Triangle Area Ratios

Competitive Mathematics · SSC CGL

4 Part31 worked examples101 practice questions

The area of a triangle is $\frac{1}{2} \times$ base $\times$ height. So the ratio of the areas of two triangles is the product of the ratio of bases and the ratio of heights. This one sentence is the root of the whole chapter: when heights are equal look at the bases, when bases are equal look at the heights, and when nothing is equal look at the angles.

In the previous chapter, “Similar Triangles and Area Ratios”, we learnt the square rule for similar triangles. Here the triangles are not similar: they simply have bases on one line, or lie between the same parallels, or share an angle.

What you see	Which method
a vertex joined to a point on the base, “BD : DC”	Base and height
medians, the centroid, joining midpoints	Midpoints and the centroid
a parallelogram, midpoints of sides, a shaded part	Parallelogram: assume a total
a point $P$ inside a parallelogram	Inner points and diagonals
a trapezium’s diagonals, four triangles	Four triangles of a trapezium
one angle $A$ , points $D$ and $E$ on its sides	The same-angle rule
small triangles made by crossing lines	Areas from weights

Start hereFirst job in every question: assume a convenient total area to avoid fractions (like 16, 24, 48, 100) and write every piece in it. Take the ratio at the end. Assuming the total does not change the answer, as only a ratio is asked.

Every method follows one order: the idea, the formula, how to recognise it, steps, solved examples, then questions to do yourself. The figures play by themselves; watch them in the order of the argument.

Base, height and midpoints

Equal height · the median

Area ratios from bases and heights, and the equal pieces made by medians and midpoints.

The base-and-height rules

Equal height · equal base · parallels

Four rules from area $= \frac{1}{2} \times$ base $\times$ height: common height, common base, between parallels and the median.

The idea

The area is $\frac{1}{2} b h$ ( $b$ = base, $h$ = height measured by the perpendicular to that base). If two triangles have the same height then $\frac{1}{2} h$ is common and the ratio is just that of the bases. In the same way, with the same base the ratio is that of the heights.

◭ Diagram

The perpendicular

h

from

A

BC

is the same for

A B D

and

A D C

, so

[A B D] : [A D C] = B D : D C = 2 : 3

∑ Formula

\frac{[ A B D ]}{[ A D C ]} = \frac{B D}{D C} and in general \frac{[ △ _{1} ]}{[ △ _{2} ]} = \frac{b _{1}}{b _{2}} \times \frac{h _{1}}{h _{2}}

Ratio of areas

$[A B D]$: the area of $△ A B D$ (square brackets mean area)
$B D, D C$: bases on the same line $BC$
$b_{1}, b_{2}$: the bases of the two triangles
$h_{1}, h_{2}$: the heights on those bases

Four rules

Rule	When	Result
1. common vertex, bases on one line	$D$ on $BC$	$[A B D] : [A D C] = B D : D C$
2. common base	both on base $BC$	area : height
3. common base, between the same parallels	both apexes on a line $l ∥ BC$	areas equal
4. the median	$D$ is the midpoint of $BC$	$[A B D] = [A D C] = \frac{1}{2} [A BC]$

Watch rule 3 in the moving figure: the apex $A$ slides anywhere on the line $l$ , the height $h$ stays the same, so the area $\frac{1}{2} \times BC \times h$ does not change.

◭ Diagram

The apex moves along

l ∥ BC

: the shape changes but the area stays the same

◭ Diagram

The median

A D

: equal bases

B D = D C

and a common height, so each part is

\frac{1}{2}

A useful consequence (rule 3): in a trapezium $A BC D$ with $A B ∥ D C$ , $[A B D] = [A BC]$ (common base $A B$ , apexes $D$ and $C$ on the line $D C ∥ A B$ ). Removing the common $[A OB]$ gives $[A O D] = [BOC]$ , where $O$ is the crossing point of the diagonals.

Watch outThe commonest slip: taking the ratio of bases when the heights are not equal. First ask: is the vertex common and are the bases on one line?

Steps

Identify the base and height of both triangles; see which one is shared.
Common height ⟹ ratio of bases; common base ⟹ ratio of heights; common base between parallels ⟹ equal.
Assume a convenient total and write the area of each piece.

Example 1 · Basic

△ A BC

D

is on

BC

with

B D : D C = 3 : 5

. The area of

△ A BC

is 96 sq cm. Find the areas of

△ A B D

and

△ A D C

Solution, step by step

1The perpendicular from $A$ to $BC$ is the same height $h$ for both triangles.
2So $[A B D] : [A D C] = B D : D C = 3 : 5$ . In all $3 + 5 = 8$ parts $= 96$ , 1 part $= 12$ .
3 $[A B D] = 3 \times 12 = 36$ and $[A D C] = 5 \times 12 = 60$ sq cm.

Answer 36 and 60 sq cm

Check:

36 + 60 = 96

. ✓

Example 2 · Standard

In trapezium

A BC D

A B ∥ D C

and the diagonals meet at

O

[A B D] = 54

and

[A OB] = 30

sq cm. Find

[A O D]

and

[BOC]

Solution, step by step

1 $△ A B D$ and $△ A BC$ share the base $A B$ . $D$ and $C$ lie on the line $D C ∥ A B$ , so the heights are equal: $[A BC] = [A B D] = 54$ .
2 $[A B D] = [A OB] + [A O D]$ ⟹ $[A O D] = 54 - 30 = 24$ .
3In the same way $[A BC] = [A OB] + [BOC]$ ⟹ $[BOC] = 54 - 30 = 24$ .

Answer $[A O D] = [BOC] = 24$ sq cm

The trick: take the common part away from equals.

Example 3 · Standard

Points

D

and

E

lie on the side

BC

△ A BC

with

B D : D E : EC = 2 : 3 : 4

. The area of

△ A BC

is 108 sq cm. Find the areas of

△ A B D

△ A D E

△ A EC

and

△ A D C

Solution, step by step

1All of them share the vertex $A$ and have bases on the line $BC$ , so the height is common. Areas are in the ratio of the bases: $2 : 3 : 4$ .
2In all $2 + 3 + 4 = 9$ parts $= 108$ , 1 part $= 12$ .
3 $[A B D] = 24$ , $[A D E] = 36$ , $[A EC] = 48$ sq cm.
4 $[A D C] = [A D E] + [A EC] = 36 + 48 = 84$ sq cm (or $\frac{7}{9} \times 108$ ).

Answer 24, 36, 48 and 84 sq cm

With many pieces write them all in parts at once.

Example 4 · Standard

Two triangles have bases 8 cm and 12 cm and the heights on them are 9 cm and 5 cm. Find the ratio of their areas.

Solution, step by step

1Here both the bases and the heights differ, so both ratios multiply.
2 $\frac{[ △ _{1} ]}{[ △ _{2} ]} = \frac{8}{12} \times \frac{9}{5} = \frac{72}{60} = \frac{6}{5}$ .
3Check: the areas are $\frac{1}{2} \times 8 \times 9 = 36$ and $\frac{1}{2} \times 12 \times 5 = 30$ , ratio $36 : 30 = 6 : 5$ . ✓

Answer $6 : 5$

Usually one thing is shared; here it was not, so both multiplied.

Example 5 · Advanced

The area of

△ A BC

is 120 sq cm.

D

is on

BC

with

B D : D C = 3 : 2

and

E

is the midpoint of

A D

. Find the area of

△ B D E

Solution, step by step

1Step one: $[A B D] : [A D C] = B D : D C = 3 : 2$ (common height), so $[A B D] = \frac{3}{5} \times 120 = 72$ .
2Step two: in $△ A B D$ , $E$ is the midpoint of $A D$ . $△ A BE$ and $△ BE D$ share the vertex $B$ and have equal bases $A E = E D$ on the line $A D$ , so they are equal.
3 $[B D E] = \frac{1}{2} \times 72 = 36$ sq cm.

Answer 36 sq cm

Two-step question: find the big parts first, then split one again with the same rule.

Try it yourself

◭ Diagram

BasicTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 2 : 3

. The area of

△ A BC

is 80 sq cm. Find the area of

△ A B D

◭ Diagram

BasicTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 5 : 3

. The area of

△ A BC

is 72 sq cm. Find the area of

△ A D C

BasicTry it yourself first

△ A BC

A D

is a median and

[A BC] = 150

sq cm. Find

[A D C]

StandardTry it yourself first

One triangle has base 15 cm and height 12 cm; another has base 25 cm and height 9 cm. Find the ratio of their areas.

StandardTry it yourself first

In trapezium

A BC D

A B ∥ D C

, the diagonals meet at

O

[A B D] = 75

and

[A OB] = 45

sq cm. Find

[A O D]

StandardTry it yourself first

The area of

△ PQR

is 36 sq cm.

