourself first

The running track of a school has two parallel straights of 80 m joined by two semicircles of inner radius 35 m. Find the length of one lap along the inner edge. (

π = \frac{22}{7}

)

BasicTry it yourself first

The running track of a college has two parallel straights of 105 m joined by two semicircles of inner radius 14 m. Find the length of one lap along the inner edge. (

π = \frac{22}{7}

)

StandardTry it yourself first

Every lane of a running track is 1.4 m wide. By how much does one lap grow when you move from lane 1 to lane 3? (

π = \frac{22}{7}

)

StandardTry it yourself first

The two straights of a running track are 84 m long and the two bends are semicircles of inner radius 28 m. The track is 3.5 m wide throughout. Find the area of the track. (

π = \frac{22}{7}

)

StandardTry it yourself first

A 400 m race is run on a track with two full bends per lap. The lanes are 1.4 m wide. Runner B's lane is 3 lanes outside runner A's. By how much should B's starting line be moved ahead of A's to make it fair? (

π = \frac{22}{7}

)

BasicTry it yourself first

A boy goes from one corner of a rectangular ground, 24 m × 32 m, to the opposite corner. How many metres does he save by going across the diagonal instead of along the two sides?

StandardTry it yourself first

A runner runs at 4 m/s from one corner of a 48 m × 64 m ground to the opposite corner. How many seconds does he save by running along the diagonal instead of along the two sides?

AdvancedTry it yourself first

The two straights of a running track are 70 m long and the two bends are semicircles of inner radius 21 m. The track is 2.8 m wide throughout. Find the area of the track. (

π = \frac{22}{7}

)

AdvancedTry it yourself first

A farmer goes from one corner of a rectangular ground, 105 m × 140 m, to the opposite corner. How many metres does he save by going across the diagonal instead of along the two sides?

Shape inside shape, and combined figures

Inscribed shapes · packing coins · shaded regions

The largest figure inside another, packing coins and belts, and areas of shaded regions.

Circle, square, equilateral triangle and hexagon inside each other

Diameter = side · diagonal = diameter ·

R = \frac{a}{3}

Circle in square, square in circle, equilateral triangle and hexagon; waste, ratios and chains of figures.

In questions on “the largest circle/square that can be cut”, “largest” means the inner figure touches the outer one — either its corners lie on the outer figure or its sides touch it. This touching gives one direct relation, and the whole question is to recognise that relation.

Circle in a square: the circle touches all four sides, so diameter = side of the square ( $2 r = a$ ).
Square in a circle: all four corners lie on the circle, so the diagonal of the square = diameter of the circle ( $a 2 = 2 R$ ); area $= \frac{d ^{2}}{2} = 2 R^{2}$ .
Equilateral triangle in a circle: the circle is its circumcircle, $R = \frac{a}{3}$ ; and the inner circle $r = \frac{a}{2 3}$ , i.e. $R = 2 r$ .
Regular hexagon in a circle: six equilateral triangles meet at the centre, so side = radius ( $a = R$ ); area $= 6 \times \frac{3}{4} R^{2} = \frac{3 3}{2} R^{2}$ .

◭ Diagram

Left: the circle touches the sides, so

2 r = a

(rose line). Right: the corners lie on the circle, so the diagonal

d = 2 R

◭ Diagram

Left: equilateral triangle — centre to vertex

R

and to a side

r

R = 2 r

. Right: the hexagon splits into six equilateral triangles, side

a = R

∑ Formula

r_{in sq.} = \frac{a}{2}, A_{sq in circle} = 2 R^{2} a_{△} = R 3, a_{hex} = R A_{hex} = \frac{3 3}{2} R^{2}

Relations for inscribed figures

$a$: side of the square / triangle / hexagon (in units)
$R$: radius of the outer circle (in units)
$r$: radius of the inner circle (in units)
$A$: area (square units)

Two more useful facts: (1) in the chain square → circle → square the area halves each time (a circle in a square then a square in that circle gives $\frac{a ^{2}}{2}$ ); (2) joining the midpoints of a square also gives exactly half. (3) The area ratio of the largest square in a circle to the circle is $2 : π$ , and of the largest circle in a square to the square is $π : 4$ .

Spot itSpot it: “largest / just fits / vertices on the circle / touches the sides”. First decide: does it touch with a side (diameter = side) or with a corner (diagonal = diameter)? Trap: taking the side as

2 R

for a square in a circle — that is the diagonal.

Draw the outer and inner figure; look at the points of contact.
Write the relation: diameter = side, or diagonal = diameter, or $a = R 3$ , or $a = R$ .
From the given measure find the other, then the area/perimeter.
Waste / left-over = outer area − inner area.

Example 25 · Basic

The largest circular disc is cut from a square plate of side 28 cm. Find the area of the disc, the area wasted and the percentage wasted. (

π = \frac{22}{7}

)

Solution, step by step

1The largest circle touches the sides of the square, so the diameter $= 28$ cm and the radius $= 14$ cm.
2Disc $= \frac{22}{7} \times 1 4^{2} = \frac{22}{7} \times 196 = 616$ sq cm; plate $= 2 8^{2} = 784$ sq cm.
3Waste $= 784 - 616 = 168$ sq cm.
4Percentage $= \frac{168}{784} \times 100 = \frac{3}{14} \times 100 \approx 21.43%$ . (Shortcut: $1 - \frac{π}{4} = 1 - \frac{11}{14} = \frac{3}{14}$ )

Answer 616 sq cm; 168 sq cm; $\approx 21.43%$

Cutting a circle from a square always wastes

\frac{3}{14} \approx 21.4%

— whatever the side.

Example 26 · Standard

The largest square is cut from a circular plate of radius 21 cm. Find the area and the side of the square and the area of plate left over. (

π = \frac{22}{7}

)

Solution, step by step

1All four corners of the square lie on the circle, so the diagonal of the square equals the diameter: $d = 2 \times 21 = 42$ cm.
2Area of the square $= \frac{d ^{2}}{2} = \frac{4 2 ^{2}}{2} = \frac{1764}{2} = 882$ sq cm; side $= 882 = 212 \approx 29.7$ cm.
3Plate $= \frac{22}{7} \times 2 1^{2} = \frac{22}{7} \times 441 = 1386$ sq cm.
4Left over $= 1386 - 882 = 504$ sq cm.

Answer 882 sq cm; side $212$ cm; 504 sq cm

The area of a square in a circle is

2 R^{2}

— straight from the diagonal, no root needed.

Example 27 · Advanced

An equilateral triangle and a regular hexagon are drawn inside a circle of radius 14 cm (both with all vertices on the circle). Find (a) the side and area of the triangle, (b) the area of the hexagon and the ratio hexagon : triangle, (c) the area of the circle outside the triangle. (

π = \frac{22}{7}

3 = 1.732

)

Solution, step by step

1(a) For an equilateral triangle $R = \frac{a}{3}$ , so $a = R 3 = 143 \approx 24.25$ cm. Area $= \frac{3}{4} a^{2} = \frac{3}{4} \times 588 = 1473 \approx 254.6$ sq cm.
2(b) The side of the hexagon $= R = 14$ cm, so its area $= \frac{3 3}{2} \times 196 = 2943 \approx 509.2$ sq cm.
3Ratio $2943 : 1473 = 2 : 1$ — the hexagon is exactly twice the triangle (true for every circle).
4(c) Circle $= \frac{22}{7} \times 196 = 616$ ; outside the triangle $= 616 - 254.6 \approx 361.4$ sq cm.

