Category Archives: Area

May 18, 2026 · 7:24 am

January 2, 2026 · 7:08 am

Geometry Puzzle (finding a fraction of an area)

Geometry Puzzles in Felt Tip: A Compilation of puzzles from 2018 – Catriona Shearer

Band $1$ and $3$ have the same area.

We want to find the area of the shaded segment.

As the dots are equally spaced, the sector’s angle is $\frac{\pi}{2} = (\frac{2\pi}{12}\times 3)$

Remember the area of a segment is $A=\frac{1}{2}r^2(\theta-sin(\theta))$ where the angle measurement is in radians.

(1) $\begin{equation*}A=\frac{1}{2}r^2(\frac{\pi}{2}-sin(\frac{\pi}{2}))=\frac{1}{2}r^2(\frac{\pi}{2}-1))=\frac{\pi r^2}{4}-\frac{r^2}{2}\end{equation*}$

We want to find the area of the shaded segment.

As the dots are equally spaced, the sector’s angle is $\frac{2\pi}{12} = (\frac{\pi}{6})$

(2) $\begin{equation*}A=\frac{1}{2}r^2(\frac{\pi}{6}-sin(\frac{\pi}{6}))=\frac{\pi r^2}{12}-\frac{r^2}{4}\end{equation*}$

The area of band $1$ is equation $1 -$ equation $2$ .

(3) $\begin{equation*}\frac{\pi r^2}{4}-\frac{r^2}{2}-(\frac{\pi r^2}{12}-\frac{r^2}{4})=\frac{\pi r^2}{6}-\frac{r^2}{4}\end{equation*}$

Band 2 consists of two congruent triangles and two congruent sectors.

$\begin{equation*}\theta=\frac{2\pi}{12}\times 5=\frac{5\pi}{6}, \alpha=\frac{\pi}{6}\end{equation}$

(4) $\begin{equation*}A=2(\frac{1}{2}r^2sin(\frac{5\pi}{6}))+2(\frac{1}{2}r^2\frac{\pi}{6})=\frac{r^2}{2}+\frac{r^2 \pi}{6}\end{equation*}$

Hence the shaded area is $2(\frac{\pi r^2}{6}-\frac{r^2}{4})+\frac{r^2}{2}+\frac{r^2 \pi}{6}=\frac{\pi r^2}{2}$

The area of the circle is $\pi r^2$

Hence the fraction of the shaded area is $=\frac{\frac{\pi r^2}{2}}{\pi r^2}=\frac{1}{2}$

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

Tagged as catriona shearer, finding an area, geometry, geometry puzzle

December 10, 2025 · 7:54 am

Geometry Problem

Problem from *50 Maths Problems with Solutions* – Ajmal KP

The blue shaded area is the area of triangles $APO$ and $AQO$ subtract the sector $POQ$ .

We can use Heron’s law to find the area of the triangle $\Delta{ABC}$

$\begin{equation*}A=\sqrt{s(s-a)(s-b)(s-c)}\end{equation}$

where $s=\frac{a+b+c}{2}$

$\begin{equation*}A=\sqrt{20(20-16)(20-10)(20-14)}=40\sqrt{3}\end{equation}$

We also know the area of triangle $\Delta{ABC}=sr$ where $r$ is the radius of the inscribed circle.

Hence, $40\sqrt{3}=20r$ and $r=2\sqrt{3}$

We know $AP=AQ, CQ=CR$ , and $BP=BR$ – tangents to a circle are congruent.

$\begin{equation*}14-x=6+x\end{equation}$

(1) $\begin{equation*}8=2x\end{equation*}$

(2) $\begin{equation*}x=4\end{equation*}$

Area $\Delta{AQO}=\frac{1}{2}10\times 2\sqrt{3}=10\sqrt{3}$

Area $\Delta{APO}=$ Area $\Delta{AQO}$

$\begin{equation*}tan(\theta)=\frac{10}{2\sqrt{3}}\end{equation}$

$\begin{equation*}\theta=70.9^{\circ}\end{equation}$

Area of sector $OPQ=\frac{2\times70.9}{360}\pi (2\sqrt{3})^2=14.8$

Blue area = $20\sqrt{3}-14.8=19.8cm^2$

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Filed under Algebra, Area, Finding an angle, Finding an area, Geometry, Heron's Law, Interesting Mathematics, Puzzles, Radius and Semi-Perimeter, Right Trigonometry, Solving Equations, Trigonometry

Tagged as geometry puzzle

December 4, 2025 · 12:52 pm

Area of a triangle from the semi-perimeter and the radius of the incircle.

$\begin{equation*}A=sr\end{equation}$

Where $s$ is the semi-perimeter, $s=\frac{a+b+c}{2}$ and $r$ is the radius of the incircle.

$AB, BC$ and $AC$ are tangents to the circle. And the radii are perpendicular to the tangents.

Add line segments $AO, CO$ and $BO$ .

$\Delta{ABC}$ is split into three triangles, $\Delta{AOB}, \Delta{AOC}$ and $\Delta{BOC}$ .

Hence Area $\Delta{ABC}=\Delta{AOB}+\Delta{AOC}+\Delta{BOC}$

$\Delta{ABC}=\frac{1}{2}cr+\frac{1}{2}br+\frac{1}{2}ar$

$\Delta{ABC}=\frac{1}{2}r(a+b+c)$

Remember $s=\frac{1}{2}(a+b+c)$

$\Delta{ABC}=sr$

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Filed under Area, Finding an area, Geometry, Interesting Mathematics, Radius and Semi-Perimeter

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

Category Archives: Area

Using Integration to find the Centroid of an Area

Geometry Puzzle (finding a fraction of an area)

Geometry Problem

Area of a triangle from the semi-perimeter and the radius of the incircle.

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

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