S

is any point on the line through

P

parallel to

QR

. Find

[SQR]

StandardTry it yourself first

Points

D

E

lie on the side

BC

△ A BC

with

B D : D E : EC = 1 : 4 : 5

[A BC] = 200

sq cm. Find

[A D E]

AdvancedTry it yourself first

The area of

△ A BC

is 210 sq cm.

D

is on

BC

with

B D : D C = 3 : 4

and

E

is on

A D

with

A E : E D = 2 : 3

. Find

[A BE]

AdvancedTry it yourself first

In trapezium

A BC D

A B ∥ D C

and

A B = 2 D C

. If

[A B D] = 48

sq cm, find the area of the trapezium.

BasicTry it yourself first

Two triangles share the base

BC = 12

cm and the heights of

A

and

A_{2}

above

BC

are 7 cm and 10 cm. Find the ratio of their areas.

AdvancedTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 4 : 3

and

[A B D] - [A D C] = 6

sq cm. Find

[A BC]

Midpoints, medians and the centroid

Equal pieces: a half, a quarter, a sixth

The four equal triangles made by joining midpoints, the six equal triangles made by the three medians and the areas that follow.

The idea

Midpoints give “equal bases” everywhere, and equal bases with a common height mean equal areas. So pictures with midpoints cut into equal pieces. Three such pictures keep coming:

1. A median $A D$ cuts the triangle into two equal halves (the previous method).

2. The midpoint triangle: if $D, E, F$ are the midpoints of $BC, C A, A B$ then joining $D E, EF, F D$ makes four congruent (equal) triangles (as $EF ∥ BC$ , $EF = \frac{1}{2} BC$ and so on). Each is $\frac{1}{4}$ .

◭ Diagram

Joining the midpoints makes four equal triangles:

[D EF] = \frac{1}{4} [A BC]

3. The centroid: the three medians meet at one point $G$ (the centroid) and cut the triangle into six equal triangles. Why? The median $A D$ gives $[A B D] = [A D C] = \frac{1}{2}$ . $G$ divides $A D$ in $2 : 1$ , so $[GB D] = [G D C] = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$ (common height, one-third of the base), and the other four are $\frac{1}{6}$ in the same way.

◭ Diagram

Three medians: six equal pieces, each

\frac{1}{6}

Region	Fraction of $[A BC]$
$[GBC], [GC A], [G A B]$	$\frac{1}{3}$ each
any one of the six small triangles ( $[GB D]$ etc.)	$\frac{1}{6}$
quadrilateral $A FGE$ (two small triangles)	$\frac{1}{3}$
the midpoint triangle $[D EF]$	$\frac{1}{4}$
$[GEF]$ ( $EF ∥ BC$ , $EF = \frac{1}{2} BC$ )	$\frac{1}{4} \times \frac{1}{3} = \frac{1}{12}$
$E$ the midpoint of median $A D$ : $[A BE]$ , $[A CE]$	$\frac{1}{4}$ each

Here: $D, E, F$ = midpoints of the sides; $G$ = the centroid; $[]$ = area.

UsefulFor any point $P$ on a median

A D

[PB D] = [PC D]

and

[A BP] = [A CP]

(as

B D = D C

). So the whole median divides the areas equally on both sides.

Steps

Mark the midpoints and medians and take the total as six or twelve equal units.
Count the asked region in those units; where a piece is a fraction, use the common-height rule.
If a ratio is asked divide the two counts.

Example 6 · Basic

The medians

A D

BE

CF

△ A BC

meet at the centroid

G

. If

[A BC] = 144

sq cm find

[GBC]

[G D C]

and the area of quadrilateral

A FGE

Solution, step by step

1The three medians cut $△ A BC$ into six equal triangles, each $\frac{144}{6} = 24$ sq cm.
2 $△ GBC$ holds two small triangles ( $GB D$ and $G D C$ ): $[GBC] = 48$ .
3 $△ G D C$ is one small triangle: $[G D C] = 24$ .
4Quadrilateral $A FGE$ holds $△ A FG$ and $△ A GE$ , i.e. two: $48$ sq cm.

Answer $[GBC] = 48$ , $[G D C] = 24$ , $[A FGE] = 48$ sq cm

Check:

[GBC] + [GC A] + [G A B] = 48 \times 3 = 144

. ✓

Example 7 · Standard

D, E, F

are the midpoints of

BC, C A, A B

[D EF] = 18

sq cm. Find

[A BC]

and the area of quadrilateral

A F D E

Solution, step by step

1Joining the midpoints makes four equal triangles, so $[A BC] = 4 \times 18 = 72$ sq cm.
2 $A F D E$ holds two small triangles ( $A FE$ and $D EF$ ): $2 \times 18 = 36$ sq cm.
3(This is $\frac{1}{2} \times 72 = 36$ too, since $A F D E$ is a parallelogram.)

Answer $[A BC] = 72$ , $[A F D E] = 36$ sq cm

The midpoint triangle is always

\frac{1}{4}

Example 8 · Standard

△ A BC

D

is the midpoint of

BC

and

E

is on

A D

with

A E : E D = 1 : 2

[A BC] = 96

sq cm. Find

[A BE]

and

[BEC]

Solution, step by step

1 $A D$ is a median: $[A B D] = \frac{1}{2} \times 96 = 48$ .
2 $△ A BE$ and $△ A B D$ share the vertex $B$ with bases $A E : A D = 1 : 3$ on the line $A D$ : $[A BE] = \frac{1}{3} \times 48 = 16$ .
3 $△ BEC$ and $△ A BC$ share the base $BC$ ; the heights are in the ratio $E D : A D = 2 : 3$ : $[BEC] = \frac{2}{3} \times 96 = 64$ .

Answer $[A BE] = 16$ , $[BEC] = 64$ sq cm

Check:

[A BE] + [A CE] = 2 \times 16 = 32

and

[BEC] = 64

, total

96

. ✓

Example 9 · Advanced

The medians

BE

and

CF

meet at the centroid

G

(

E, F

the midpoints of

C A, A B

[A BC] = 60

sq cm. Find

[GEF]

and the ratio

[GEF] : [BCEF]

Solution, step by step

1 $[GBC] = \frac{1}{3} \times 60 = 20$ (the three triangles at $G$ are equal).
2The line $EF$ joining the midpoints: $EF ∥ BC$ and $EF = \frac{1}{2} BC$ . In $△ GEF$ and $△ GCB$ , $∠ GEF = ∠ GCB$ and $∠ GFE = ∠ GBC$ (alternate angles), so they are similar with multiplier $\frac{1}{2}$ .
3The square rule (previous chapter): $[GEF] = (\frac{1}{2})^{2} \times 20 = 5$ sq cm.
4The quadrilateral $BCEF = [A BC] - [A FE] = 60 - 15 = 45$ ( $A FE$ is $\frac{1}{4}$ , the midpoint triangle). So $[GEF] : [BCEF] = 5 : 45 = 1 : 9$ .

Answer $[GEF] = 5$ sq cm; ratio $1 : 9$

Similarity and area rules worked together;

\frac{1}{12} : \frac{3}{4} = 1 : 9

Example 10 · Advanced

A D

is a median and

P

lies on it with

A P : P D = 3 : 1

[A BC] = 80

sq cm. Find

[PBC]

[A BP]

and

[A CP]

Solution, step by step

1 $△ PBC$ and $△ A BC$ share the base $BC$ . Their heights are from $P$ and $A$ to $BC$ , in the ratio $P D : A D$ ( $P$ , $A$ , $D$ are collinear).
2 $A D = A P + P D = 3 + 1 = 4$ parts, so $P D : A D = 1 : 4$ and $[PBC] = \frac{1}{4} \times 80 = 20$ sq cm.
3The remaining $80 - 20 = 60$ sq cm splits equally between $[A BP]$ and $[A CP]$ (a point on the median): $30$ each.

Answer $[PBC] = 20$ , $[A BP] = [A CP] = 30$ sq cm

With

A P : P D = 2 : 1

we get

P = G

and

[PBC] = \frac{1}{3}

. ✓

Try it yourself

◭ Diagram

BasicTry it yourself first

The medians

A D

BE

CF

△ A BC

meet at the centroid

G

and

[A BC] = 120

sq cm. Find the area of

△ GBC

BasicTry it yourself first

D, E, F

are the midpoints of

BC, C A, A B

and

[D EF] = 25

sq cm. Find

[A BC]

BasicTry it yourself first

D, E, F

are the midpoints of

BC, C A, A B

and

[A BC] = 84

sq cm. Find the area of quadrilateral

B D EF

◭ Diagram

StandardTry it yourself first

The medians

A D

BE

CF

△ A BC

meet at the centroid

G

and

[A BC] = 72

sq cm. Find the area of quadrilateral

A FGE

StandardTry it yourself first

△ A BC

D

is the midpoint of

BC

and

E

is on

A D

with

A E : E D = 1 : 2

[BEC] = 60

sq cm. Find

[A BC]

StandardTry it yourself first

G

is the centroid. The line through

G

parallel to

BC

meets

A B

and

A C

P

and

Q

[A BC] = 81

sq cm. Find

[A PQ]

AdvancedTry it yourself first

The medians

A D

BE

CF

meet at the centroid

G

. Find

[GEF] : [A FGE]

AdvancedTry it yourself first

A D

is a median and

P

is on it with

A P : P D = 2 : 3

[A BC] = 150

sq cm. Find

[A BP]

AdvancedTry it yourself first

D, E

lie on

BC

with

B D = D E = EC

F

is the midpoint of

A C

[A BC] = 60

sq cm. Find

[D EF]

AdvancedTry it yourself first

The area of

△ A BC

is 60 sq cm.