Answer (a) $143$ cm; $1473$ (≈254.6) sq cm (b) $2943$ ; 2 : 1 (c) ≈361.4 sq cm

In ratio questions the

3

cancels, and so does the radius: the answer 2 : 1 comes straight out.

BasicTry it yourself first

The largest circular disc is cut from a square plate of side 42 cm. Find the area of the disc. (

π = \frac{22}{7}

)

StandardTry it yourself first

How much area is wasted when the largest circular disc is cut from a square tin sheet of side 56 cm? (

π = \frac{22}{7}

)

BasicTry it yourself first

The largest square is cut from a circular plate of radius 25 cm. Find the area of the square.

StandardTry it yourself first

How much of a circular plate of radius 35 cm is left over (wasted) after the largest square is cut from it? (

π = \frac{22}{7}

)

StandardTry it yourself first

An equilateral triangle is drawn inside a circle of radius 9 cm with all three vertices on the circle. Find its side.

StandardTry it yourself first

The side of an equilateral triangle is 24

3

cm. Find the radius of the largest circle drawn inside it.

StandardTry it yourself first

A regular hexagon with all six vertices on the circle is drawn inside a circle of radius 13 cm. Find the perimeter of the hexagon.

AdvancedTry it yourself first

Find the area of the equilateral triangle inscribed (vertices on the circle) in a circle of radius 6 cm.

AdvancedTry it yourself first

The midpoints of the four sides of a square of side 40 cm are joined to make a second square; then the midpoints of that give a third, and so on, 4 times in all. Find the area of the last square.

AdvancedTry it yourself first

The largest circle is drawn inside a square of side 36 cm and the largest square is drawn inside that circle. Find the area of the inner square.

Largest square or rectangle in a triangle or semicircle

s = \frac{bh}{b + h}

· greatest rectangle = ½ triangle ·

a = \frac{2 r}{5}

A square standing on the base of a triangle, the right-angle corner square, the greatest rectangle, a square in a semicircle and a circle in a right triangle.

Here two corners of the inner figure lie on the boundary of the outer one, and the relation comes from similar triangles or Pythagoras. Three situations come again and again.

The largest square standing on the base of a triangle (base $b$ , height $h$ , the two top corners on the two sides): the small triangle left above the square is similar to the whole one — base $s$ , height $h - s$ . So $\frac{s}{b} = \frac{h - s}{h}$ ⟹ $s = \frac{bh}{b + h}$ .
A square at the right angle of a right triangle: one leg is the base and the other the height, so $s = \frac{pq}{p + q}$ . The greatest rectangle standing on the base: it has half the height and also half the base as width, so its area is exactly half the triangle.
The largest square in a semicircle (base on the diameter): the top corner is on the circle, so $r^{2} = a^{2} + (\frac{a}{2})^{2}$ ⟹ $a = \frac{2 r}{5}$ , area $= \frac{4 r ^{2}}{5}$ .

◭ Diagram

Left: the square (orange) with a similar small triangle above it. Right: a square in a semicircle; from the centre to the top corner is

r

, and

r^{2} = a^{2} + (a /2)^{2}

◭ Diagram

Change the height of the rectangle: its area is greatest at height

h /2

, where it is half the triangle

∑ Formula

s = \frac{bh}{b + h}, s = \frac{pq}{p + q} A_{rect max} = \frac{bh}{4} a^{2} = \frac{4 r ^{2}}{5}, r_{in} = \frac{p + q - c}{2}

Largest figures in a triangle and a semicircle

$b, h$: base of the triangle and its height (in units)
$p, q, c$: the two legs and the hypotenuse of a right triangle
$s, a$: side of the square (in units)
$r$: radius of the semicircle; $r_{in}$ = radius of the incircle in a right triangle

Spot itSpot it: “the largest square/rectangle is cut” and the outer figure is a triangle or semicircle. Traps: (1) the right-angle square and a square on the hypotenuse are different —

\frac{pq}{p + q}

is only for the right-angle corner; (2) the side of a square in a semicircle is

\frac{2 r}{5}

, not

r

Draw the figure; mark which corners of the square/rectangle lie on the boundary.
In a triangle: $s = \frac{bh}{b + h}$ (the side the square stands on is the base). In a semicircle: $r^{2} = \frac{5 a ^{2}}{4}$ .
Find the area; for “left over” subtract from the outer figure.

Example 28 · Basic

A triangular metal plate has a base of 30 cm and a height of 20 cm on it. Find the side and the area of the largest square standing on the base.

Solution, step by step

1Let the side of the square be $s$ . The small triangle above is similar to the whole: base $s$ , height $20 - s$ .
2Similarity: $\frac{s}{30} = \frac{20 - s}{20}$ ⟹ $20 s = 600 - 30 s$ ⟹ $50 s = 600$ ⟹ $s = 12$ cm.
3By the formula: $\frac{bh}{b + h} = \frac{30 \times 20}{50} = 12$ ✓. Area $= 1 2^{2} = 144$ sq cm.

Answer side 12 cm; area 144 sq cm

Check:

s <

the smaller measure (20) ✓; the square must fit inside the triangle (area 300), and 144 < 300 ✓.

Example 29 · Standard

The right-angle sides of a triangular corner shelf are 42 cm and 56 cm (hypotenuse 70 cm). Find the largest square that can sit at the right-angle corner and the area of the shelf left after that.

Solution, step by step

1At the right-angle corner take one leg as the base and the other as the height: $s = \frac{pq}{p + q} = \frac{42 \times 56}{98} = \frac{2352}{98} = 24$ cm.
2Area of the square $= 2 4^{2} = 576$ sq cm.
3Shelf (triangle) $= \frac{1}{2} \times 42 \times 56 = 1176$ sq cm.
4Left $= 1176 - 576 = 600$ sq cm.

Answer square of side 24 cm; 600 sq cm left

The left-over part consists of three small triangles — but subtracting is the fastest.

Example 30 · Advanced

The largest square, with one side on the diameter, is cut from a semicircular plate of radius 35 cm. Find the area of the square and the area of plate left over. (

π = \frac{22}{7}

)

Solution, step by step

1Let the side of the square be $a$ . Its base extends $\frac{a}{2}$ on both sides of the centre and its height is $a$ . The top corner is on the circle, so its distance from the centre is $r = 35$ .
2Pythagoras: $3 5^{2} = a^{2} + (\frac{a}{2})^{2} = \frac{5 a ^{2}}{4}$ ⟹ $a^{2} = \frac{4 \times 1225}{5} = 980$ .
3Area of the square $= 980$ sq cm (side $= 145 \approx 31.3$ cm).
4Semicircle $= \frac{1}{2} \times \frac{22}{7} \times 1225 = 1925$ ; left $= 1925 - 980 = 945$ sq cm.

Answer 980 sq cm; 945 sq cm left

Trap: taking the side as

r

\frac{r}{2}

; the correct relation is

a^{2} = \frac{4 r ^{2}}{5}

BasicTry it yourself first

A triangular metal plate has a base of 24 cm and a height of 12 cm on it. Find the side of the largest square standing on the base (its top corners touch the two sides of the triangle).

