Category Archives: Area

September 30, 2025 · 3:54 am

Geometry Problem

*Geometry Snacks* by Ed Southall and Vincent Pantaloni

I started by trisecting another side of the triangle

This makes it clearer that the two lines are parallel

Which means the two angles labelled above are corresponding and therefore congruent.

Let the side length be $x$ .

The area of the equilateral triangle is

(1) $\begin{equation*}A=\frac{1}{2}x^2 sin(60)=\frac{\sqrt{3}x^2}{4}\end{equation*}$

$\begin{equation*}cos(60)=\frac{y}{\frac{2x}{3}}\end{equation}$

$\begin{equation*}y=\frac{x}{3}\end{equation}$

Area of right triangle

(2) $\begin{equation*}A=(\frac{1}{2})(\frac{2x}{3})(\frac{x}{3})sin(60)=\frac{\sqrt{3}x^2}{18}\end{equation*}$

The fraction of the area is

$\begin{equation*}=\frac{\frac{\sqrt{3}x^2}{18}}{\frac{\sqrt{3}x^2}{4}}=\frac{2}{9}\end{equation}$

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Filed under Algebra, Area, Area of Triangles (Sine), Finding an area, Geometry, Puzzles, Right Trigonometry, Simplifying fractions, Trigonometry

Tagged as geometry problem, geometry snacks

September 9, 2025 · 11:19 am

Linear Transformation (Rotation) Question

The unit square is rotated about the origin by $45^\circ$ anti-clockwise.
(a) Find the matrix of this transformation.
(b) Draw the unit square and its image on the same set of axes.
(c) Find the area of the over lapping region.

Remember the general rotation matrix is

$\begin{equation*}\begin{bmatrix}cos(\theta)&-sin(\theta)\\sin(\theta)&cos(\theta)\end{bmatrix}\end{equation}$

Hence

$\begin{equation*}\begin{bmatrix}cos(45)&-sin(45)\\sin(45)&cos(45)\end{bmatrix}=\begin{bmatrix}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}\end{equation}$

The unit square has co-ordinates

$\begin{bmatrix}0 & 1 & 1& 0\\0&0&1&1\end{bmatrix}$

Transform the unit square

$\begin{bmatrix}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}0 & 1 & 1& 0\\0&0&1&1\end{bmatrix}=\begin{bmatrix}0&\frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}}\\0&\frac{1}{\sqrt{2}}&\frac{2}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}$

The overlapping area is the area of $\Delta ADC$ – the area of $\Delta EFC$

We know $\angle{FCE}=45^\circ$ because the diagonal of a square bisects the angle.

We know $\angle{EFC}$ is a right angle as it’s on a straight line with the vertex of a square.

Hence $\Delta EFC$ is isosceles.

$\overline{AC}=\sqrt{1^2+1^2}=\sqrt{2}$ and $\overline{AF}=1$ , hence $\overline{FC}=\sqrt{2}-1$

$A_{\Delta ADC}=\frac{1}{2}(1)(1)=\frac{1}{2}$

$A_{\Delta ECF}=\frac{1}{2}(\sqrt{2}-1)(\sqrt{2}-1)$

Area of shaded region = $\frac{1}{2}-\frac{1}{2}(3-2\sqrt{2})=\sqrt{2}-1$

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Filed under Area, Co-ordinate Geometry, Finding an area, Geometry, Matrices, Transformations, Year 11 Specialist Mathematics

Tagged as matrix tranformations, rotations

September 1, 2025 · 3:50 am

Area Problem

Two rectangular garden beds have a combined area of $40m^2$ . The larger bed has twice the perimeter of the smaller and the larger side of the smaller bed is equal to the smaller side of the larger bed. If the two beds are not similar, and if all edges are a whole number of metres, what is the length, in metres, of the longer side of the larger bed?
AMC 2007 S.14

Let’s draw a diagram

From the information in the question, we know

(1) $\begin{equation*}xy+xz=40\end{equation*}$

and

$\begin{equation*}2x+2y=4x+4z\end{equation}$

$\begin{equation*}x+y=2x+2z\end{equation}$

(2) $\begin{equation*}y=x+2z\end{equation*}$

Equation $1$ becomes

$\begin{equation*}x(x+3z)=40\end{equation}$

As the sides are whole numbers, consider the factors of 40.