D

is the midpoint of

BC

E

A D

and

F

A E

. Find

[A BF]

Parallelograms and trapeziums

Assume a total · four triangles

Pieces of a parallelogram, the inner-point rules, and the four triangles made by the diagonals of a trapezium.

Pieces of a parallelogram: assume a total

Midpoints · shaded parts · fractions

Areas of the triangles made by joining midpoints and vertices of a parallelogram, by assuming the total.

The idea

Let the parallelogram $A BC D$ have area $T$ . A diagonal cuts it into two equal triangles: $[A BC] = \frac{T}{2}$ . More: for any point $P$ on the side $C D$ , $△ A BP$ has base $A B$ and its apex $P$ on the line $C D ∥ A B$ , so its height is the height of the parallelogram and $[A BP] = \frac{1}{2} A B \cdot h = \frac{T}{2}$ .

Now the trick for midpoint questions: assume $T$ to be a number that makes every piece a whole number (like 16, 32, 64). Then write each triangle’s area as “fraction of base × fraction of height × $\frac{T}{2}$ ” and take the ratio of shaded to unshaded at the end.

A handy rule: when a triangle has an angle of the parallelogram (like angle $C$ in $△ ECF$ ), its area is $\frac{1}{2} \times$ (fraction of one side) $\times$ (fraction of the other side) $\times T$ . (Multiplying two such fractions is the “same-angle rule”, a separate method in this chapter.)

◭ Diagram

BE : EC = 1 : 3

and

F

the midpoint: four pieces

\frac{1}{8}, \frac{1}{4}, \frac{3}{16}, \frac{7}{16}

(fractions of

T

)

Situation ( $T$ = the parallelogram)	Area
triangle $A BC$ from a diagonal	$\frac{1}{2} T$
any point $P$ on $C D$ : $[A BP]$	$\frac{1}{2} T$
$E$ the midpoint of $BC$ : $[A BE]$ , $[A EC]$	$\frac{1}{4} T$
$E, F$ the midpoints of $BC, C D$ : $[ECF]$	$\frac{1}{8} T$
the same: $[A EF]$ (take three corners from $T$ )	$\frac{3}{8} T$
the quadrilateral of the four side midpoints	$\frac{1}{2} T$

Here: $T$ = the area of the parallelogram; $E, F, P$ = the given points.

Two pretty figures

Joining the midpoints of the four sides makes an inner parallelogram of area $\frac{T}{2}$ : the four corner triangles are $\frac{1}{8} T$ each ( $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} T$ ), together $\frac{T}{2}$ .

◭ Diagram

Four corner triangles of

\frac{1}{8}

each, the inner quadrilateral

\frac{1}{2}

Two lines from a vertex cut the diagonal in three equal parts: if $E, F$ are the midpoints of $BC, C D$ then $A E$ and $A F$ cut the diagonal $B D$ at $P, Q$ with $BP = PQ = Q D$ . Because in $△ A BC$ both $A E$ and $BO$ ( $O$ the crossing of the diagonals) are medians, $P$ is the centroid and $BP = \frac{2}{3} BO = \frac{1}{3} B D$ ; the same for $Q$ . Then $[A PQ] = \frac{1}{3} [A B D] = \frac{1}{6} T$ .

◭ Diagram

BP = PQ = Q D

and

[A PQ] = \frac{1}{6} T

CheckWhen shaded : unshaded is asked write both in “parts” and add: the sum must equal the total

T

. That is your check.

Steps

Assume the area $T$ so that all fractions become whole numbers.
Write every corner triangle as “ $\frac{1}{2} \times$ fraction of a side $\times$ fraction of the side $\times T$ ”; a triangle on a common base between the parallels as $\frac{T}{2}$ or a fraction of it.
Get the middle piece by subtracting from the total.

Example 11 · Basic

The parallelogram

A BC D

has area 96 sq cm and

E

is the midpoint of

BC

. Find

[A BE]

[A E D]

and

[A EC]

Solution, step by step

1 $[A BC] = \frac{T}{2} = 48$ (from the diagonal $A C$ ).
2 $E$ is the midpoint, so $[A BE] = [A EC] = \frac{1}{2} \times 48 = 24$ sq cm (common vertex $A$ , equal bases $BE = EC$ ).
3 $△ A E D$ has base $A D$ and its apex $E$ on the line $BC ∥ A D$ , so $[A E D] = \frac{T}{2} = 48$ sq cm.

Answer $[A BE] = 24$ , $[A EC] = 24$ , $[A E D] = 48$ sq cm

Check:

24 + 48 + 24 = 96

(the three together make

T

). ✓

Example 12 · Standard

The parallelogram

A BC D

has area 64 sq cm.

E

is on

BC

with

BE : EC = 1 : 3

and

F

is the midpoint of

C D

. Find

[A BE]

[A D F]

[ECF]

and

[A EF]

Solution, step by step

1 $BE = \frac{1}{4} BC$ . $△ A BE$ and $△ A BC$ share the vertex $A$ with bases $BE : BC = 1 : 4$ : $[A BE] = \frac{1}{4} \times 32 = 8$ .
2 $F$ is the midpoint: $△ A D F$ and $△ A D C$ share the vertex $A$ with bases $D F : D C = 1 : 2$ : $[A D F] = \frac{1}{2} \times 32 = 16$ .
3 $△ ECF$ : angle $C$ shared, $CE = \frac{3}{4} CB$ and $CF = \frac{1}{2} C D$ . $[ECF] = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{2} \times 64 = 12$ (the rule $\frac{1}{2} \cdot$ fraction $\cdot$ fraction $\cdot T$ ).
4 $[A EF] = 64 - 8 - 16 - 12 = 28$ sq cm.

Answer $8, 16, 12$ and $[A EF] = 28$ sq cm

Check by adding all four:

8 + 16 + 12 + 28 = 64

. ✓ (formula

\frac{7}{16} \times 64 = 28

)

Example 13 · Standard

A parallelogram has area 90 sq cm. Find the area of the quadrilateral made by joining the midpoints of its four sides and of each corner triangle.

Solution, step by step

1Each corner triangle (like $△ A PS$ ): angle $A$ shared, both sides halved. Area $= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times T = \frac{T}{8} = 11.25$ .
2The four corners together are $4 \times 11.25 = 45$ sq cm.
3The inner quadrilateral is $90 - 45 = 45$ sq cm, exactly half.

Answer Inner quadrilateral 45 sq cm; each corner 11.25 sq cm

For any quadrilateral (not just a parallelogram) the midpoint quadrilateral is exactly half.

Example 14 · Advanced

The parallelogram

A BC D

has area 72 sq cm.

E, F

are the midpoints of

BC, C D

and

A E

A F

meet the diagonal

B D

P, Q

. Find

BP : PQ : Q D

and

[A PQ]

Solution, step by step

1The diagonals $A C$ and $B D$ bisect each other at $O$ . In $△ A BC$ , $A E$ is a median ( $E$ is a midpoint) and so is $BO$ ( $O$ is a midpoint).
2The two medians meet at $P$ , so $P$ is the centroid: $BP = \frac{2}{3} BO = \frac{2}{3} \cdot \frac{1}{2} B D = \frac{1}{3} B D$ . Likewise in $△ A D C$ , $Q$ is the centroid: $D Q = \frac{1}{3} B D$ .
3So $BP : PQ : Q D = 1 : 1 : 1$ .
4 $△ A PQ$ and $△ A B D$ share the vertex $A$ with base $PQ = \frac{1}{3} B D$ : $[A PQ] = \frac{1}{3} [A B D] = \frac{1}{3} \times 36 = 12$ sq cm.

Answer $1 : 1 : 1$ and $[A PQ] = 12$ sq cm

The centroid’s

2 : 1

fact came in indirectly:

\frac{T}{6} = 12

. ✓

Try it yourself

BasicTry it yourself first

The parallelogram

A BC D

has area 120 sq cm and

E

is the midpoint of

BC

. Find

[A E D]

BasicTry it yourself first

The parallelogram

A BC D

has area 84 sq cm and

P

is any point on the side

C D

. Find

[A BP]

BasicTry it yourself first

A parallelogram has area 150 sq cm. Find the area of the quadrilateral formed by joining the midpoints of its four sides.