StandardTry it yourself first

The sides forming the right angle of a right triangle are 21 cm and 28 cm (hypotenuse 35 cm). Find the area of the largest square at the right-angle corner (two sides of the square along the legs).

StandardTry it yourself first

A rectangle of the greatest possible area, standing on the base, is cut from a triangle of base 36 cm and height 28 cm. Find the area of that rectangle.

StandardTry it yourself first

The largest square, with one side on the diameter, is cut from a semicircular plate of radius 15 cm. Find the area of the square.

StandardTry it yourself first

A round pot is to be fitted as large as possible into a triangular corner stand whose right-angle sides are 24 cm and 32 cm. Find the radius of the pot (the incircle).

StandardTry it yourself first

A triangular metal plate has a base of 42 cm and a height of 28 cm on it. Find the side of the largest square standing on the base (its top corners touch the two sides of the triangle).

AdvancedTry it yourself first

How much of a semicircular plate of radius 70 cm is left after the largest square with a side on the diameter is cut from it? (

π = \frac{22}{7}

)

AdvancedTry it yourself first

The sides forming the right angle of a right triangle are 63 cm and 84 cm (hypotenuse 105 cm). Find the area of the largest square at the right-angle corner (two sides of the square along the legs).

AdvancedTry it yourself first

A round pot is to be fitted as large as possible into a triangular corner stand whose right-angle sides are 20 cm and 21 cm. Find the radius of the pot (the incircle).

Packing coins and belts

Figure of centres · belt = perimeter +

2 π r

Touching circles: the enclosing circle, a tight belt, cutting discs from a sheet (square and zigzag rows) and the empty gap between them.

When equal circles of radius $r$ touch each other, the distance between the centres of two neighbours is $r + r = 2 r$ . So the centres together form a simple figure (a triangle, a square, a straight line) whose every side is $2 r$ . All the work is done on this “figure of the centres”.

Six round one: the centre of an outer circle is $2 r$ from the middle centre, so the circle enclosing all seven has radius $2 r + r = 3 r$ .
A tight belt: the straight parts = the perimeter of the figure of centres; the curved arcs together make exactly one full circle ( $2 π r$ ). So belt = perimeter of the figure of centres $+ 2 π r$ .
Cutting discs from a sheet: in square rows every disc needs $2 r$ , so the count $= ⌊ L /2 r ⌋ \times ⌊ B /2 r ⌋$ . In zigzag rows the centres of neighbouring rows are $r 3$ apart (the height of an equilateral triangle), so less width is needed and sometimes more discs fit.

◭ Diagram

Seven coins: from the middle to an outer centre

2 r

(green), then

r

more (orange), so the enclosing circle has radius

R = 3 r

◭ Diagram

A tight belt round three touching circles: three straights (each

2 r

) and three arcs which together make one full circle

◭ Diagram

The same sheet: 8 discs in square rows, 11 in zigzag rows. Zigzag rows fit more in a narrow width

∑ Formula

R = 3 r (7 coins) belt = P_{centres} + 2 π r N = ⌊ \frac{L}{2 r} ⌋ ⌊ \frac{B}{2 r} ⌋

Rules for packing

$r$: radius of every coin / pipe / disc (in units)
$R$: radius of the circle enclosing the seven coins
$P_{centres}$: perimeter of the figure joining the centres (side $2 r$ ): triangle $6 r$ , square $8 r$ , $n$ in a row $4 r (n - 1)$
$L, B$: length and breadth of the sheet (same unit)
$N$: number of discs that can be cut in square rows

Spot itSpot it: “touching circles / coins / pipes”, “tight belt / rope”, “discs from a sheet”. Traps: (1) using

π r

or two circumferences for the belt (the arcs always total

36 0^{\circ}

); (2) counting a fractional disc in a sheet; (3) giving the second zigzag row as many discs as the first (it is often one fewer).

Draw the figure of the centres (every side $2 r$ ) and find its perimeter or distance.
Belt: perimeter $+ 2 π r$ . Enclosing circle: distance to the outermost centre $+ r$ .
Sheet: count whole diameters in both directions; for zigzag rows check the width $2 r + (k - 1) r 3$ .
Empty/waste = outer area − (count × $π r^{2}$ ).

Example 31 · Basic

Seven coins of radius 3.5 cm lie with one in the middle and six round it, each touching its neighbours. Find the radius of the smallest circle holding all seven and the area of it not covered by the coins. (

π = \frac{22}{7}

)

Solution, step by step

1The centre of an outer coin is $2 r = 7$ cm from the middle centre and its edge is $3.5$ cm further; radius of the enclosing circle $R = 3 r = 10.5$ cm.
2Enclosing circle $= \frac{22}{7} \times 10. 5^{2} = \frac{22}{7} \times 110.25 = 346.5$ sq cm.
3Seven coins $= 7 \times \frac{22}{7} \times 3. 5^{2} = 22 \times 12.25 = 269.5$ sq cm.
4Not covered $= 346.5 - 269.5 = 77$ sq cm. (Shortcut: $2 π r^{2} = 2 \times \frac{22}{7} \times 12.25 = 77$ ✓)

Answer $R = 10.5$ cm; not covered 77 sq cm

The seven coins cover only

\frac{7}{9}

of the enclosing circle — this ratio is the same for every

r

Example 32 · Standard

Three pipes of radius 7 cm are laid touching each other in a triangle and tied with a tight band. (a) Find the minimum length of the band. (b) If four such pipes are laid in a square pattern, how long is the band? (

π = \frac{22}{7}

)

Solution, step by step

1(a) The three centres form an equilateral triangle of side $2 r = 14$ cm; straight parts $= 3 \times 14 = 42$ cm.
2The three arcs together make one full circle (each $12 0^{\circ}$ , three make $36 0^{\circ}$ ): $2 π r = 2 \times \frac{22}{7} \times 7 = 44$ cm.
3Band $= 42 + 44 = 86$ cm.
4(b) In a square the four centres form a square of side $14$ cm: straight parts $4 \times 14 = 56$ ; the arcs are again $4 \times 9 0^{\circ} = 36 0^{\circ}$ , i.e. $44$ . Band $= 56 + 44 = 100$ cm.

Answer (a) 86 cm (b) 100 cm

However many circles there are, as long as the belt goes round them on the outside the arcs total exactly

36 0^{\circ}

Example 33 · Advanced

Circular discs of radius 7 cm are to be cut from a sheet of 70 cm × 42 cm. (a) How many discs fit in square rows and how much sheet area is wasted? (b) Can zigzag rows give more discs? (

π = \frac{22}{7}

3 = 1.732

)

Solution, step by step

1(a) Diameter $= 14$ cm. Along the length $\frac{70}{14} = 5$ discs, along the breadth $\frac{42}{14} = 3$ ; total $5 \times 3 = 15$ .
2Area covered $= 15 \times \frac{22}{7} \times 49 = 15 \times 154 = 2310$ ; sheet $= 70 \times 42 = 2940$ ; wasted $= 2940 - 2310 = 630$ sq cm.
3(b) Zigzag rows: the first row holds 5 discs (70 cm). The second row is shifted by half a diameter and needs $5 \times 14 + 7 = 77$ cm, but the sheet is 70 cm, so only 4 discs.
4Width of the rows: $14 + (k - 1) \times 73 = 14 + (k - 1) \times 12.12$ . For $k = 3$ : $14 + 24.25 = 38.25 \leq 42$ ✓ (for $k = 4$ : $50.4 > 42$ ). Three rows: $5 + 4 + 5 = 14$ .
5 $14 < 15$ : on this sheet zigzag rows are not better; square rows give the maximum (15).