$1, 2, 4, 5, 8, 10, 20, 40$

Remember $z<x<y$

$x$	$x+3z$	$z$	$y$	Perimeter Large	Perimeter Small	Comment
$2$	$20$	$6$				$x$ must be greater than $z$
$4$	$10$	$2$	$8$	$2(4+8)=24$	$2(2+4)=12$	This one works
$5$	$8$	$1$	$7$	$2(5+7)=24$	$2(5+1)=12$	This one also works
$8$	$10$	$\frac{2}{3}$				$z$ not a whole number
$10$	$4$	$z<0$				Not possible
$20$	$2$	$z<0$				Not possible
$40$	$1$	$z<0$				Not possible

There are two possibilities

The large garden bed could be $4$ by $8$ and the smaller $4$ by $2$ (Area $40$ Perimeters $24$ and $12$ )

The large garden bed could be $5$ by $7$ and the smaller $5$ by $1$ (Area $40$ Perimeters $24$ and $12$ )

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Filed under Area, Interesting Mathematics, Measurement, Puzzles, Solving Equations, Year 8 Mathematics

Tagged as area problem

April 23, 2025 · 9:30 am

Problem Solving

I am came across this problem and was fascinated. It’s from this book

At first I went straight to the 14-sided polygon, and tried to draw the diamonds (parallelograms), but then I thought let’s start smaller and see if there is a pattern.

Clearly a square contains 1 diamond (itself).

Pentagon

It’s not possible with a pentagon.

Hexagon

A hexagon has 6 diamonds

Septagon

I am guessing it’s not possible to fill a regular 7-sided shape with diamonds

It’s not possible with odd numbers of sides. Regular polygons with an odd number of sides have no parallel sides, so we can’t cover it with rhombi (which have opposite sides parallel).

Octagon

An octagon has 6 diamonds.

We know a decoagon has 10 diamonds (from the question)

Let’s put together what we know

$n$	$4$	$6$	$8$	$10$
Diamonds	$1$	$3$	$6$	$10$

These are the triangular numbers, so when $n=12$ the number of diamonds is $15$ , and for $n=14$ it’s $21$ .

We can work out a rule for calculating the number of diamonds given the number of sides.

Because the difference in the $n$ values is not $1$ , I am going to get $n$ and $D$ in terms of $k$ and then combine the two equations.

From the above table, $n=2k+2$

We know this rule is quadratic as the second difference is constant, hence

$D=\frac{1}{2}k^2+bk+c$

$\begin{equation*}1=\frac{1}{2}+b+c\end{equation}$

(1) $\begin{equation*}\frac{1}{2}=b+c\end{equation*}$

$\begin{equation*}3=\frac{1}{2}2^2+2b+c\end{equation}$

(2) $\begin{equation*}1=2b+c\end{equation*}$

Solve simultaneously, subtract equation $1$ from equation $2$

(3) $\begin{equation*}\frac{1}{2}=b\end{equation*}$

Substitute for $b=\frac{1}{2}$ into equation $1$

$\begin{equation*}\frac{1}{2}=\frac{1}{2}+c\end{equation}$

$c=0$ , therefore $D=\frac{1}{2}k^2+\frac{1}{2}k$

We know $n=2k+2$ hence $k=\frac{n-2}{2}$

Hence $D=\frac{1}{2}(\frac{n-2}{2})^2+\frac{1}{2}(\frac{n-2}{2})$

$D=\frac{1}{8}(n^2-4n+4)+\frac{1}{4}(n-2)=\frac{n^2}{8}-\frac{n}{2}+\frac{1}{2}+\frac{n}{4}-\frac{1}{2}=\frac{n^2}{8}-\frac{n}{4}$

$\begin{equation*}D=\frac{n^2}{8}-\frac{n}{4}\end{equation}$

Let’s test our rule for $n=14$

$\begin{equation*}D=\frac{14^2}{8}-\frac{14}{4}=\frac{49}{2}-\frac{7}{2}=\frac{42}{2}=21\end{equation}$

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Filed under Area, Geometry, Interesting Mathematics, Puzzles, Quadratics

Tagged as Finding a quadratic rule, regular polygons, rhombus, tessellations

April 21, 2025 · 11:08 am

Area/Geometry Problem

This problem is from The Geometry Forum Problem of the Week June 1996

In triangle ABC, AC=18 and D is the point on AC for which AD=5. Perpendiculars drawn from D to AB and CB have lengths of 4 and 5 respectively. What is the area of triangle ABC?