◭ Diagram

StandardTry it yourself first

The parallelogram

A BC D

has area 72 sq cm.

E

is on

BC

with

BE : EC = 1 : 2

and

F

is on

C D

with

CF : F D = 1 : 1

. Find the area of

△ A EF

◭ Diagram

StandardTry it yourself first

The parallelogram

A BC D

has area 64 sq cm.

E

is on

BC

with

BE : EC = 1 : 3

and

F

is on

C D

with

CF : F D = 1 : 3

. Find the area of

△ A EF

StandardTry it yourself first

The parallelogram

A BC D

has area 160 sq cm, its diagonals meet at

O

and

E

is on

A B

with

A E : EB = 1 : 3

. Find

[OEB]

StandardTry it yourself first

In the parallelogram

A BC D

M

is the midpoint of

A B

and

N

C D

. The quadrilateral

A MCN

has area 36 sq cm. Find

[A BC D]

AdvancedTry it yourself first

The parallelogram

A BC D

has area 108 sq cm.

E, F

are the midpoints of

BC, C D

and

A E, A F

meet the diagonal

B D

P, Q

. Find

[A PQ]

AdvancedTry it yourself first

A BC D

is a parallelogram,

E

is the midpoint of

BC

and

F

is on

C D

with

CF : F D = 1 : 2

. Find

[A EF] : [ECF]

StandardTry it yourself first

The parallelogram

A BC D

has area 144 sq cm and

P

is on

A B

with

A P : PB = 1 : 2

. Find the larger of

[A P D]

and

[PBC D]

Inner points and diagonals

Sum of opposite triangles · diagonals

The rule for the four triangles made by a point inside a parallelogram and the triangles made by the diagonals of any quadrilateral.

The idea

Take any point $P$ inside the parallelogram $A BC D$ and join it to all four corners. Four triangles appear: $P A B, PBC, PC D, P D A$ . Drop perpendiculars from $P$ to $A B$ and $C D$ ; their lengths add up to the distance $h$ between the parallels $A B$ and $C D$ . So $[P A B] + [PC D] = \frac{1}{2} A B (h_{1} + h_{2}) = \frac{1}{2} A B \cdot h = \frac{T}{2}$ (as $A B = C D$ ). The same holds for $[PBC] + [P D A]$ .

◭ Diagram

Wherever

P

moves, blue + green is always

\frac{1}{2} [A BC D]

∑ Formula

[P A B] + [PC D] = [PBC] + [P D A] = \frac{1}{2} [A BC D]

A point

P

inside a parallelogram

$[P A B]$: the area of the triangle made by $P$ , $A$ and $B$
$[A BC D]$: the area $T$ of the whole parallelogram

Direct use: if two opposite triangles (like $P A B$ and $PC D$ ) are given the whole area is $T = 2 ([P A B] + [PC D])$ ; if three are given the fourth follows: $[P D A] = [P A B] + [PC D] - [PBC]$ .

The diagonals

In a parallelogram the diagonals bisect each other at $O$ , so the four triangles are equal ( $\frac{T}{4}$ each) and any line through $O$ halves the parallelogram (a half-turn about the centre puts one part onto the other).

In any quadrilateral whose diagonals $A C$ and $B D$ meet at $O$ , four triangles appear. $△ A OB$ and $△ BOC$ share the height (from $B$ to $A C$ ), so $[A OB] : [BOC] = A O : OC$ . The same ratio holds for $[A O D] : [D OC]$ . The result is the diagonal-product rule:

∑ Formula

[A OB] \times [CO D] = [A O D] \times [BOC] \frac{[ A OB ]}{[ BOC ]} = \frac{A O}{OC}

The diagonals of a quadrilateral

$O$: the crossing point of the diagonals
$A O, OC$: the two pieces of the diagonal $A C$

LinkThe four triangles of a trapezium (

a^{2} : ab : ab : b^{2}

) are a special case of this rule:

a^{2} \cdot b^{2} = ab \cdot ab

. See the next method.

Steps

Check whether the quadrilateral is a parallelogram (inner-point rule) or general (diagonal product).
Form the opposite pairs of triangles: $(P A B, PC D)$ and $(PBC, P D A)$ .
Match the sum of each pair with $\frac{T}{2}$ .

Example 15 · Basic

For a point

P

inside the parallelogram

A BC D

[P A B] = 14

[PC D] = 22

and

[PBC] = 20

sq cm. Find

[A BC D]

and

[P D A]

Solution, step by step

1The opposite pair $P A B$ and $PC D$ : $[P A B] + [PC D] = 14 + 22 = 36 = \frac{T}{2}$ , so $T = 72$ sq cm.
2The other pair also gives $\frac{T}{2} = 36$ : $[PBC] + [P D A] = 36$ .
3 $[P D A] = 36 - 20 = 16$ sq cm.

Answer $[A BC D] = 72$ , $[P D A] = 16$ sq cm

Check:

14 + 20 + 22 + 16 = 72

. ✓

Example 16 · Standard

The parallelogram

A BC D

has area 120 sq cm. Its diagonals meet at

O

and a line through

O

meets

A B

E

and

C D

F

. Find the area of quadrilateral

A EF D

Solution, step by step

1 $O$ is the centre of the parallelogram. A half-turn about $O$ sends $A \to C$ , $D \to B$ and $E \to F$ ( $EF$ passes through the centre).
2So the quadrilateral $A EF D$ turns onto $CFEB$ : the two parts are congruent and equal.
3 $[A EF D] = \frac{1}{2} \times 120 = 60$ sq cm.

Answer 60 sq cm

Every line through the centre halves the parallelogram, however tilted.

Example 17 · Standard

The diagonals of a quadrilateral

A BC D

meet at

O

with

A O : OC = 2 : 3

and

BO : O D = 1 : 2

[A OB] = 16

sq cm. Find

[BOC]

[A O D]

and

[CO D]

Solution, step by step

1 $△ A OB$ and $△ BOC$ share the height (from $B$ to $A C$ ): $[A OB] : [BOC] = A O : OC = 2 : 3$ , so $[BOC] = 16 \times \frac{3}{2} = 24$ .
2 $△ A OB$ and $△ A O D$ share the height (from $A$ to $B D$ ): $[A OB] : [A O D] = BO : O D = 1 : 2$ , so $[A O D] = 32$ .
3 $△ A O D$ and $△ CO D$ : $[A O D] : [CO D] = A O : OC = 2 : 3$ , so $[CO D] = 32 \times \frac{3}{2} = 48$ .
4Check (diagonal product): $[A OB] \times [CO D] = 16 \times 48 = 768$ and $[A O D] \times [BOC] = 32 \times 24 = 768$ . ✓

Answer $[BOC] = 24$ , $[A O D] = 32$ , $[CO D] = 48$ sq cm

When both

A O : OC

and

BO : O D

are given, all four triangles follow from one.

Example 18 · Advanced

For a point

P

inside the parallelogram

A BC D

[P A B] = 12

[PBC] = 20

and

[PC D] = 30

sq cm. Find

[P D A]

and

[A BC D]

. Then, if

P

were on the diagonal

B D

with

BP : P D = 1 : 3

and

[A BC D] = 96

, find

[P A B]

Solution, step by step

1The sums of opposite pairs are equal: $[P A B] + [PC D] = [PBC] + [P D A]$ , i.e. $12 + 30 = 20 + [P D A]$ .
2 $[P D A] = 22$ sq cm and $T = 2 \times (12 + 30) = 84$ sq cm.
3Second part: the diagonal $B D$ cuts the parallelogram into two equal triangles, $[A B D] = \frac{96}{2} = 48$ .
4 $△ P A B$ and $△ A B D$ share the vertex $A$ with bases $BP : B D = 1 : 4$ on the line $B D$ : $[P A B] = \frac{1}{4} \times 48 = 12$ sq cm.

Answer $[P D A] = 22$ , $T = 84$ sq cm; and the second part $12$ sq cm

Check:

12 + 20 + 30 + 22 = 84

. ✓

Try it yourself

BasicTry it yourself first

For a point

P

inside the parallelogram

A BC D

[P A B] = 18

and

[PC D] = 26

sq cm. Find

[A BC D]

◭ Diagram

BasicTry it yourself first

The parallelogram

A BC D

has area 150 sq cm and an inner point

P

gives

[P A B] = 40

[PBC] = 35

. Find

[PC D]

StandardTry it yourself first

The diagonals of a quadrilateral

A BC D

meet at

O

with

A O : OC = 3 : 2

and

[A OB] = 27

sq cm. Find

[BOC]

StandardTry it yourself first

The diagonals of a quadrilateral

A BC D

meet at

O

[A OB] = 12

[BOC] = 18

and

[CO D] = 27

sq cm. Find

[A O D]

StandardTry it yourself first

For a point

P

inside the parallelogram

A BC D

[P A B] = 19

[PBC] = 21

and

[P D A] = 15

sq cm. Find

[PC D]

AdvancedTry it yourself first

A parallelogram has area 160 sq cm and an inner point

P

has

[P A B] : [PC D] = 3 : 2

and

[PBC] : [P D A] = 5 : 3

. Find

[P D A]

AdvancedTry it yourself first

The parallelogram

A BC D

has area 96 sq cm and

P

is on the diagonal

B D

with

BP : P D = 1 : 3

. Find

[PC D]

StandardTry it yourself first

For a point

P

inside the rectangle

A BC D

[P A B] = 15

[PBC] = 21

and

[P D A] = 9

sq cm. Find

[PC D]

The four triangles of a trapezium

The four regions made by the diagonals: a², ab, ab, b²

The diagonals of a trapezium cut it into four triangles whose areas are in the ratio $a^{2} : ab : ab : b^{2}$ .