Answer (a) 15 discs; 630 sq cm wasted (b) no (zigzag gives 14)

Count both ways for every sheet — zigzag wins on a narrow sheet, not here.

BasicTry it yourself first

Seven coins of radius 7 cm are placed with one coin in the middle and six round it, each touching the middle coin and its two neighbours. Find the area of the smallest circular tray holding all seven that is not covered by coins. (

π = \frac{22}{7}

)

BasicTry it yourself first

Three pipes in a triangle (each touching the other two), each of radius 28 cm, are tied with a tight belt. Find the minimum length of the belt. (

π = \frac{22}{7}

)

StandardTry it yourself first

Four pipes in a row, each of radius 3.5 cm, are tied with a tight belt. Find the minimum length of the belt. (

π = \frac{22}{7}

)

StandardTry it yourself first

Four pipes in a square arrangement (touching their neighbours), each of radius 21 cm, are tied with a tight belt. Find the minimum length of the belt. (

π = \frac{22}{7}

)

StandardTry it yourself first

Four equal coins of radius 14 cm lie in a square pattern, each touching its two neighbours. Find the area of the empty gap in the middle of the four. (

π = \frac{22}{7}

)

StandardTry it yourself first

Circular discs of radius 7 cm are cut from a metal sheet of 84 cm × 56 cm in square rows (every row and column straight). What is the greatest number of discs?

StandardTry it yourself first

How much area of a 98 cm × 70 cm sheet is wasted when the greatest number of discs of radius 7 cm is cut from it in square rows? (

π = \frac{22}{7}

)

AdvancedTry it yourself first

Discs of radius 7 cm are cut from a 56 cm × 40 cm sheet in “zigzag rows” (the second row sits in the gaps of the first; neighbouring rows have their centres

r 3

apart). What is the greatest number of discs? (

3 = 1.732

)

AdvancedTry it yourself first

Five pipes in a row, each of radius 7 cm, are tied with a tight belt. Find the minimum length of the belt. (

π = \frac{22}{7}

)

Area of a shaded region

Add and subtract: petal, crescent, flower

Breaking a shaded region into a sum and difference of simple figures: circle in square, petal, flower, crescents (lunes), triangle in a semicircle and a part of a ring.

The pictures of “area of the shaded region” in SSC look the most frightening, but the solution is always one story: shaded part = (what is whole) − (what is not there), or a sum of pieces. The method — (1) see the whole figure; (2) see which pieces are to be removed or added; (3) apply the simple formula to each piece; (4) add and subtract. And for a piece that is covered twice (like a petal) use: sum of the two pieces − the whole figure = the doubly covered part.

◭ Diagram

Pattern 1: square − circle. The four corner pieces (orange)

= a^{2} - π r^{2}

, where

r = \frac{a}{2}

◭ Diagram

Pattern 2: the petal. The two quarter circles (blue, green) cover the square and the petal (orange) is covered twice: petal

= \frac{1}{4} π a^{2} + \frac{1}{4} π a^{2} - a^{2}

◭ Diagram

Pattern 3: crescents (lunes). The four crescents (orange) left outside the circumcircle of the square have a total area exactly equal to the area

a^{2}

of the square

∑ Formula

A_{square - circle} = a^{2} (1 - \frac{π}{4}) A_{petal} = a^{2} (\frac{π}{2} - 1) A_{crescents} = A_{square} or A_{triangle}

The three main patterns

$a$: side of the square (in units); in the petal the quarter circles also have radius $a$
$A_{square - circle}$: the four corners outside the largest circle in a square (square units)
$A_{petal}$: the common part of two quarter circles (square units)
$A_{crescents}$: the crescents left outside the circle: four on a square = the square; two on a right triangle = the triangle

Two more common forms: the flower — draw semicircles inwards on the four sides of a square; they cover the square and the flower in the middle is covered twice: flower $= 4 \times \frac{1}{2} π (\frac{a}{2})^{2} - a^{2} = a^{2} (\frac{π}{2} - 1)$ . A part of a ring $= \frac{θ}{360} π (R^{2} - r^{2})$ . A triangle in a semicircle: semicircle $-$ triangle.

Spot itSpot it: the shaded region has curved and straight boundaries mixed. Traps: (1) misreading the radius of a quarter circle (side of the square vs half the side); (2) adding a doubly covered part straight; (3) using

\frac{22}{7}

when the radius is not a multiple of 7. Trick: turn the picture round — sometimes shaded pieces swap places and make a simple figure.

Identify the whole figure and write its measures (what gives the radius?).
Decide: shaded = whole − something, or a sum of pieces, or a doubly covered part.
Find the area of every piece by a simple formula.
Add and subtract; the answer must be smaller than the whole figure — check.

Example 34 · Basic

A square handkerchief of side 14 cm has a circular embroidery in the middle touching all four sides. Find the total area of the four corners left outside the circle. (

π = \frac{22}{7}

)

Solution, step by step

1The circle touches the sides of the square, so its diameter $= 14$ cm and radius $= 7$ cm.
2Circle $= \frac{22}{7} \times 49 = 154$ sq cm; square $= 1 4^{2} = 196$ sq cm.
3Four corners $= 196 - 154 = 42$ sq cm.

Answer 42 sq cm

Formula:

a^{2} (1 - \frac{π}{4}) = 196 \times \frac{3}{14} = 42

✓.

Example 35 · Standard

In a square ABCD of side 14 cm two quarter circles of radius 14 cm, centred at A and C, are drawn inside the square. Find the area of the petal common to both. (

π = \frac{22}{7}

)

Solution, step by step

1Each quarter circle has area $= \frac{1}{4} \times \frac{22}{7} \times 1 4^{2} = \frac{1}{4} \times 616 = 154$ sq cm.
2Together the two quarter circles cover the whole square (nothing is left outside both), but the petal is in both, i.e. covered twice.
3So the petal $=$ sum of the two $-$ the square $= 154 + 154 - 196 = 112$ sq cm.

Answer 112 sq cm

Formula

a^{2} (\frac{π}{2} - 1) = 196 \times \frac{4}{7} = 112

✓ (

\frac{π}{2} - 1 = \frac{11}{7} - 1 = \frac{4}{7}

Example 36 · Advanced

A circle through the four vertices of a square of side 14 cm is drawn, and semicircles are drawn outwards on all four sides as diameters. Find the total area of the four crescent-shaped parts left outside the circle. (

π = \frac{22}{7}

)

Solution, step by step

1Four semicircles (radius 7 cm): $4 \times \frac{1}{2} \times \frac{22}{7} \times 49 = 308$ sq cm.
2The circumcircle of the square: diagonal $= 142$ , radius $= 72$ , area $= \frac{22}{7} \times 98 = 308$ sq cm — the same.
3Four semicircles = four crescents + (the four segments inside the circle). And circle = square + the same four segments. Both are 308, so the four crescents = the square.
4Crescents $= 196$ sq cm. (Check: segments $= 308 - 196 = 112$ ; crescents $= 308 - 112 = 196$ ✓)

Answer 196 sq cm

This pattern is called the lunes of Hippocrates: the answer is always the area of the square — no radii need to be found.