I put together a diagram (in Geogebra)

Add points P and Q

Triangle APD and triangle DQC are right angled. Using pythagoras, $AP=3$ and $QC=12$

$BQDP$ is a cyclic quadrilateral and $BD$ is the diameter. I am not sure if this is useful, but it is good to notice.

$\begin{equation*}sin(A+B+C)=sin(180)=0\end{equation}$

$\begin{equation*}sin((A+C)+B)=sin(A+C)cosB+sinBcos(A+C)=0\end{equation}$

$\begin{equation*}cosB(sinAcosC+sinCcosA)+sinB(cosAcosC-sinAsinC)=0\end{equation}$

$\begin{equation*}cosB(\frac{4}{5}\times\frac{12}{13}+\frac{5}{13}\times\frac{3}{5})+sinB(\frac{3}{5}\times\frac{12}{13}-\frac{4}{5}\times\frac{5}{13})=0\end{equation}$

$\begin{equation*}cosB(\frac{48}{65}+\frac{15}{65})+sinB(\frac{36}{65}-\frac{20}{65})=0\end{equation}$

$\begin{equation*}\frac{63}{65}cosB+\frac{16}{65}sinB=0\end{equation}$

$\begin{equation*}63cosB+16sinB=0\end{equation}$

$\begin{equation*}63+16tanB=0\end{equation}$

$\begin{equation*}tanB=\frac{-63}{16}\end{equation}$

If $tanB=\frac{-63}{16}$ then $sinB=\frac{63}{65}$

Now,

$\begin{equation*}\frac{y+12}{sinA}=\frac{18}{sinB}\end{equation}$

$\begin{equation*}y+12=\frac{4}{5}(18)\frac{65}{63}\end{equation}$

$\begin{equation*}y+12=\frac{104}{7}\end{equation}$

Hence the Area is

$\begin{equation*}A=\frac{1}{2}(18)(\frac{104}{7})sinC\end{equation}$

$\begin{equation*}A=\frac{1}{2}(18)(\frac{104}{7})\frac{5}{13}\end{equation}$

$\begin{equation*}A=\frac{360}{7}=51.43\end{equation}$

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Filed under Area, Finding an area, Geometry, Identities, Non-Right Trigonometry, Pythagoras, Trigonometry

Tagged as area, area of triangles, problem solving

April 3, 2025 · 7:28 am

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Find the area of an $n-$ sided regular polygon if you know the side length, $l$ .

Find the $h$ of the triangle in terms of $l$ and theta.

$tan(\theta)=\frac{\frac{l}{2}}{h}$

$h=\frac{l}{2tan(\theta)}$

Remember the area of a triangle is $A=\frac{1}{2}bh$

Hence, $A=\frac{1}{2} l \times \frac{l}{2tan(\theta)}=\frac{l^2}{4tan(\theta)}$

And $\theta=\frac{360}{2n}=\frac{180}{n}$

Therefore $A=\frac{l^2}{4tan(\frac{180}{n})}$

There are $n$ triangles in an $n-$ sided polygon

(1) $\begin{equation*}A=\frac{nl^2}{4tan(\frac{180}{n})}\end{equation*}$

A=\frac{6\times10^2}{4tan(\frac{180}{6})}

Finding the area of a polygon if you know the inradius or the apothem

The apothem and the inradius are the same. It is the radius of the incircle.

Find the area of the triangle in terms of $a$ and theta.

