Category Archives: Geometry

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 6, 2025 · 9:50 am

Clock (Geometry Question)

At what time after 2pm will the minute hand over take the hour hand?

Let the distance the minute hand moves be $x^\circ$ . Then the distance the hour hand moves is $x-60^\circ$ (The angle between the hands at 2pm is $60^\circ$ )

The rate the hour hand moves is $R_H=\frac{30}{60}=\frac{1}{2}^\circ$ per minute, and the rate the minute hand moves is $R_M=\frac{360}{60}=6^\circ$ per minute.

The time is the distance divided by the rate,

$\begin{equation*}\frac{x}{6}=\frac{x-60}{\frac{1}{2}}\end{equation}$

$\begin{equation*}\frac{x}{2}=6x-360\end{equation}$

$\begin{equation*}360=\frac{11x}{2}\end{equation}$

$\begin{equation*}x=65\frac{5}{11}^\circ\end{equation}$

As the minute hand moves $65\frac{5}{11}^\circ$ , the change in time is $65\frac{5}{11}^\circ \div 6=10$ minutes and $54$ seconds. Hence the time is $2:11$ pm

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Filed under Finding an angle, Geometry, Puzzles

Tagged as geometry clock question

May 24, 2025 · 10:02 am

Circle Geometry Problem

In the diagram, points $A, B, C, D$ and $Q$ lie on a circle centre $O$ , radius $6$ cm and diameter $BQ, \angle{ABQ}=50^\circ, AB$ is parallel to $DO$ and point $P$ lies on diameter $BQ$ such that $OP=DP=4$ cm.

(a) Find $\angle{BCD}$

(b) Determine the length of $PC$ .

$\angle{BOD}=180^\circ-50^\circ=130^\circ$ (Co-interior angles in parallel lines are supplementary.)

$\angle{BCD}=\frac{1}{2}\times 130=65^\circ$ (Angles subtended by the same arc. The angle at the centre is twice the angle at the circumference.)

$\angle{BCD}=65^\circ.$

Let $PC=x$

From the intersecting chord theorem

$\begin{equation*}4\times x=2\times 10\end{equation}$

$\begin{equation*}4x=20\end{equation}$

$\begin{equation*}x=5\end{equation}$

$PC=5cm$

A chord $AB$ of a circle $O$ is extended to $C$ . The straight line bisecting $\angle{OAB}$ meets the circle at $E$ . Let $\angle{BAE}=x$ . Prove that $EB$ bisects $\angle{OBC}$ .

$\angle{BAO}=2x$ ( $AE$ bisects $\angle {BAO}$ )

$\Delta AOB$ is isosceles ( $AO=B0$ radii of the circle)

$\angle{ABO}=2x$ (Equal angles in isosceles triangle)

Therefore $\angle {AOB}=180^\circ-4x$ (angle sum of a triangle)

$\angle {BEA}=90^\circ-2x$ (angle at the circumference is half angle at the centre)

$\angle{ABE}=180^\circ-x-(90^\circ-2x)=90^\circ+x$ (angle sum of a triangle)

$\angle{CBE}=180^\circ-(90^\circ+x)=90^\circ-x$ (angles on a straight line)

$\angle{OBE}=90^\circ+x-2x=90^\circ-x$

$\angle{OBE}=\angle{CBE}$

Hence, $BE$ bisects $\angle{OBC}$

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Filed under Circle Theorems, Geometry, Uncategorized, Year 11 Specialist Mathematics

Tagged as circle geometry, circle theorems

May 3, 2025 · 3:34 am

Geometry Puzzle

Another puzzle from this book

Two ladders are propped up vertically in a narrow passageway between two vertical buildings. The ends of the ladders are 8 metres and 4 metres above the pavement.
Find the height above the ground, $T$ ,

$\Delta ABC \sim \Delta TEC$ and $\Delta DCB \sim \Delta TEB$ (Angle Angle Similarity)

Hence,

(1) $\begin{equation*}\frac{h}{8}=\frac{x}{x+y}\end{equation*}$

(2) $\begin{equation*}\frac{h}{4}=\frac{y}{x+y}\end{equation*}$

From equation $1$ $h=\frac{8x}{x+y}$ and from $2$ $h=\frac{4y}{x+y}$

Hence, $\frac{8x}{x+y}=\frac{4y}{x+y}$

Therefore, $8x=4y$ and $y=2x$

From equation $1$

$\begin{equation*}h=\frac{8x}{3x}=\frac{8}{3}\end{equation}$

Hence $T$ is $2 \frac{2}{3}$ m above the ground.

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Filed under Geometry, Puzzles, Similarity, Simplifying fractions, Solving Equations

Tagged as Puzzle

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 6, 2025 · 1:51 am

Using a Vector Method to Find an Angle Bisector

Points $A$ and $B$ are defined by the position vectors $\mathbf{a}=3\mathbf{i}+4\mathbf{j}$ and $\mathbf{b}=12\mathbf{i}+5\mathbf{j}$ .

Find a vector that bisects $\angle{AOB}$ .

If we think about how we add vectors using the parallelogram rule

Finding the resultant vector using the parallelogram rule

we can take advantage of the geometric properties of parallelograms (or of a rhombus).

If $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, then the parallelogram is a rhombus, and the diagonal (i.e the resultant) bisects the angle.

We need to find the sum of the unit vectors.

$|\mathbf{a}|=\sqrt{3^2+4^2}=5$

$\therefore \hat{\mathbf{a}}=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}$

$|\mathbf{b}|=\sqrt{12^2+5^2}=13$

$\therefore \hat{\mathbf{b}}=\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}$

The vector that bisects $\angle{AOB}$ is

$\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}+\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}=\frac{99}{65}\mathbf{i}+\frac{64}{65}\mathbf{j}$

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Filed under Geometry, Vectors, Year 11 Specialist Mathematics

Tagged as bisecting an angle, vectors, Year 11 Specialist Mathematics

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

March 18, 2025 · 6:04 am

Intersecting Secant Theorem

$CD$ is a tangent to the circle.

Prove $c^2=a(a+b)$

I am going to add two chords to the circle

$\angle{BDC}=\angle{CAD}$ (angles in alternate segments are congruent)

$\angle{BCD}=\angle{DCA}$ (shared angle)

$\therefore \Delta BDC\cong \Delta{DAC}$ (AA)

Hence

$\frac{DC}{AC}=\frac{BC}{DC}$ (Corresponding sides in similar triangles)

$\frac{c}{a+b}=\frac{a}{c}$

$\therefore c^2=a(a+b)$

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Filed under Circle Theorems, Geometry, Year 11 Specialist Mathematics

Tagged as circle geometry, Intersecting Secants Theorem

Category Archives: Geometry

Geometry Problem

Linear Transformation (Rotation) Question

Clock (Geometry Question)

Circle Geometry Problem

Geometry Puzzle

Problem Solving

Pentagon

Hexagon

Septagon

Octagon

Area/Geometry Problem

Using a Vector Method to Find an Angle Bisector

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

Intersecting Secant Theorem

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