BasicTry it yourself first

A circle touching all four sides is drawn inside a square of side 42 cm. Find the total area of the four shaded corner pieces outside the circle. (

π = \frac{22}{7}

)

StandardTry it yourself first

Quarter circles of radius 10.5 cm are cut off at the four corners of a square handkerchief of side 21 cm. Find the area of the part left in the middle. (

π = \frac{22}{7}

)

StandardTry it yourself first

Two quarter circles of radius 21 cm are drawn inside a square of side 21 cm, centred at two opposite corners. Find the area of the petal common to both. (

π = \frac{22}{7}

)

StandardTry it yourself first

Semicircles are drawn inside a square garden of side 28 cm using each side as a diameter. Find the total area of the “flower” (the parts covered twice where the semicircles overlap). (

π = \frac{22}{7}

)

StandardTry it yourself first

A triangle is drawn in a semicircle of radius 7 cm with the diameter as its base and the apex at the midpoint of the arc (an isosceles right triangle). Find the shaded area of the semicircle outside the triangle. (

π = \frac{22}{7}

)

StandardTry it yourself first

The sides of a right triangle are 12 cm, 16 cm and the hypotenuse 20 cm. Semicircles are drawn outwards on all three sides as diameters. Find the total area of the two crescent-shaped (shaded) parts of the small semicircles that lie outside the hypotenuse semicircle.

AdvancedTry it yourself first

A circle through the four vertices of a square of side 18 cm is drawn, and semicircles are drawn outwards on each side as diameter. Find the total area of the four crescents that lie outside the circle.

AdvancedTry it yourself first

A flat circular ring has outer radius 21 cm and inner radius 14 cm. Find the area of the piece of the ring cut by a central angle of

4 5^{\circ}

. (

π = \frac{22}{7}

)

AdvancedTry it yourself first

Two quarter circles of radius 35 cm are drawn inside a square of side 35 cm, centred at two opposite corners. Find the area of the petal common to both. (

π = \frac{22}{7}

)

AdvancedTry it yourself first

The sides of a right triangle are 9 cm, 40 cm and the hypotenuse 41 cm. Semicircles are drawn outwards on all three sides as diameters. Find the total area of the two crescent-shaped (shaded) parts of the small semicircles that lie outside the hypotenuse semicircle.

Changes in dimensions and reshaping

Percent change · similar figures · a wire in a new shape

Percent change in area and perimeter when lengths, breadths and radii change, ratios of areas, and bending a wire into another shape.

Percent change in dimensions and ratio of areas

Multiply the factors ·

x + y + \frac{x y}{100}

k^{2}

Change of area when the measures of a rectangle, square, triangle or circle change, the condition to keep the area the same, and similar figures.

Area is the product of two measures ( $l \times b$ in a rectangle, $\frac{1}{2} bh$ in a triangle, side $\times$ side in a square, $π r \times r$ in a circle), so when the measures change by percents the area changes by the product of the factors. A rise of $x %$ has factor $1 + \frac{x}{100}$ , a fall of $y %$ has $1 - \frac{y}{100}$ . Do not add percents — multiply factors.

◭ Diagram

The old rectangle (dashed) turns into the new one (orange); length

+ 20%

, breadth

- 10%

: area factor

1.2 \times 0.9 = 1.08

∑ Formula

net change = x + y + \frac{x y}{100} % square / circle: 2 x + \frac{x ^{2}}{100} % similar: area ratio = k^{2}

Percent change in area

$x, y$: percent changes of the two measures (rise positive, fall negative)
$2 x + \frac{x ^{2}}{100}$: when both measures change alike: the side of a square or the radius of a circle changes by $x %$
$k$: ratio of corresponding sides of two similar figures

Similar figures are bigger and smaller copies of each other: every linear measure (side, perimeter, diagonal, radius, height) changes in the same ratio $k$ , but the area becomes $k^{2}$ times (a factor $k$ in each direction). So: side doubled gives 4 times the area; tripled gives 9 times; and if the area rises by 25% the side is $1.25$ times. Keeping the area the same: if the length has factor $f$ the breadth must have factor $\frac{1}{f}$ ; if the length rises by $x %$ the breadth must fall by $\frac{100 x}{100 + x} %$ .

◭ Diagram

Side

k

times: the count of unit squares is

k^{2}

times — 1, 4, 9

Figure	Area	Factor
Rectangle	$l \times b$	$(1 + \frac{x}{100}) (1 + \frac{y}{100})$
Square	$a^{2}$	$(1 + \frac{x}{100})^{2}$
Triangle	$\frac{1}{2} bh$	factor of base × factor of height
Circle	$π r^{2}$	$(1 + \frac{x}{100})^{2}$
Perimeter of a rectangle	$2 (l + b)$	with real lengths: $\frac{p f _{1} + q f _{2}}{p + q}$

Letters in the table: $x, y$ = percent changes, $a$ = side of a square, $r$ = radius, $p : q$ = ratio of length to breadth, $f_{1}, f_{2}$ = factors of length and breadth. A perimeter is a sum (not a product), so percents do not simply multiply in it — take real numbers for the length and breadth.

Spot itSpot it: “increased/decreased by …% … percent change in area”, “similar … ratio”. Traps: (1) writing

x + y

(the

\frac{x y}{100}

part is lost); (2) raising by

x %

and lowering by

x %

does not return the original (

1.2 \times 0.8 = 0.96

, a

4%

fall); (3) taking the change of area of a square/circle as

2 x

Write the factor of every measure ( $1 \pm \frac{x}{100}$ ).
Multiply the factors of the measures in the area formula (constants like $\frac{1}{2}$ and $π$ do not change).
Change $= (product - 1) \times 100%$ . For a perimeter, the new sum from real lengths.
Reverse question (measure from the area): square root or division of the factor.

Example 37 · Basic

The length of a rectangular field is increased by 20% and the breadth by 10%. By what percent does the area increase?

Solution, step by step

1Area $= l \times b$ , so the factors are $1 + \frac{20}{100} = 1.2$ and $1 + \frac{10}{100} = 1.1$ .
2The new area is $1.2 \times 1.1 = 1.32$ times the old; increase $= (1.32 - 1) \times 100 = 32%$ .
3By the formula: $x + y + \frac{x y}{100} = 20 + 10 + \frac{200}{100} = 32$ ✓.

Answer 32%

Trap: saying

20 + 10 = 30%

— the shared part of two rises,

2%

, is lost.

Example 38 · Standard

(a) The length of a rectangle is increased by 30% and the breadth decreased by 10% — what is the change in area? (b) If the length is increased by 25%, by what percent must the breadth be decreased to keep the area the same?