$tan(\theta)=\frac{\frac{l}{2}}{a}$

$l=2atan(\theta)$

$A=\frac{1}{2}bh$

$A=\frac{1}{2}2atan(\theta)a=a^2tan(\theta)$

And $\theta=\frac{180}{n}$

Hence for an $n-$ sided polygon

(2) $\begin{equation*}A=na^2tan(\frac{180}{n})\end{equation*}$

Finding the area of a regular polygon given the circumradius

The circumradius is the radius of the circumscribed circle ( $R$ in the diagram above)

Remember the area of triangle formula

$A=\frac{1}{2}absin(\theta)$

$A=\frac{1}{2}R^2sin(\theta)$

$\theta=\frac{360}{n}$

Hence, $A=\frac{1}{2}R^2sin(\frac{360}{n})$

Hence, for an $n-$ sided polygon

(3) $\begin{equation*}A=\frac{nR^2sin(\frac{360}{n})}{2}\end{equation*}$

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Filed under Area, Area of Triangles (Sine), Finding an area, Non-Right Trigonometry, Regular Polygons, Right Trigonometry, Year 11 Mathematical Methods

Tagged as area of regular polygons, area of triangles, trig

February 2, 2025 · 7:46 am

Trigonometric Limits

$\lim\limits_{x \to 0}\frac{sin(x)}{x}=?$

Remember $cos(x)=\frac{OA}{OB}=\frac{OA}{1}$ , hence $OA=cos(x)$ and the co-ordinate of $A$ is $(cos(x), 0)$ .

$sin(x)=\frac{AB}{OB}=\frac{AB}{1}$ , hence $AB=sin(x)$ and the co-ordinate of $B$ is $(cos(x), sin(x))$

And from the definition of $tan(x)$ we know $D$ is the point $(1, tan(x))$

Consider the areas of triangle $OAB$ , sector $OBC$ , and triangle $OCD$ .

We know from inspection of the above diagram that

Area $OAB<$ Area $OCB<$ Area $OCD$

Which means,

$\frac{1}{2}b_1 h_1<\frac{1}{2}r^2x<\frac{1}{2}b_2 h_2$

We can ignore all of the halves.

$cos(x)sin(x)<x<(1)tan(x)$

Remember $tan(x)=\frac{sin(x)}{cos(x)}$

$cos(x)sin(x)<x<\frac{sin(x)}{cos(x)}$

Divide everything by $sin(x)$ (as we are in the first quadrant we know $sin(x)>0$ , so we don’t need to worry about the inequality)

$cos(x)<\frac{x}{sin(x)}<\frac{1}{cos(x)}$

Invert everything and change the direction of the inequalities)

$\frac{1}{cos(x)}>\frac{sin(x)}{x}>cos(x)$

I am going to rewrite it as follows

$cos(x)<\frac{sin(x)}{x}<\frac{1}{cos(x)}$

because I like to use less thans rather than greater thans.

Now what happens as $x$ tends to $0$ ?

$cos(0)=1$

$1<\frac{sin(x)}{x}<\frac{1}{1}$

Hence by the squeeze theorem $\lim\limits_{x \to 0}\frac{sin(x)}{x}=1$

Now we know this limit, we are going to use it to find $\lim\limits_{x \to 0}\frac{1-cos(x)}{x}$

Multiply by $\frac{1+cos(x)}{1+cos(x)}$

$\lim\limits_{x \to 0}\frac{1-cos(x)}{x}\times \frac{1+cos(x)}{1+cos(x)}$

$\lim\limits_{x \to 0}\frac{1-cos^2(x)}{x(cos(x)+1)}$

$\lim\limits_{x \to 0}\frac{sin^2(x)}{x(cos(x)+1)}$

$\lim\limits_{x \to 0}\frac{sin(x)}{x}\times sin(x)(cos(x)+1)}$

$\lim\limits_{x \to 0}\frac{sin(x)}{x}\times \lim\limits_{x \to 0}sin(x)(cos(x)+1)$

If we evaluate the limits,

$(1)(sin(0)(cos(0)+1)=1\times 0 \times 2=0$

Hence, $\lim\limits_{x \to 0}\frac{1-cos(x)}{x}=0$

In the next post we are going to use these limits to differentiate sine and cosine functions.