The idea

In a trapezium $A BC D$ with $A B ∥ D C$ , $A B = a$ , $D C = b$ , the diagonals meet at $O$ . By the bow-tie of the previous chapter $△ O A B \sim △ OC D$ with ratio $a : b$ , so by the square rule $[O A B] : [OC D] = a^{2} : b^{2}$ . Also $A O : OC = a : b$ and $△ A O D$ , $△ CO D$ share the height (from $D$ to $A C$ ), so $[A O D] : [CO D] = a : b$ . Taking $[CO D] = b^{2}$ units gives $[A O D] = ab$ and $[A OB] = a^{2}$ . And by the rule of the previous methods $[A O D] = [BOC]$ too.

◭ Diagram

The four triangles:

a^{2}

(blue),

b^{2}

(green) and two of

ab

(orange); total

(a + b)^{2}

∑ Formula

[A OB] : [A O D] : [BOC] : [CO D] = a^{2} : ab : ab : b^{2} [A BC D] = (a + b)^{2} units

The diagonals of a trapezium

$a, b$: the lengths of the parallel sides $A B$ and $D C$
$O$: the crossing point of the diagonals
$units$: one unknown unit (worked out in sq cm later)

If given	then one unit is	result
$[CO D]$ and $a : b$	$\frac{[ CO D ]}{b ^{2}}$	the other three from $a^{2}, ab, ab$
$[A OB]$ and $[CO D]$	$[A O D] = [A OB] [CO D]$	total $= ([A OB] + [CO D])^{2}$
$[A OB]$ and $[A O D]$	$a : b = [A OB] : [A O D]$	$[CO D] = \frac{[ A O D ] ^{2}}{[ A OB ]}$

Here: $[A OB]$ , $[CO D]$ , $[A O D]$ , $[BOC]$ = the areas of the four triangles; the square root is taken when the result is a perfect square.

RememberA way to remember: “the top one is

b^{2}

, the bottom one

a^{2}

, and the two at the sides are the middle term

ab

”. The four add up to

(a + b)^{2}

, a perfect square, so the total is always a perfect-square number of units.

Steps

Identify the parallel sides $a, b$ (or their ratio).
Write the four triangles as $a^{2} : ab : ab : b^{2}$ .
From the given area get the value of one unit and write the asked region.

Example 19 · Basic

In trapezium

A BC D

A B ∥ D C

A B = 6

cm,

D C = 4

cm and

[CO D] = 32

sq cm. Find

[A OB]

[A O D]

[BOC]

and the area of the whole trapezium.

Solution, step by step

1 $a = 6, b = 4$ . The four triangles are $a^{2} : ab : ab : b^{2} = 36 : 24 : 24 : 16$ , i.e. (divide by 4) $9 : 6 : 6 : 4$ .
2 $[CO D]$ is 4 parts $= 32$ sq cm, so 1 part $= 8$ .
3 $[A OB] = 9 \times 8 = 72$ , $[A O D] = [BOC] = 6 \times 8 = 48$ sq cm.
4The trapezium $= 72 + 48 + 48 + 32 = 200$ sq cm ( $25$ parts).

Answer $72, 48, 48$ and a total of $200$ sq cm

Check:

25 = (3 + 2)^{2}

parts (ratio

3 : 2

). ✓

Example 20 · Standard

In trapezium

A BC D

A B ∥ D C

and the diagonals meet at

O

[A OB] = 81

and

[CO D] = 49

sq cm. Find

[BOC]

and the area of the whole trapezium.

Solution, step by step

1 $[A OB] : [CO D] = a^{2} : b^{2} = 81 : 49$ , so $a : b = 9 : 7$ (square root).
2 $a^{2} = 81$ sq cm, so the unit is $1$ and $[BOC] = ab = 9 \times 7 = 63$ sq cm.
3The whole trapezium $= 81 + 49 + 63 + 63 = 256$ sq cm $= (9 + 7)^{2}$ .

Answer $[BOC] = 63$ , total $256$ sq cm

Direct shortcut:

[BOC] = [A OB] \times [CO D] = 81 \times 49 = 63

Example 21 · Standard

The diagonals of trapezium

A BC D

(

A B ∥ D C

) meet at

O

[A OB] = 45

and

[A O D] = 30

sq cm. Find

[CO D]

and the area of the whole trapezium.

Solution, step by step

1 $[A OB] : [A O D] = a^{2} : ab = a : b$ , so $a : b = 45 : 30 = 3 : 2$ .
2 $[A OB] : [CO D] = a^{2} : b^{2} = 9 : 4$ , so $[CO D] = 45 \times \frac{4}{9} = 20$ sq cm.
3 $[BOC] = [A O D] = 30$ . Total $= 45 + 30 + 30 + 20 = 125$ sq cm.

Answer $[CO D] = 20$ sq cm; total $125$ sq cm

Check:

[A OB] \times [CO D] = 45 \times 20 = 900 = 30 \times 30

. ✓

Example 22 · Advanced

The parallel sides of a trapezium are in the ratio

2 : 3

and its area is 125 sq cm. Find the areas of the four triangles made by the diagonals and

[A BC]

(

A B

the shorter parallel side).

Solution, step by step

1 $a : b = 2 : 3$ ( $A B = a$ the shorter). The four triangles are $a^{2} : ab : ab : b^{2} = 4 : 6 : 6 : 9$ , in all $25$ parts.
2 $25$ parts $= 125$ , so 1 part $= 5$ sq cm.
3 $[A OB] = 20$ , $[A O D] = [BOC] = 30$ , $[CO D] = 45$ sq cm.
4 $[A BC] = [A OB] + [BOC] = 20 + 30 = 50$ sq cm. (Check: $[A BC] = \frac{A B}{A B + D C} \times 125 = \frac{2}{5} \times 125 = 50$ .)

Answer $20, 30, 30, 45$ sq cm; $[A BC] = 50$

Cutting the trapezium by the diagonal

A C

gives

△ A BC : △ A C D = a : b

. ✓

Try it yourself

◭ Diagram

BasicTry it yourself first

In trapezium

A BC D

A B ∥ D C

A B = 8

cm,

D C = 5

cm, the diagonals meet at

O

and

[CO D] = 50

sq cm. Find

[A OB]

◭ Diagram

BasicTry it yourself first

In trapezium

A BC D

A B ∥ D C

A B = 6

cm,

D C = 9

cm, the diagonals meet at

O

and

[A OB] = 40

sq cm. Find

[CO D]

◭ Diagram

BasicTry it yourself first

In trapezium

A BC D

A B ∥ D C

A B = 2

cm,

D C = 5

cm, the diagonals meet at

O

and

[A OB] = 12

sq cm. Find

[CO D]

StandardTry it yourself first

In a trapezium the diagonals make

[A OB] = 16

and

[CO D] = 25

sq cm. Find

[A O D]

◭ Diagram

StandardTry it yourself first

In trapezium

A BC D

A B ∥ D C

A B = 3

cm,

D C = 1

cm, the diagonals meet at

O

and the trapezium has area 160 sq cm. Find

[A OB]

StandardTry it yourself first

In a trapezium the diagonals make

[A OB] = 16

and

[A O D] = [BOC] = 24

sq cm. Find the area of the whole trapezium.

AdvancedTry it yourself first

In trapezium

A BC D

A B ∥ D C

and

A B = 3 D C

. The diagonals meet at

O

and

[BOC] = 18

sq cm. Find the area of the trapezium.

AdvancedTry it yourself first

In trapezium

A BC D

A B ∥ D C

. The diagonals meet at

O

[A OB] : [CO D] = 9 : 4

and the diagonal

A C = 40

cm. Find

A O

AdvancedTry it yourself first

The parallel sides of a trapezium are 10 cm and 6 cm. The line joining the midpoints of the non-parallel sides cuts it into two trapeziums. Find the ratio of the areas of the upper (shorter side) one to the lower one.