Solution, step by step

1(a) Factors $1.3$ and $0.9$ ; product $1.3 \times 0.9 = 1.17$ , a $17%$ increase. (Formula: $30 - 10 + \frac{30 \times ( - 10 )}{100} = 20 - 3 = 17$ ✓)
2(b) The factor of the length is $1.25$ . For the same area the factor of the breadth $= \frac{1}{1.25} = 0.8$ .
3The breadth falls by $1 - 0.8 = 0.2$ , i.e. $20%$ . (Formula: $\frac{100 x}{100 + x} = \frac{2500}{125} = 20$ ✓)

Answer (a) 17% increase (b) 20% decrease

A 25% rise is not undone by a 25% fall: the fall is always the fraction

\frac{x}{100 + x}

, here

\frac{1}{5}

Example 39 · Advanced

The area of a square plot increases by 125%. By what percent do its side, perimeter and diagonal increase? If the fence cost ₹40 per m and ₹6,400 for the old length, what will the new fence cost?

Solution, step by step

1The new area is $1 + \frac{125}{100} = 2.25$ times the old. Area $\propto$ (side) $^{2}$ , so the factor of the side $= 2.25 = 1.5$ .
2The side rises by $50%$ . The perimeter ( $4 a$ ) and the diagonal ( $a 2$ ) are in proportion to the side — they also rise by $50%$ (all linear measures change by the same factor).
3Old perimeter $= \frac{6400}{40} = 160$ m; new $= 160 \times 1.5 = 240$ m.
4Cost of the new fence $= 240 \times 40 = 9600$ rupees (or $6400 \times 1.5 = 9600$ ).

Answer side, perimeter, diagonal: 50% increase; cost ₹9,600

Match 125% area rise ↔ 50% side rise:

1. 5^{2} = 2.25

— remember it (125 ↔ 50, 44 ↔ 20, 69 ↔ 30, 21 ↔ 10).

BasicTry it yourself first

The length of a rectangular screen is increased by 15% and the breadth is increased by 20%. Find the percent change in the area.

BasicTry it yourself first

Each side of a square plot is lengthened by 30%. By what percent does the area of the plot change?

StandardTry it yourself first

The length of a rectangular field is decreased by 10% and the breadth is decreased by 20%. Find the percent change in the area.

StandardTry it yourself first

The area of a square tile is to be increased by 44%. By what percent must its side be increased?

StandardTry it yourself first

The length of a rectangle is increased by 60%. By what percent must the breadth be decreased to keep the area the same?

StandardTry it yourself first

The radius of a round pizza is increased by 20%. By what percent does its area increase?

StandardTry it yourself first

Corresponding sides of two similar figures are in the ratio 3 : 5. The smaller has an area of 63 sq cm. Find the area of the larger.

AdvancedTry it yourself first

The base of a triangle is increased by 40% and the height on it is decreased by 25%. Find the percent change in the area.

AdvancedTry it yourself first

The length and breadth of a rectangle are in the ratio 3 : 2. The length is increased by 20% and the breadth decreased by 10%. Find the percent change in the perimeter.

Bending a wire into another shape

Equal perimeter: the circle holds most · sum-product method

Bending a wire/rope from a rectangle into a square or circle, the shape with the largest area, and sides from area and perimeter.

However a wire is bent, its length does not change — it is the perimeter of every shape (the circumference of a circle). So “the wire was bent from a square into a circle” means: equal perimeter, different shape. For an equal perimeter the enclosed area depends on the shape, and a lovely fact is that the circle encloses the most area, then the square, then any rectangle (the longer and thinner, the less).

◭ Diagram

One wire of length

P = 12

units: a

4 \times 2

rectangle encloses 8 square units, the square 9, the circle about 11.5. The order: rectangle < square < circle

∑ Formula

A_{□} = \frac{P ^{2}}{16}, A_{◯} = \frac{P ^{2}}{4 π} l + b = \frac{P}{2}, (l - b)^{2} = (l + b)^{2} - 4 l b

For a wire of length

P

$P$: length of the wire = perimeter of every shape (in units)
$A_{□}, A_{◯}$: area of the square and of the circle made from the same wire (square units)
$l, b$: length and breadth of the rectangle

Sides of a rectangle from area and perimeter: if the half-perimeter $l + b = S$ and the area $l b = A$ are given, find the difference from $(l - b)^{2} = S^{2} - 4 A$ ; then $l = \frac{S + ( l - b )}{2}$ , $b = \frac{S - ( l - b )}{2}$ . Rectangle to square: the side of the square $= \frac{l + b}{2}$ (the mean), and the area gain is exactly $(\frac{l - b}{2})^{2}$ . For a fixed sum the product is greatest when the two numbers are equal (a square); with whole numbers, the closest pair.

Spot itSpot it: “a wire/rope/thread bent into … shape”, “area … and perimeter given, find the sides?”. Traps: (1) do not forget to link the wire to the circumference

2 π r

for a circle; (2) when a wire goes from a square to a circle its length stays the same, its area does not; (3) the largest square cut from a rectangular sheet has the shorter side as its side.

Find the length $P$ of the wire (the perimeter of the given shape).
In the new shape treat $P$ as the perimeter/circumference and find the measure: square $a = \frac{P}{4}$ ; circle $r = \frac{P}{2 π}$ .
Find the areas and compare.
Rectangle sides: from $S = \frac{P}{2}$ and $l b$ use $(l - b)^{2} = S^{2} - 4 A$ .

Example 40 · Basic

A wire 132 cm long is bent first into a square and then into a circle. Find both areas and their difference. (

π = \frac{22}{7}

)

Solution, step by step

1Square: the wire is the perimeter $= 132$ , side $= \frac{132}{4} = 33$ cm, area $= 3 3^{2} = 1089$ sq cm.
2Circle: the wire is the circumference: $2 π r = 132$ ⟹ $r = \frac{132 \times 7}{2 \times 22} = 21$ cm.
3Area of the circle $= \frac{22}{7} \times 2 1^{2} = \frac{22}{7} \times 441 = 1386$ sq cm.
4Difference $= 1386 - 1089 = 297$ sq cm — the circle encloses more.

Answer 1089 and 1386 sq cm; difference 297

Check: a circle always encloses about

\frac{4}{π} \approx 1.27

times the area of the square:

1089 \times 1.27 \approx 1386

✓.

Example 41 · Standard

A rectangular garden has an area of 168 sq m and a perimeter of 52 m. Find its length and breadth.

Solution, step by step

1Half-perimeter $l + b = \frac{52}{2} = 26$ and product $l b = 168$ .
2To find the difference: $(l - b)^{2} = (l + b)^{2} - 4 l b = 2 6^{2} - 4 \times 168 = 676 - 672 = 4$ ⟹ $l - b = 2$ .
3From sum 26 and difference 2: $l = \frac{26 + 2}{2} = 14$ m and $b = \frac{26 - 2}{2} = 12$ m.
4Check: $14 \times 12 = 168$ ✓ and $2 (14 + 12) = 52$ ✓.

Answer 14 m × 12 m

Another way: two numbers with sum 26 and product 168 — the factor pair

14 \times 12

shows them directly.

Example 42 · Advanced

The sides of a rectangular wire frame are in the ratio 5 : 3 and its perimeter is 176 cm. By how much more area does the same wire enclose when bent into a circle? (

π = \frac{22}{7}

)

Solution, step by step

1Sides $5 x, 3 x$ : perimeter $2 (5 x + 3 x) = 16 x = 176$ ⟹ $x = 11$ . The sides are 55 cm and 33 cm.
2Area of the rectangle $= 55 \times 33 = 1815$ sq cm.
3Circle: circumference $= 176$ ⟹ $r = \frac{176 \times 7}{44} = 28$ cm; area $= \frac{22}{7} \times 784 = 2464$ sq cm.
4Extra area $= 2464 - 1815 = 649$ sq cm.