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Filed under Area, Area of Triangles (Sine), Calculus, Identities, Trigonometry, Year 12 Mathematical Methods

Tagged as trig limits, trigonometric limits

December 14, 2024 · 6:56 am

Volume and Surface Area of a Conical Frustrum

My best attempt at drawing a Frustrum in Geogebra.

We have a truncated cone,

(1) $\begin{equation*}V=\frac{1}{3}\pi R_2^2(h_1+h_2)-\frac{1}{3}\pi R_1^2h_1\end{equation*}$

We are unlikely to know $h_1$ . Can we get $h_1$ in terms that we do know (i.e. $R_1, R_2, h_2$ )?

Think of similar triangles

$\Delta ABC \sim \Delta ADE (AA)$

$R_1 \parallel R_2$

$\angle {C}=\angle{E}$ (Corresponding Angles in Parallel Lines)

$\angle {B}=\angle {D}$ (Corresponding Angles in Parallel Lines)

Therefore

$\begin{equation*}\frac{h_1}{R_1}=\frac{h_1+h_2}{R_2}\end{equation}$

Rearrange to make $h_1$ the subject.

$\begin{equation*}h_1=\frac{h_2R_1}{R_2-R_1}\end{equation}$

Substitute into equation $(1)$

$\begin{equation*}V=\frac{1}{3} \pi((R_2^2(\frac{h_2R_1}{R_2-R_1})+h_2)-R_1^2(\frac{h_2R_1}{R_2-R_1}))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi (R_2^2h_2+\frac{R_2^2h_2R_1}{R_2-R_1}-\frac{R_1^2h_2R_1}{R_2-R_1})\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi (R_2^2h_2+\frac{R_2^2h_2R_1-R_1^2h_2R_1}{R_2-R_1})\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi (R_2^2h_2+\frac{h_2R_1}{R_2-R_1}(R_2^2-R_1^2))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi h_2(R_2^2+\frac{R_1}{R_2-R_1}(R_2-R_1)(R_2+R_1))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi h_2(R_2^2+R_1(R_2+R_1))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi h_2(R_2^2+R_1 R_2+R_1^2)\end{equation}$

Now let’s think about the surface area.

The surface area of a cone is $A=\pi r^2+\pi rs$ where $s$ is the slant height of the cone.

Once again, we need to subtract the ‘missing’ part of the cone.

(2) $\begin{equation*}A=\pi R_2^2+\pi R_2(s_1+s_2)+ \pi R_1^2- \pi R_1s_1\end{equation*}$

We don’t need to subtract the circle of the top cone because it is the top of the frustrum, but we do need to add it on.

Using similar triangles again

$\begin{equation*}\frac{s_1}{R_1}=\frac{s_1+s_2}{R_2}\end{equation}$

$\begin{equation*}s_1=\frac{s_2R_1}{R_2-R_1}\end{equation}$

Substitute into equation $(2)$

$\begin{equation*}A=\pi(R_2^2+R_2(\frac{s_2R_1}{R_2-R_1}+s_2)+\pi R_1^2-R_1(\frac{s_2R_1}{R_2-R_1}))\end{equation}$

$\begin{equation*}A=\pi(R_2^2+R_2s_2+R_1^2+\frac{s_2R_1}{R_2-R_1}(R_2-R_1))\end{equation}$

$\begin{equation*}A=\pi(R_2^2+R_2s_2+R_1s_2+R_1^2)\end{equation}$

$\begin{equation*}A =\pi (R_1^2+s_2(R_1+R_2)+R_2^2)\end{equation}$

And hence the curved surface area is $\pi s_2(R_1+R_2)$ .

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Filed under Area, Area of Frustrum, Interesting Mathematics, Measurement, Volume of Frustrum

Tagged as frustrum, surface area, volume

October 13, 2024 · 4:59 am

Tricky Area Ratio Question

A circle of radius $r$ , is inscribed within an isosceles triangle $ABC$ . $CA=CB=5r$ . Given that $\angle{ACB}$ is acute, find the ratio of the area of the circle to that of triangle $ABC$ .

Mathematics Specialist 3AB Question15 page 55

I came across this question while searching for an area of sectors and segments question.

Here’s a diagram

We know $AC, AB and BC$ are tangents to the circle. Because the triangle is isosceles, the distance from $A$ to the circle is the same as the distance from $B$ to the circle.

