2D Mensuration: Polygons, Circles and Sectors

Exterior and interior angles, the sum of interior angles, the number of diagonals and finding the sides from them.

Walk round a polygon (say the fence of a field). At every corner you have to turn a little. When you are back at the start facing the way you began, you have turned exactly one full turn, i.e. $360^{\circ }$. The angle turned at a corner is the exterior angle. So the exterior angles of any (convex) polygon always add up to $360^{\circ }$.

In a regular polygon all exterior angles are equal, so each exterior angle $=\frac{360^{\circ }}{n}$. At every corner the interior and exterior angles together make a straight line ($180^{\circ }$), so each interior angle $=180^{\circ }-\frac{360^{\circ }}{n}$.

Sum of the interior angles: draw diagonals from one vertex to all the others. The polygon splits into $(n-2)$ triangles, each with an angle sum of $180^{\circ }$. From that vertex $n-3$ diagonals start (leaving out itself and its two neighbours). All $n$ vertices together give $n(n-3)$, but every diagonal was counted twice — so halve it.

Letters in the table: $n$ = sides (with the polygon's name), $i$ and $e$ = one interior and one exterior angle of the regular polygon, $D$ = diagonals. Memorise these rows: $108,120,135,144,150$ come up again and again. The sum of the interior angles for each row follows directly from $(n-2)\times 180$.

1. Decide whether the polygon is regular. If it is, every angle is equal.
2. Come to the exterior angle first: $e=\frac{360}{n}$ or $e=180-i$. This is the shortest road to the number of sides.
3. For a sum use $(n-2)\times 180$; for diagonals $\frac{n(n-3)}{2}$.
4. Check the answer: $n$ must be a whole number, 3 or more.

Area, perimeter, radii and diagonals of a hexagon and its link with circles.

The nicest property of a regular hexagon: joining the centre to the six corners cuts it into six equal equilateral triangles. Reason: the six equal angles at the centre are $\frac{360}{6}=60^{\circ }$ each and the two sides (centre to corner) are equal, so the other two angles are also $60^{\circ }$. It means: the distance $R$ from the centre to every corner equals the side $a$ of the hexagon.

Everything now follows from the equilateral triangle. One triangle has area $\frac{\sqrt{3}}{4}{a}^2$, so six have $\frac{3\sqrt{3}}{2}{a}^2$. The perpendicular from the centre to a side (the apothem) is the height of the equilateral triangle: $r=\frac{\sqrt{3}}{2}a$. The distance between two opposite sides is $2r=\sqrt{3}a$ and the longest diagonal (through the centre) is $2a$.

1. Find the side $a$ of the hexagon (given, or from the area / perimeter / radius / distance).
2. Think of what is asked in terms of the six triangles: area $=6\times \frac{\sqrt{3}}{4}{a}^2$.
3. In a circle question first decide whether the circle passes through the corners ($R=a$) or touches the sides ($r=\frac{\sqrt{3}}{2}a$).

Making an octagon from a square: side, area, the link with the square and the waste.

The best way to remember a regular octagon (like a stop sign) is to see it as a square with its four corners cut off. A right isosceles triangle comes off each corner. If its equal sides are $x$ the hypotenuse is $x\sqrt{2}$, and this hypotenuse is one side $a$ of the octagon: $a=x\sqrt{2}$, i.e. $x=\frac{a}{\sqrt{2}}$.

Side of the square: $S=a+x+x=a+2\cdot \frac{a}{\sqrt{2}}=a+\sqrt{2}a=a(1+\sqrt{2})$. Area: take the four corners out of the square. The four corners have area $4\times \frac{1}{2}{x}^2=2x^2=2\cdot \frac{a^2}{2}=a^2$. So the octagon $=S^2-a^2=a^2(1+\sqrt{2}{)}^2-a^2=a^2(3+2\sqrt{2}-1)=a^2(2+2\sqrt{2})=2(1+\sqrt{2})a^2$.

Letters in the table: $a$ = side of the octagon, $S$ = side of the original square. $\sqrt{2}-1$ and $\sqrt{2}+1$ are reciprocals (product 1), so $\frac{1}{1+\sqrt{2}}=\sqrt{2}-1$.

1. Draw the cut-off right triangle: equal sides $x$, hypotenuse $a=x\sqrt{2}$.
2. Link $a$, $x$ and $S$ by $S=a+2x$: given any one, the others follow.
3. For the area use $2(1+\sqrt{2})a^2$ or “square $-$ four corners”; both give the same answer.

Area by cutting into $n$ triangles, the apothem and circumradius, and comparing polygons.

Joining the centre of any regular polygon to all its corners gives $n$ equal isosceles triangles. Each has base equal to the side $a$ and height equal to the perpendicular distance from the centre to a side, called the apothem $r$. The angle of each triangle at the centre is $\frac{360^{\circ }}{n}$.

Total area $=n\times \frac{1}{2}ar=\frac{1}{2}(na)r=\frac{1}{2}\times P\times r$, where $P=na$ is the perimeter. The same works for a circle (think of it as a polygon with endless sides): perimeter $2\pi r$ and apothem $r$ give $\frac{1}{2}\times 2\pi r\times r=\pi r^2$. The right formula!

The right triangle linking apothem and circumradius: $r$ from the centre to the midpoint of a side, $\frac{a}{2}$ from that midpoint to a corner, and $R$ from the centre to the corner. Pythagoras: $R^2=r^2+{\left(\frac{a}{2}\right)}^2$. Using the angle $\frac{180^{\circ }}{n}$ gives $r=\frac{a}{2}\cot \frac{180^{\circ }}{n}$ and $R=\frac{a}{2\sin \frac{180^{\circ }}{n}}$. You need not work these out each time: the values are easy for $n=3,4,6$ and for the rest the numbers are given in the question.