StandardTry it yourself first

In trapezium

A BC D

A B ∥ D C

and

A B : D C = 5 : 3

. The diagonals meet at

O

and

[CO D] = 45

sq cm. Find

[A O D]

Angles and crossing lines

Shared angle · weights

The same-angle rule and the areas of triangles made by crossing lines.

The same-angle rule

The product of two side fractions

When two triangles share (or equal, or supplement) an angle: the ratio of areas is the ratio of the products of the sides around it.

The idea

The area of any triangle is $\frac{1}{2} ab sin C$ , where $a, b$ are the sides around the angle $C$ . [Why? The height is $b sin C$ and the base is $a$ .] Now take two triangles sharing the angle $A$ : $D$ on $A B$ and $E$ on $A C$ . $[A D E] = \frac{1}{2} A D \cdot A E sin A$ and $[A BC] = \frac{1}{2} A B \cdot A C sin A$ . In the division $\frac{1}{2} sin A$ cancels.

◭ Diagram

The angle

θ

is shared. The perpendiculars

p

and

q

from

E

and

C

A B

have

p : q = A E : A C

; so

[A D E] : [A BC] = \frac{A D \cdot p}{A B \cdot q} = \frac{A D \cdot A E}{A B \cdot A C}

∑ Formula

\frac{[ A D E ]}{[ A BC ]} = \frac{A D}{A B} \times \frac{A E}{A C}

The same-angle rule

$A$: the shared angle (the vertex $A$ of both triangles)
$A D, A E$: the sides around angle $A$ in the small triangle
$A B, A C$: the same sides in the big triangle

So the ratio of areas = the fraction of one side × the fraction of the other (part : whole for both). If $D$ and $E$ lie on the extensions of the sides a fraction can exceed 1, but the rule still works.

Supplementary angles

If two triangles have supplementary angles ( $θ$ and $18 0^{\circ} - θ$ ) then $sin θ = sin (18 0^{\circ} - θ)$ and the same rule holds: the ratio of areas is the ratio of the products of the sides around those angles. This happens for triangles at adjacent angles of a parallelogram.

◭ Diagram

θ

and

18 0^{\circ} - θ

have the same sine, so

[A BC] : [PQR] = (A B \cdot A C) : (PQ \cdot PR)

Situation	Ratio
angle $A$ shared, $D$ on $A B$ , $E$ on $A C$	$\frac{A D}{A B} \cdot \frac{A E}{A C}$
equal angles: $∠ A = ∠ P$	$\frac{A B \cdot A C}{PQ \cdot PR}$
supplementary angles: $∠ A + ∠ P = 18 0^{\circ}$	$\frac{A B \cdot A C}{PQ \cdot PR}$
two sides $a, b$ and the angle $C$ between	$\frac{1}{2} ab sin C$

Useful values: $sin 3 0^{\circ} = \frac{1}{2}$ , $sin 4 5^{\circ} = \frac{1}{2}$ , $sin 6 0^{\circ} = \frac{3}{2}$ , $sin 9 0^{\circ} = 1$ , and supplementary angles give the same value ( $sin 15 0^{\circ} = \frac{1}{2}$ ).

LinkThis rule shines when $D E ∥ BC$ is not given. If it is parallel the square rule of the previous chapter works too (and both give the same answer: with parallel lines

\frac{A D}{A B} = \frac{A E}{A C}

, so the product becomes a square).

Steps

Find the shared (or equal/supplementary) angle and write both pairs of sides around it.
Form part : whole fractions in both and multiply.
Multiply the area of the big triangle by this ratio; subtract for the quadrilateral.

Example 23 · Basic

The area of

△ A BC

is 150 sq cm.

D

is on

A B

with

A D : D B = 2 : 3

and

E

is on

A C

with

A E : EC = 1 : 4

. Find

[A D E]

and

[BCE D]

Solution, step by step

1Angle $A$ is shared, so $\frac{[ A D E ]}{[ A BC ]} = \frac{A D}{A B} \times \frac{A E}{A C}$ .
2Part : whole: $\frac{A D}{A B} = \frac{2}{5}$ ( $A D : A B = 2 : 5$ ) and $\frac{A E}{A C} = \frac{1}{5}$ .
3The ratio is $\frac{2}{5} \times \frac{1}{5} = \frac{2}{25}$ ; $[A D E] = 150 \times \frac{2}{25} = 12$ sq cm.
4 $[BCE D] = 150 - 12 = 138$ sq cm.

Answer $[A D E] = 12$ , $[BCE D] = 138$ sq cm

Use

A D : A B

, not

A D : D B

Example 24 · Standard

△ A BC

D

is on

A B

with

A D : D B = 1 : 2

and

E

is on

A C

with

A E : EC = 3 : 1

. Find

[A D E] : [BCE D]

Solution, step by step

1 $\frac{A D}{A B} = \frac{1}{3}$ and $\frac{A E}{A C} = \frac{3}{4}$ (the whole side $A C = 3 + 1 = 4$ parts).
2 $[A D E] : [A BC] = \frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ , i.e. $1 : 4$ .
3The quadrilateral is $4 - 1 = 3$ parts, so $[A D E] : [BCE D] = 1 : 3$ .

Answer $1 : 3$

(Here

D E ∥ BC

is not true, yet the rule works.)

Example 25 · Standard

△ A BC

and

△ PQR

∠ B + ∠ Q = 18 0^{\circ}

A B = 6

BC = 10

PQ = 4

and

QR = 15

(cm). Find

[A BC] : [PQR]

Solution, step by step

1The angles are supplementary, so $sin B = sin Q$ and $[A BC] = \frac{1}{2} \cdot 6 \cdot 10 sin B$ , $[PQR] = \frac{1}{2} \cdot 4 \cdot 15 sin Q$ .
2The ratio is $\frac{6 \times 10}{4 \times 15} = \frac{60}{60} = 1$ .

Answer $1 : 1$ (equal areas)

Triangles with supplementary angles can have equal areas though their shapes differ a lot.

Example 26 · Standard

△ A BC

A B = 8

cm,

A C = 15

cm and

∠ A = 3 0^{\circ}

. Find

[A BC]

. If

D

is the midpoint of

A B

and

E

is on

A C

with

A E = 6

cm find

[A D E]

too.

Solution, step by step

1 $[A BC] = \frac{1}{2} \cdot A B \cdot A C \cdot sin A = \frac{1}{2} \times 8 \times 15 \times \frac{1}{2} = 30$ sq cm.
2 $A D = 4$ and $A E = 6$ with the same angle: $[A D E] = \frac{1}{2} \times 4 \times 6 \times \frac{1}{2} = 6$ sq cm.
3Check by the rule: $\frac{A D}{A B} \cdot \frac{A E}{A C} = \frac{1}{2} \cdot \frac{6}{15} = \frac{1}{5}$ and $30 \times \frac{1}{5} = 6$ . ✓

Answer $[A BC] = 30$ , $[A D E] = 6$ sq cm

With a

3 0^{\circ}

angle the area is

\frac{1}{4} ab

Example 27 · Advanced

The area of

△ A BC

is 80 sq cm.

D

is on

A B

with

A D = \frac{1}{2} A B

and

E

is a point on

A C

produced beyond

C

with

A E = \frac{3}{2} A C

. Find

[A D E]

Solution, step by step

1Angle $A$ is still shared ( $E$ lies on the ray $A C$ ), so the same rule applies.
2 $\frac{[ A D E ]}{[ A BC ]} = \frac{A D}{A B} \cdot \frac{A E}{A C} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$ .
3 $[A D E] = \frac{3}{4} \times 80 = 60$ sq cm.

Answer 60 sq cm

A fraction can exceed 1; it only means that

E

is on the extension.

Try it yourself

◭ Diagram

BasicTry it yourself first

△ A BC

D

is on

A B

and

E

A C

with

A D : D B = 3 : 2

A E : EC = 2 : 3

[A BC] = 100

sq cm. Find

[A D E]

BasicTry it yourself first

△ A BC

D, E

are the midpoints of

A B, A C

and

[A BC] = 64

sq cm. Find

[A D E]

BasicTry it yourself first

The area of

△ A BC

is 90 sq cm.

D

is on

A B

with

A D = \frac{1}{3} A B

and

E

A C

with

A E = \frac{1}{2} A C

. Find

[A D E]

◭ Diagram

StandardTry it yourself first

△ A BC

D

is on

A B

and

E

A C

with

A D : D B = 2 : 1

A E : EC = 1 : 1

[A BC] = 120

sq cm. Find the area of quadrilateral

BCE D

StandardTry it yourself first

△ A BC

D

is on

A B

with

A D : D B = 3 : 1

and

E

is the midpoint of

A C

. Find

[A D E] : [BCE D]

StandardTry it yourself first

△ A BC

and

△ PQR

∠ B + ∠ Q = 18 0^{\circ}

A B = 9

BC = 8

PQ = 6

QR = 16

(cm). Find

[A BC] : [PQR]

StandardTry it yourself first

Two sides of a triangle are 14 cm and 10 cm and the angle between them is

3 0^{\circ}

. Find its area.