Answer 649 sq cm

Bend the same wire into a square: side 44, area 1936 — rectangle (1815) < square (1936) < circle (2464): the same order.

BasicTry it yourself first

A wire 88 cm long is first bent into a square and then straightened and bent into a circle. By how much more area does the circle enclose than the square? (

π = \frac{22}{7}

)

StandardTry it yourself first

A rectangular garden has an area of 240 sq m and a perimeter of 62 m. Find its longer side.

StandardTry it yourself first

A wire bent into a rectangle of 30 cm × 14 cm is straightened and bent into a square. By how much does the enclosed area increase?

StandardTry it yourself first

The sides of a rectangular wire frame are in the ratio 8 : 3 and its perimeter is 44 cm. By how much more area is enclosed when the same wire is bent into a circle? (

π = \frac{22}{7}

)

BasicTry it yourself first

The largest square is cut from a rectangular sheet of 48 cm × 30 cm. Find the area of the strip left over.

StandardTry it yourself first

A rectangular flower bed is to be made with 62 m of wire so that its length and breadth are whole numbers of metres. Find the greatest possible area of the bed.

AdvancedTry it yourself first

A rectangular garden has an area of 300 sq m and a perimeter of 70 m. Find its longer side.

AdvancedTry it yourself first

A wire 264 cm long is first bent into a square and then straightened and bent into a circle. By how much more area does the circle enclose than the square? (

π = \frac{22}{7}

)

AdvancedTry it yourself first

A wire bent into a rectangle of 45 cm × 19 cm is straightened and bent into a square. By how much does the enclosed area increase?

Summary, mixed practice, revision

Every method in one place

A table of all methods, the shortcuts of the notes and the ones SSC adds, exam-style mixed questions and revision.

Summary of all methods

When · what · how

Every method: how to recognise the question and which formula or shortcut to use.

Method	Recognise	Key formula / shortcut
Path round a rectangle	path / border / frame; outside or inside	$2 w (l + b) \pm 4 w^{2}$ ; $8 w$
Crossing paths	two paths through the middle	$w (l + b - w)$ ; $(l - w_{2}) (b - w_{1})$
Circular path	round park / well / pond	$π w (2 r + w)$ ; $2 π w$
Fencing	fence, gate, posts	$(P - g) s c$ ; closed $\frac{P}{d}$ , open $\frac{L}{d} + 1$
Carpet and tiles	width, square tiles, border	$\frac{l b}{w}$ ; $\frac{l b}{t ^{2}}$ ; $2 (p + q) - 4$ ; HCF
Walls	whitewash / paint / paper	$2 (l + b) h -$ openings
Grazing	animal tied by a rope	$\frac{θ}{360} π r^{2}$ ; $\frac{3}{4} π R^{2} + \frac{1}{2} π (R - a)^{2}$
Track, diagonal	lap, lane, start; corner to corner	$2 s + 2 π r$ ; $2 π w$ ; $a + b - d$
Shape inside shape	largest circle / square / triangle	$2 r = a$ ; $d = 2 R$ ; $a = R 3$ ; hexagon $a = R$
In a triangle, semicircle	largest square / rectangle	$\frac{bh}{b + h}$ ; $\frac{bh}{4}$ ; $\frac{4 r ^{2}}{5}$
Packing, belt	touching circles, belt, sheet	$R = 3 r$ ; $P_{c} + 2 π r$ ; $⌊ \frac{L}{2 r} ⌋ ⌊ \frac{B}{2 r} ⌋$
Shaded region	curved + straight boundary	$a^{2} (1 - \frac{π}{4})$ ; $a^{2} (\frac{π}{2} - 1)$ ; crescents = square/triangle
Percent change	measures change by %, similar	$x + y + \frac{x y}{100}$ ; $2 x + \frac{x ^{2}}{100}$ ; $k^{2}$
Reshaping a wire	a wire is bent into …	$\frac{P ^{2}}{16}$ ; $\frac{P ^{2}}{4 π}$ ; $(l - b)^{2} = S^{2} - 4 A$

Letters in the table: $l, b$ = length and breadth, $w$ = width of a path/carpet, $w_{1}, w_{2}$ = widths of two paths, $r, R$ = radii, $P$ = perimeter / length of wire, $g$ = gate, $s$ = strands (or a straight), $c$ = rate, $d$ = distance between posts, $L$ = open fence, $p, q$ = rows/columns (or legs), $t$ = side of a tile, $θ$ = angle, $a$ = side, $h$ = height, $b$ = base, $S$ = half-perimeter, $x, y$ = percents, $k$ = ratio of sides.

Shortcut	Formula	Source
Largest square at a right angle	$\frac{pq}{p + q}$	notes
Largest square in a semicircle	$\frac{2 r}{5}$	notes
Six circles round one	$R = 3 r$	notes
The tight-belt rule	$P_{centres} + 2 π r$	notes (in our words)
Incircle of a right triangle	$\frac{p + q - c}{2}$	notes
Crescents = triangle / square	total crescents $=$ triangle or square	notes
Path = $w \times$ middle line	$w P_{mid}$	added for SSC
Outer − inner perimeter	$8 w$	added for SSC
Lawn left after crossing paths	$(l - w_{2}) (b - w_{1})$	added for SSC
Posts: closed / open	gaps / gaps $+ 1$	added for SSC
Largest tile	HCF	added for SSC
Grazing angle	inside $θ$ ; outside $36 0^{\circ} - θ$	added for SSC
Gap per lane	$2 π w$	added for SSC
Greatest rectangle in a triangle	half the area	added for SSC
Circle → square → square	halves each time	added for SSC
Similar figures	area $k^{2}$	added for SSC

Here: the letters are those of the table above; $P_{centres}$ = perimeter of the figure joining the centres of touching circles, $P_{mid}$ = length of the line through the exact middle of the path, $θ$ = the angle through which the rope swings.

Order of thinkingIn every question ask three things: (1) which simple figures are there, and are they added or subtracted? (2) is the edge (perimeter) asked or the surface (area)? (3) what is the last step — cost, percent, ratio or count? Only then choose the formula.

Mixed practice (exam style)

All methods of 2D mensuration mixed

Mixed questions like SSC CGL/CHSL/CPO — triangles, quadrilaterals, circles and every method of this chapter.

These questions do not name the method — as in the exam, first spot which figures are there and what is added or subtracted. They include triangle, quadrilateral and circle questions too (from parts 1 and 2 of this topic). Every question has a full hidden solution; look at the hint if you are stuck.

BasicTry it yourself first

A triangular garden has sides 10 m, 17 m and 21 m. Find its area.

BasicTry it yourself first

A rectangular school ground measures 48 m × 32 m. A path 3 m wide runs all round it on the outside. Find the area of the path.

BasicTry it yourself first

An equilateral triangle has a perimeter of 48 cm. Find its area.

StandardTry it yourself first

Through the middle of a 100 m × 70 m field run a road 6 m wide parallel to the length and a road 4 m wide parallel to the breadth. Find the total area of the roads.