Letters in the table: $n$ = sides; $r$, $R$ and $A$ are given for side $a=1$ (for any other side multiply $r,R$ by $a$ and the area $A$ by $a^2$). No need to learn these by heart — just see that as $n$ grows the shape nears a circle ($r$ and $R$ come closer).

Letters in the table: $n$ = sides of the polygon inscribed in a circle of radius $R$; the area is in units of $R^2$; each value is $\frac{n}{2}\sin \frac{360^{\circ }}{n}$, i.e. $n$ triangles with two sides $R$ and an angle $\frac{360}{n}$ between them. As the sides grow the area climbs to that of the circle ($\pi R^2$). Remember both facts: in the same circle more sides ⟹ more area; for equal perimeter also more sides ⟹ more area, and the circle is the largest.

1. Write down whichever of side $a$, perimeter $P=na$ and apothem $r$ is given.
2. Area wanted: $\frac{1}{2}Pr$. $r$ or $P$ wanted: turn the formula round.
3. To link a corner and a side use the right triangle $\left(r,\frac{a}{2},R\right)$.

Where the circle formulas come from, which value of $\pi$ to use and how to find the radius.

All points of a circle are at the same distance $r$ (the radius) from the centre. The longest chord, through the centre, is the diameter $d=2r$, and the length of the boundary is the circumference $C$. Take any circle and divide its circumference by its diameter — the answer is always the same, about $3.14159$, which we call $\pi$. So the diameter fits round the circumference exactly three times and a little (about $0.14$) more.

Here is why the area formula works. Cut the circle into many round rings. Pull each ring straight to make a strip: the ring of radius $\rho$ gives a strip of length $2\pi \rho$. Stack the strips from smallest to biggest and you get a triangle whose base is the outermost ring, $2\pi r$, and whose height is $r$. Area of the triangle $=\frac{1}{2}\times 2\pi r\times r=\pi r^2$.

Letters in the table: $r$ = radius (cm or m), $C$ = circumference, $A$ = area, with $\pi =\frac{22}{7}$. The same values come up again and again ($7\to 44,154$; $14\to 88,616$; $21\to 132,1386$), so learn them — they help directly in reverse questions (circumference or area given, radius asked).

1. Find out whether the measure given is the radius or the diameter; if it is the diameter halve it to get $r$.
2. Decide: length of the boundary ($2\pi r$) or the surface inside ($\pi r^2$).
3. Choose $\pi$: $\frac{22}{7}$ if $r$ or $d$ is a multiple of 7, otherwise the value in the question.
4. In a reverse question solve the formula for $r$: $r=\frac{C}{2\pi }$ or $r=\sqrt{\frac{A}{\pi }}$.

Area and perimeter of half and quarter circles (do not forget the straight edge), and shapes joined to rectangles.

A semicircle is half a circle and a quadrant a quarter. The area is straightforward: $\frac{1}{2}\pi r^2$ and $\frac{1}{4}\pi r^2$. The perimeter is where the common mistake happens: the boundary of a semicircle has the arc $\pi r$ and the straight edge (diameter) $2r$, and the boundary of a quadrant has the arc $\frac{1}{2}\pi r$ and two radii $2r$. Straight edges are part of the boundary too!

Now shapes made from them. A window: a rectangle ($w\times h$) with a semicircle on top whose diameter is the width $w$ of the rectangle. A track (stadium): a rectangle of length $l$ and width $2r$ with a semicircle at each short end; the two semicircles together make one full circle, so the arcs total $2\pi r$ and the area gains $\pi r^2$.

1. Split the figure into familiar parts: rectangles, circles, semicircles; make a rough sketch.
2. Area: add (or subtract) the areas of the parts.
3. Perimeter: count only the outer boundary. Edges buried between two parts (joints) are not counted; open edges (a diameter) are.

Arc length, sector area and perimeter, clock and pizza questions and finding the angle or radius.

A full circle is a turn of $360^{\circ }$. If a radius turns through ${\theta }^{\circ }$ about the centre its tip covers exactly the fraction $\frac{\theta }{360}$ of the circumference — this is the arc. And the region the radius sweeps is the same fraction $\frac{\theta }{360}$ of the area of the circle — this is the sector. Both formulas come from this one idea: a fraction of the whole.

Letters in the table: $\theta$ = angle at the centre (in degrees), fraction = $\frac{\theta }{360}$, $l$ = length of the arc, $A$ = area of the sector, $r$ = radius. Remember the fractions of these few angles ($\theta =90\to \frac{1}{4}$ and so on).

Another handy form: a sector is like many thin triangles with bases on the arc and height $r$, so $A=\frac{1}{2}\times l\times r$. (Put $l=2\pi r$ for the whole circle and you get $\pi r^2$.) The length of the boundary has both radii besides the arc: $P=l+2r$. If the angle is in radians then $l=r\theta$ and $A=\frac{1}{2}{r}^2\theta$ — but in SSC the angle usually comes in degrees.

Letters in the table: $m$ = minutes passed. The minute hand turns $360^{\circ }$ in 60 minutes, so $6^{\circ }$ per minute; the hour hand turns $360^{\circ }$ in 12 hours, i.e. $30^{\circ }$ per hour and $0.5^{\circ }$ per minute.

1. Find the angle $\theta$ (on a clock $6m$ degrees; for a pizza $\frac{360}{n}$).
2. Simplify the fraction $\frac{\theta }{360}$.
3. Multiply that fraction by the circumference $2\pi r$ (arc) or by $\pi r^2$ (area).
4. If the boundary is asked do not forget to add both radii.