AdvancedTry it yourself first

The area of

△ A BC

is 70 sq cm.

D

is on

A B

with

A D = \frac{2}{5} A B

and

E

is on

A C

produced with

A E = \frac{5}{2} A C

. Find

[A D E]

AdvancedTry it yourself first

△ A BC

D

is on

A B

with

A D : A B = 3 : 5

and

E

is on

A C

such that

[A D E] = \frac{1}{2} [A BC]

. Find

A E : A C

StandardTry it yourself first

△ A BC

D

is on

A B

with

A D : A B = 1 : 3

and

E

A C

with

A E : A C = 2 : 3

[A D E] = 14

sq cm. Find

[A BC]

Crossing lines and areas: from weights

areas : weights at the vertices

When two lines cross inside a triangle: the areas of the triangles they make, by two common-height steps or directly from mass points.

The idea

In the previous chapter crossing lines gave length ratios (the mass points $A, B, C$ ). The same data now gives areas. If $E$ is a point on $A D$ ( $D$ on $BC$ ), the three triangles $EBC$ , $EC A$ , $E A B$ come out like this: $△ EBC$ and $△ A BC$ share the base $BC$ and their heights are in the ratio $E D : A D$ , so $[EBC] = \frac{E D}{A D} [A BC]$ . The remaining region $[EC A] + [E A B]$ is cut by the line $A D$ in the ratio $[A BE] : [A CE] = B D : D C$ .

◭ Diagram

B D : D C = 2 : 3

A E : E D = 3 : 4

: weights

A = 20, B = 9, C = 6

and the three triangles are

20 : 9 : 6

(out of

35

parts)

∑ Formula

[EBC] : [EC A] : [E A B] = A : B : C (weights at A, B, C)

The three triangles at an inner point

E

of crossing lines

$A, B, C$: balancing weights at the vertices (by the trick of the previous chapter)
$[EBC]$: the area of the triangle made by $E$ with $B, C$ (the one opposite $A$ )

Remember: the heavier the weight at a vertex, the bigger the triangle opposite it (a heavy vertex sits near $E$ so the opposite base lies far). Here $A = 20$ is the heaviest, so $[EBC] = 20$ parts is the largest.

Given	First step	Then
$B D : D C$ and $A E : E D$	$[EBC] = \frac{E D}{A D} [A BC]$ (common base)	split the rest in $B D : D C$
weights $A, B, C$	total $= A + B + C$ parts	each triangle gets the weight opposite it
two crossing lines $A D$ , $CE$ meeting at $F$	$[FBC] : [FC A] : [F A B]$ from weights	$B D FE = [FB D] + [FBE]$ in pieces

TrickAlways find a region like the quadrilateral

B D FE

by cutting it into two triangles:

[B D FE] = [FB D] + [FBE]

, each a part of a bigger triangle by the common-height rule.

Steps

Get the weights $A, B, C$ from the given ratios (the mass-point method of the previous chapter).
Write the three triangles at the inner point in the ratio $A : B : C$ and find a part from the total.
Write a compound region as a sum of parts of these triangles.

Example 28 · Standard

The area of

△ A BC

is 140 sq cm.

D

is on

BC

with

B D : D C = 3 : 4

and

E

is on

A D

with

A E : E D = 2 : 3

. Find

[EBC]

[A BE]

and

[A CE]

Solution, step by step

1 $△ EBC$ and $△ A BC$ share the base $BC$ . The heights of $E$ and $A$ above $BC$ are in the ratio $E D : A D = 3 : 5$ ( $A D = 2 + 3 = 5$ parts).
2 $[EBC] = \frac{3}{5} \times 140 = 84$ sq cm.
3In the remaining $140 - 84 = 56$ the line $A D$ splits $[A BE] : [A CE] = B D : D C = 3 : 4$ (common apex and bases $B D$ , $D C$ scaled alike).
4 $[A BE] = \frac{3}{7} \times 56 = 24$ and $[A CE] = \frac{4}{7} \times 56 = 32$ sq cm.

Answer $84, 24, 32$ sq cm

Check by weights:

B = 4, C = 3, D = 7

A = \frac{21}{2}

; doubled:

A = 21, B = 8, C = 6

⟹

21 : 8 : 6

(35 parts),

\frac{21}{35} \times 140 = 84

. ✓

Example 29 · Standard

The same figure:

B D : D C = 2 : 3

A E : E D = 3 : 4

and

[A BC] = 70

sq cm. Find

[EBC]

[EC A]

[E A B]

from the weights.

Solution, step by step

1Weights from the previous chapter: $B = 3, C = 2, D = 5$ and $A = \frac{20}{3}$ ; times 3: $A = 20, B = 9, C = 6$ .
2 $[EBC] : [EC A] : [E A B] = A : B : C = 20 : 9 : 6$ , in all $35$ parts $= 70$ sq cm, 1 part $= 2$ .
3 $[EBC] = 40$ , $[EC A] = 18$ , $[E A B] = 12$ sq cm.

Answer $40, 18, 12$ sq cm

Check (common base):

[EBC] = \frac{4}{7} \times 70 = 40

E D : A D = 4 : 7

. ✓

Example 30 · Advanced

△ A BC

D

is the midpoint of

BC

and

E

is on

A B

with

A E : EB = 2 : 1

A D

and

CE

meet at

F

. Find the area of quadrilateral

B D FE

when

[A BC] = 120

sq cm.

Solution, step by step

1Weights: $D$ midpoint ⟹ $B = C = 1$ , $D = 2$ . $A E : EB = 2 : 1$ ⟹ $A : B = 1 : 2$ ; with $B = 1$ , $A = \frac{1}{2}$ . Double all: $A = 1, B = 2, C = 2$ .
2The three triangles at $F$ : $[FBC] : [FC A] : [F A B] = A : B : C = 1 : 2 : 2$ , 5 parts, 1 part $= 24$ : $[FBC] = 24$ , $[FC A] = 48$ , $[F A B] = 48$ .
3The quadrilateral $B D FE = [FB D] + [FBE]$ . $D$ is a midpoint: $[FB D] = \frac{1}{2} [FBC] = 12$ .
4 $△ FBE$ and $△ F A B$ share the vertex $F$ with bases $BE : B A = 1 : 3$ : $[FBE] = \frac{1}{3} \times 48 = 16$ .
5 $[B D FE] = 12 + 16 = 28$ sq cm (that is $\frac{7}{30}$ of the whole).

Answer 28 sq cm

Break the quadrilateral into two triangles, each a part of a known triangle.

Example 31 · Advanced

B D : D C = 2 : 3

and

A E : E D = 3 : 4

(

D

BC

E

A D

BE

produced meets

A C

F

. If

[A BC] = 130

sq cm find

[A BF]

and

[BFC]

Solution, step by step

1By the trick of the previous chapter $A F : FC = 3 : 10$ , i.e. $A F : A C = 3 : 13$ .
2 $△ A BF$ and $△ A BC$ share the vertex $B$ with bases $A F$ , $A C$ on one line: $[A BF] = \frac{3}{13} \times 130 = 30$ sq cm.
3 $[BFC] = 130 - 30 = 100$ sq cm.

Answer $[A BF] = 30$ , $[BFC] = 100$ sq cm

The length method and the common-height rule together give the area.

Try it yourself

◭ Diagram

StandardTry it yourself first

D

is on

BC

with

B D : D C = 2 : 3

and

E

is on

A D

with

A E : E D = 1 : 2

[A BC] = 150

sq cm. Find

[EBC]

◭ Diagram

StandardTry it yourself first

D

is on

BC

with

B D : D C = 2 : 3

and

E

is on

A D

with

A E : E D = 1 : 2

[A BC] = 150

sq cm. Find

[A BE]

StandardTry it yourself first

At an inner point

E

△ A BC

the balancing weights are

A = 5

B = 3

C = 2

and

[A BC] = 100

sq cm. Find

[EC A]

◭ Diagram

StandardTry it yourself first

D

is on

BC

with

B D : D C = 1 : 1

and

E

is on

A D

with

A E : E D = 3 : 1

[A BC] = 80

sq cm. Find

[A BE]

AdvancedTry it yourself first

B D : D C = 3 : 1

and

A E : E D = 1 : 1

(

D

BC

E

A D

BE

produced meets

A C

F

. If

[A BC] = 140

sq cm find

[A BF]

AdvancedTry it yourself first

△ A BC

D

is the midpoint of

BC

and

E

is on

A B

with

A E : EB = 1 : 3

A D

and

CE

meet at

F

. What fraction of

[A BC]

is the quadrilateral

B D FE

AdvancedTry it yourself first

△ A BC

D

is the midpoint of

BC

and

E

is on

A B

with

A E : EB = 2 : 1

A D

and

CE

meet at

F

. If

[A BC] = 120

sq cm find

[FC A]

StandardTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 1 : 2

and

E

is the midpoint of

A C

A D

and

BE

meet at

P

[A BC] = 80

sq cm. Find

[P A B]

AdvancedTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 3 : 2

and

E

is on

A D

with

A E : E D = 1 : 1

[A BC] = 100

sq cm. Find

[A CE]

Summary, mixed practice, revision

Every method in one place

A one-page table of all methods, exam-style mixed questions and revision cards.