StandardTry it yourself first

The edge of a round table measures 176 cm round. Find the area of its top. (

π = \frac{22}{7}

)

StandardTry it yourself first

A circular park has radius 49 m. Inside, along its edge, runs a footpath 7 m wide. Find the area of the footpath. (

π = \frac{22}{7}

)

StandardTry it yourself first

A plant nursery measures 96 m × 54 m. It is to be fenced with 3 strands of barbed wire, leaving a 6 m wide gate open. If the wire costs ₹11 per m, find the total cost.

BasicTry it yourself first

A rectangular screen has a diagonal of 26 cm and one side of 10 cm. Find its area.

StandardTry it yourself first

A prayer room measures 10.5 m × 8 m. It is to be covered with carpet 75 cm wide, sold at ₹60 per metre of length. Find the total cost.

StandardTry it yourself first

A regular polygon has 15 sides. Find each interior angle.

StandardTry it yourself first

A cow is tied by a 28 m rope at a corner outside a rectangular house (every wall is longer than the rope). How much area can it graze? (

π = \frac{22}{7}

)

StandardTry it yourself first

The sides of a right triangle containing the right angle are 9 cm and 12 cm. Find the length of the perpendicular from the right-angle vertex to the hypotenuse.

StandardTry it yourself first

The running track of a city stadium has two parallel straights of 120 m joined by two semicircles of inner radius 21 m. Find the length of one lap along the inner edge. (

π = \frac{22}{7}

)

StandardTry it yourself first

How much area is wasted when the largest circular disc is cut from a square piece of cloth of side 70 cm? (

π = \frac{22}{7}

)

BasicTry it yourself first

The parallel sides of a trapezium-shaped field are 19 m and 31 m and the distance between them is 12 m. Find its area.

StandardTry it yourself first

A classroom measures 10 m × 7.5 m × 3.6 m and has 1 opening of 2 m × 1.5 m and 2 openings of 1.2 m × 1 m. Find the cost of painting its four inside walls once at ₹16 per sq m (per coat).

StandardTry it yourself first

The sides forming the right angle of a right triangle are 84 cm and 112 cm (hypotenuse 140 cm). Find the area of the largest square at the right-angle corner (two sides of the square along the legs).

StandardTry it yourself first

A slice of a round pizza of radius 28 cm is cut with a central angle of

4 5^{\circ}

. Find its area. (

π = \frac{22}{7}

)

StandardTry it yourself first

A circle touching all four sides is drawn inside a square of side 56 cm. Find the total area of the four shaded corner pieces outside the circle. (

π = \frac{22}{7}

)

StandardTry it yourself first

A floor of 1,485 cm × 1,155 cm is to be covered, without cutting, with the largest possible square tiles. How many tiles are needed?

StandardTry it yourself first

The length of a rectangular poster is increased by 40% and the breadth is decreased by 15%. Find the percent change in the area.

StandardTry it yourself first

The wheel of a bicycle has a radius of 35 cm. How many revolutions does it make in covering 11,000 m? (

π = \frac{22}{7}

)

StandardTry it yourself first

Three pipes in a triangle (each touching the other two), each of radius 10.5 cm, are tied with a tight belt. Find the minimum length of the belt. (

π = \frac{22}{7}

)

StandardTry it yourself first

A quadrilateral plot has a diagonal of 32 m, and the perpendiculars to it from the other two vertices are 10 m and 6 m. Find the area of the plot.

StandardTry it yourself first

A postman goes from one corner of a rectangular ground, 33 m × 56 m, to the opposite corner. How many metres does he save by going across the diagonal instead of along the two sides?

◭ Diagram

StandardTry it yourself first

Two quarter circles of radius 28 cm are drawn inside a square of side 28 cm, centred at two opposite corners. Find the area of the petal common to both. (

π = \frac{22}{7}

)

AdvancedTry it yourself first

A rectangular garden measures 44 m × 30 m. An even path runs round it on the outside, and the outer boundary of the path is 48 m longer than the garden's perimeter. Find the area of the path.

AdvancedTry it yourself first

A goat is tied by a 42 m rope at a corner of the outer wall of a square water tank of side 28 m. How much area can it graze? (

π = \frac{22}{7}

)

StandardTry it yourself first

Each side of a square plot is shortened by 20%. By what percent does the area of the plot change?

AdvancedTry it yourself first

How much area of a 112 cm × 84 cm sheet is wasted when the greatest number of discs of radius 7 cm is cut from it in square rows? (

π = \frac{22}{7}

)

AdvancedTry it yourself first

How much of a circular plate of radius 49 cm is left over (wasted) after the largest square is cut from it? (

π = \frac{22}{7}

)

StandardTry it yourself first

A wire 308 cm long is first bent into a square and then straightened and bent into a circle. By how much more area does the circle enclose than the square? (

π = \frac{22}{7}

)

AdvancedTry it yourself first

The sides of a right triangle are 15 cm, 20 cm and the hypotenuse 25 cm. Semicircles are drawn outwards on all three sides as diameters. Find the total area of the two crescent-shaped (shaded) parts of the small semicircles that lie outside the hypotenuse semicircle.

BasicTry it yourself first

The radii of two circles are in the ratio 5 : 8. Find the ratio of their areas.

AdvancedTry it yourself first

A 400 m race is run on a track with two full bends per lap. The lanes are 1.75 m wide. Runner B's lane is 2 lanes outside runner A's. By how much should B's starting line be moved ahead of A's to make it fair? (

π = \frac{22}{7}

)

StandardTry it yourself first

A rectangular garden has an area of 360 sq m and a perimeter of 76 m. Find its longer side.

StandardTry it yourself first

Square tiles are laid in a courtyard in 26 rows with 17 tiles in each row. The tiles along the outer edge are red. How many red tiles are there?

StandardTry it yourself first

Corresponding sides of two similar figures are in the ratio 4 : 7. The smaller has an area of 96 sq cm. Find the area of the larger.

StandardTry it yourself first

A flat circular ring has outer radius 28 cm and inner radius 21 cm. Find the area of the piece of the ring cut by a central angle of

9 0^{\circ}

. (

π = \frac{22}{7}

)

BasicTry it yourself first

A kite-shaped piece has diagonals of 18 cm and 24 cm (perpendicular to each other). Find its area.

BasicTry it yourself first

A regular polygon has 9 sides. How many diagonals does it have?

Revision

Cards · common traps

Cards to memorise formulas and shortcuts, and the most common exam mistakes.

Forgetting the corner squares — $+ 4 w^{2}$ for an outside path, $- 4 w^{2}$ for an inside one.
Counting the crossing twice — subtract it once for crossing paths.
Taking the diameter as the radius — wells, pipes, coins, round tables.
Mixing cm and m — carpet/paper width in cm, rate per metre.
Adding +1 on a closed loop (or not adding on an open line) — the commonest trap with posts.
Taking an outside corner as ¼ — at an outside corner the rope swings $27 0^{\circ}$ .
Counting two semicircles as two circles — in tracks and belts the arcs make one circle.
Adding percents — multiply factors; a rise of $x %$ followed by a fall of $x %$ does not return the original.

PracticeTen minutes a day: make up and solve one question for each row of the summary table. Redo the practice questions of any method that keeps tripping you.