$CP$ is perpendicular to $AB$ (because the triangle is isosceles).

Because it is proportional, i.e. $r$ and $5r$ , we can let $r=1$

Let $h=CP$

(1) $\begin{equation*}h=\sqrt{25-a^2}\end{equation*}$

but $h$ also equals

(2) $\begin{equation*}h=1+\sqrt{(5-a)^2+1}\end{equation*}$

Set equation $1$ equal to equation $2$

$\begin{equation*}\sqrt{25-a^2}={1+\sqrt{(5-a)^2+1}\end{equation}$

Square both sides

$\begin{equation*}25-a^2=1+(5-a)^2+1+2\sqrt{(5-a)^2+1}\end{equation}$

Expand and simplify

$\begin{equation*}25-a^2=2+25-10a+a^2+2\sqrt{(5-a)^2+1}\end{equation}$

$\begin{equation*}0=2-10a+2a^2+2\sqrt{(5-a)^2+1}\end{equation}$

Divide by $2$

$\begin{equation*}0=1-5a+a^2+\sqrt{(5-a)^2+1}\end{equation}$

$\begin{equation*}-\sqrt{(5-a)^2+1}=a^2-5a+1\end{equation}$

Square both sides

$\begin{equation*}(5-a)^2+1=a^4-5a^3+a^2-5a^3+25a^2-5a+a^2-5a+1\end{equation}$

$\begin{equation*}25-10a+a^2+1=a^4-10a^3+27a^2-10a+1\end{equation}$

$\begin{equation*}0=a^4-10a^3+26a^2-25 \end{equation}$

I solved this using a graphics calculator

$a=-0.8434, a=1.3068, a=4.5367, a=5$

We can reject $a=-0.8434$ , $a=4.5367$ , and $a=5$ . If $a=5$ , there isn’t a triangle, and if $a=4.5367$ $\angle{ACB}$ is not acute.

Hence the area of triangle $ABC=a\times h$

$\begin{equation*}A_{\Delta}=1.3068\times\sqrt{25-1.3068^2}\end{equation}$

(3) $\begin{equation*}A_{\Delta}=6.3069\end{equation*}$

(4) $\begin{equation*}A_{\circ}=\pi\end{equation*}$

Hence, equation $4$ divided by equation $3$ is

$\begin{equation*}\frac{\pi}{6.3069}\approx 0.5\end{equation}$

Perhaps I approached this question in the wrong way. Is there an easier process?

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Filed under Area, Geometry, Pythagoras

Tagged as area, inscribed circle, isosceles triangles, tangents to circles

March 29, 2024 · 6:20 am

Three Circles – Area Problem

This is question 5 from the UK Maths Trust Senior Challenge October 2023.

I have tackled this in three ways; using non-right trig to find the area, Heron’s Law, and the Shoelace Formula.

Method 1

Use the area of a triangle formula

Use the cosine rule to find cosθ.

Once we have cosθ, we can find sinθ.

Hence the area is,

Method 2

Use Heron’s law.

Heron’s law is a way of calculating area of a triangle from the lengths of the three sides of the triangle.

This is my preferred method – simple and direct.

Method 3

Shoelace formula (Gauss’s Area formula)

We need to allocate each of the vertices a co-ordinate.

The co-ordinates are listed in an anti-clockwise direction.

This is probably a bit over the top, but once you get the hang of it, it’s very easy.

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Filed under Area, Area of Triangles (Sine), Heron's Law, Non-Right Trigonometry, Shoelace Forumla, Trigonometry

Tagged as Heron's law, Shoelace Formula, UK Senior Maths Challenge

Category Archives: Area

Geometry Problem

Linear Transformation (Rotation) Question

Area Problem

Problem Solving

Pentagon

Hexagon

Septagon

Octagon

Area/Geometry Problem

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Finding the area of a polygon if you know the inradius or the apothem

Finding the area of a regular polygon given the circumradius

Trigonometric Limits

Volume and Surface Area of a Conical Frustrum

Tricky Area Ratio Question

Three Circles – Area Problem

Method 1

Method 2

Method 3

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