Summary of all methods

Recognise · formula

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

Method	Recognise	Key formula
Base and height	common vertex, bases on one line; between parallels	$[A B D] : [A D C] = B D : D C$ ; $\frac{b _{1} h _{1}}{b _{2} h _{2}}$
Midpoints, centroid	median, midpoint triangle	$\frac{1}{2}$ ; $\frac{1}{4}$ (four pieces); $\frac{1}{6}$ (six pieces); $[GEF] = \frac{1}{12}$
Parallelogram: assume a total	midpoints, shaded parts	$[A BP] = \frac{T}{2}$ ; $[ECF] = \frac{1}{2} \cdot$ frac $\cdot$ frac $\cdot T$ ; inner quadrilateral $\frac{T}{2}$
Inner points, diagonals	a point $P$ inside; diagonals at $O$	$[P A B] + [PC D] = \frac{T}{2}$ ; $[A OB] [CO D] = [A O D] [BOC]$
Four triangles of a trapezium	trapezium + diagonals	$a^{2} : ab : ab : b^{2}$ ; $(a + b)^{2}$
The same angle	shared, equal or supplementary angle	$\frac{A D}{A B} \cdot \frac{A E}{A C}$ ; $\frac{1}{2} ab sin C$
Crossing lines	$B D : D C$ , $A E : E D$ given	$[EBC] = \frac{E D}{A D} [A BC]$ ; $A : B : C$

Letters: $[]$ = area; $T$ = the area of the parallelogram; $a, b$ = the parallel sides of a trapezium; $b_{1}, h_{1}$ etc. = bases and heights; $A, B, C$ = weights at the vertices.

OrderEvery question: assume a total → spot a common height/base/angle → write the pieces in parts → check that they add up to the total.

Mixed practice

Exam-style questions

Questions from every method, with no hint about which one to use.

The method is no longer named. First ask: a common height or base? midpoints? a parallelogram or a trapezium? a shared angle? Then choose the method.

BasicTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 7 : 5

and

[A B D] = 63

sq cm. Find

[A BC]

BasicTry it yourself first

G

is the centroid and

[GBC] = 26

sq cm. Find

[A BC]

BasicTry it yourself first

In trapezium

A BC D

A B ∥ D C

and

[A BC] = 45

sq cm. Find

[A B D]

BasicTry it yourself first

Points

D, E

on the side

BC

△ A BC

have

B D : D E : EC = 3 : 2 : 5

[A BC] = 200

sq cm. Find

[A D E]

BasicTry it yourself first

D, E, F

are the midpoints of

BC, C A, A B

and

[A FE] = 17

sq cm. Find

[A BC]

StandardTry it yourself first

△ A BC

D

is the midpoint of

BC

and

E

is on

A D

with

A E : E D = 1 : 3

[A BC] = 96

sq cm. Find

[A BE]

BasicTry it yourself first

The parallelogram

A BC D

has area 200 sq cm and

P

is any point on

A B

. Find

[PC D]

BasicTry it yourself first

The midpoints of the four sides of a 12 cm × 10 cm rectangle are joined to make a rhombus. Find its area.

StandardTry it yourself first

For a point

P

inside the parallelogram

A BC D

[P A B] = 10

[PBC] = 14

and

[PC D] = 26

sq cm. Find

[P D A]

StandardTry it yourself first

The diagonals of a quadrilateral meet at

O

[A OB] = 6

[BOC] = 9

and

[A O D] = 10

sq cm. Find

[CO D]

StandardTry it yourself first

In a trapezium

A B ∥ D C

A B = 10

cm,

D C = 4

cm and the diagonals make

[CO D] = 20

sq cm. Find

[A OB]

StandardTry it yourself first

In a trapezium

A B ∥ D C

and

A B = 2 D C

. Its area is 54 sq cm. Find

[A O D]

(

O

the crossing of the diagonals).

StandardTry it yourself first

The diagonals of a trapezium make

[A OB] = 100

and

[CO D] = 36

sq cm. Find the area of the trapezium.

StandardTry it yourself first

The area of

△ A BC

is 180 sq cm.

D

is on

A B

with

A D = \frac{1}{3} A B

and

E

A C

with

A E = \frac{2}{3} A C

. Find

[A D E]

StandardTry it yourself first

△ A BC

D

is on

A B

with

A D : D B = 1 : 3

and

E

is the midpoint of

A C

. Find

[A D E] : [BCE D]

StandardTry it yourself first

△ A BC

∠ A = 12 0^{\circ}

A B = 5

A C = 6

; in

△ PQR

∠ P = 6 0^{\circ}

PQ = 3

PR = 10

. Find

[A BC] : [PQR]

StandardTry it yourself first

Two sides of a triangle are 12 cm and 9 cm with an angle of

3 0^{\circ}

between them. Find the area.

StandardTry it yourself first

D

is on

BC

with

B D : D C = 1 : 2

and

E

is on

A D

with

A E : E D = 2 : 1

[A BC] = 90

sq cm. Find

[EBC]

StandardTry it yourself first

An inner point

E

has weights

A = 2

B = 3

C = 7

and

[A BC] = 96

sq cm. Find

[EBC]

AdvancedTry it yourself first

G

is the centroid. The line through

G

parallel to

BC

meets

A B, A C

P, Q

[A PQ] = 28

sq cm. Find

[A BC]

AdvancedTry it yourself first

The quadrilateral formed by the midpoints of the four sides of

A BC D

has area 42 sq cm. Find

[A BC D]

AdvancedTry it yourself first

The parallelogram

A BC D

has area 90 sq cm, the diagonals meet at

O

and

E

is any point on

C D

. Find

[E A B] + [O A B]

AdvancedTry it yourself first

△ A BC

D

is the midpoint of

BC

and

E

the midpoint of

A D

BE

produced meets

A C

F

[A BC] = 90

sq cm. Find

[A BF]

AdvancedTry it yourself first

△ A BC

D

is the midpoint of

BC

E

A D

and

F

BE

[A BC] = 64

sq cm. Find

[D EF]

AdvancedTry it yourself first

G

is the centroid and

[A GB] = 22

sq cm. Find

[A BC] + [G D C]

(

D

the midpoint of

BC

AdvancedTry it yourself first

The parallelogram

A BC D

has area 96 sq cm.

E

is on

A B

with

A E = \frac{1}{4} A B

and

F

A D

with

A F = \frac{1}{2} A D

. Find

[A EF]

StandardTry it yourself first

Two triangles have bases 9 cm and 15 cm with heights 10 cm and 6 cm on them. Find the ratio of their areas.

AdvancedTry it yourself first

In trapezium

A BC D

A B ∥ D C

[A OB] = 36

sq cm and

[A O D] = 24

sq cm. Find

[A BC D]

AdvancedTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 2 : 1

and

E

is on

A C

with

A E : EC = 1 : 2

A D

and

BE

meet at

P

. Find

[A PB] : [A BC]

StandardTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 5 : 3

and

[A D C] = 27

sq cm. Find

[A B D]

StandardTry it yourself first

The parallel sides of a trapezium are 10 cm and 6 cm. The mid-segment cuts it into two trapeziums. Find the ratio of the upper to the lower area.

StandardTry it yourself first

△ A BC

D

is on

A B

with

A D : A B = 2 : 5

and

E

A C

with

A E : A C = 3 : 4

[A BC] = 200

sq cm. Find

[A D E]

AdvancedTry it yourself first

The parallelogram

A BC D

has area 144 sq cm and

P

is any point on

A B

. Find

[PC D] + [P A D] + [PBC]

Revision

Cards · common traps

Cards to memorise formulas and the commonest exam mistakes.

Taking the ratio of bases when the height is not common. First spot what is shared.
Mixing up $A D : D B$ and $A D : A B$ (the same-angle rule and the square rule need part : whole).
Not assuming a total and getting tangled in fractions. Assume 16, 24, 48, 100 and so on.
The inner-point rule holds only for a parallelogram; for a general quadrilateral use the diagonal product.
In a trapezium the two side triangles are equal ( $ab$ ); the top and bottom ones are the squares.
Trying to get a quadrilateral like $B D FE$ directly. Break it into two triangles.
Placing weights the wrong way (the heavier end sits nearer) and forgetting that $[EBC]$ goes with the weight at $A$ .

PracticeTen minutes a day: read a card, say the answer, then check. Redo the practice questions of the method behind any card that trips you.