Racquel Sanderson

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 24, 2025 · 6:28 am

Christmas Cartesian Co-ordinates

I found a co-ordinate puzzle here.

Finished image

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Filed under Co-ordinate Geometry

Tagged as Christmas Maths, Coordinate picture, Rudolf

December 19, 2025 · 8:10 am

Algebra Time Question

This question is from Challenging Problems in Algebra

It’s the type of question students hate – “Who talks like that?”

Let $t$ be the number of hours from noon.

$\begin{equation*}\frac{t}{8}+6-\frac{t}{4}=t\end{equation}$

$\begin{equation*}6=\frac{9t}{8}\end{equation}$

$\begin{equation*}t=\frac{48}{9}=5\frac{1}{3}\end{equation}$

Hence the time is 5:20pm

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Filed under Algebra, Puzzles, Simplifying fractions, Solving Equations

Tagged as algebra, solving linear equations, time

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

November 26, 2025 · 10:57 am

Trigonometric Equation

Solve $cos(4\theta)+cos(2\theta)+cos(\theta)=0$ for $0\le \theta \le\pi$

Remember the identity

(1) $\begin{equation*}cos(A)+cos(B)=2cos(\frac{A+B}{2})cos(\frac{A-B}{2})\end{equation*}$

Hence

$\begin{equation*}cos(4\theta)+cos(2\theta)=2cos(3\theta)cos(\theta)\end{equation}$

Now I have

$\begin{equation*}2cos(3\theta)cos(\theta)+cos(\theta)=0\end{equation}$

$\begin{equation*}cos(\theta)(2cos(3\theta)+1)=0\end{equation}$

$cos(\theta)=0$ or $cos(3\theta)=\frac{-1}{2}$

$\theta=\frac{\pi}{2}$

$cos(3\theta)=-\frac{1}{2}$ for $0 \le \theta \le 3\pi$

$3\theta=\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}$

$\theta=\frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}$

Hence $\theta =\frac{\pi}{2},\frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}$

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Filed under Identities, Quadratic, Solving Equations, Solving Trig Equations, Trigonometry, Year 11 Specialist Mathematics

Tagged as solving trig equations, sum to product trig identity

November 22, 2025 · 6:10 am

I usually choose to use synthetic division when factorising polynomials, but I know some teachers are unhappy when their students do this. So for completeness, here is my PDF for Polynomial Long Division.

Polynomial Long Division Download

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Filed under Algebra, Cubics, Factorisation, Factorising, Factorising, Polynomials, Quadratics, Solving, Solving, Solving Equations, Year 11 Mathematical Methods

Tagged as polynomial long division, solving cubic equations

November 16, 2025 · 5:54 am

Geometry Circle Question

In the diagram below, $A, B, C$ and $D$ lie on the circle with centre $O$ . If $\angle{DBC} = 41^{\circ}$ and $\angle{ACD} = 53^{\circ}$ , determine with reasoning $\angle{BAC}$ and $\angle{AOB}$

We know $OA=OB=OD$ – radii of the circle.

Which means, $\Delta{AOB}$ is isosceles and $\angle{OAB}=\angle{OBA}$ – equal angles isosceles triangle.

$\angle{AOD}=2\angle{ACD}$ – angle at the centre twice the angle at the circumference.

$\angle{AOB}=106^{\circ}$

This means $\angle{AOB}=74^{\circ}$ – angles on a straight line are supplementary

$\angle{OAD}=\angle{ODA}=37^{\circ}$ – equal angles isosceles triangle and the angle sum of a triangle.

$\angle{DBA}=\angle{DCA}=53^{\circ}$ – angle at the circumference subtended by the same arc are congruent.

$\angle{CAD}=\angle{CBD}=41^{\circ}$ – angles at the circumference subtended by the same arc are congruent.

$\angle{OAB}=53^{\circ}$ – equal angle isosceles triangle

Hence $\angle{BAC}=12^{\circ}$

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Filed under Circle Theorems, Finding an angle, Geometry, Year 11 Specialist Mathematics

Tagged as circle theorems, geometry, Year 11 Specialist Mathematics

November 11, 2025 · 10:19 am

Completing the Square

$x^2+6x-4 \rightarrow (x+3)^2-13$

Completing the square is useful to

sketch parabolas.
solve quadratics.
factorising quadratics
finding the centre and radius version of the equation of a circle.

When completing the square we take advantage of perfect squares. For example, $(x+3)^2=(x+3)(x+3)=x^2+6x+9$

$6=2\times 3$ and $9=3\times 3$

Example 1

Put $x^2+8x-5$ into completed square form.

What perfect square has an $8x$ term?

$(x+4)^2=x^2+8x+16$

We don’t want $+16$ , we want $-5$ , so subtract $16+5$

$x^2+8x-5=(x+4)^2-21$

$x^2+bx+c=(x+\frac{b}{2})^2-(\frac{b}{2})^2+c$

What about a non-monic quadratic? For example,

$2x^2+12x+11$

Factorise the $2$

$2(x^2+6x+\frac{11}{2})$

And continue as before

$2[(x+3)^2-9+\frac{11}{2}]=2[(x+3)^2-\frac{18}{2}+\frac{11}{2}]=2[(x+3)^2-\frac{7}{2}]=2(x+3)^2-7$

Example 2

$y=2x^2+7x-5$

$2(x^2+\frac{7}{2}x-\frac{5}{2})$

$2[(x+\frac{7}{4})^2-(\frac{7}{4})^2-\frac{5}{2}]$

$2[(x+\frac{7}{4})^2-\frac{49}{16}-\frac{40}{16}]$

$2[(x+\frac{7}{4})^2-\frac{89}{16}]$

$2(x+\frac{7}{4})^2-\frac{89}{8}$

$ax^2+bx+c=a(x+\frac{b}{2a})^2-a((\frac{b}{2a})^2+\frac{c}{a})$

Casio Classpad e-activity

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Filed under Algebra, Arithmetic, Classpad Skills, Completing the Square, Fractions, Quadratic, Quadratics, Year 10 Mathematics, Year 9 Mathematics

Tagged as completing the square, quadratics

November 5, 2025 · 6:57 am

Equation of a Circle and Geometry Question

A circle has equation $x^2+y^2+4x-6y=36$
(a) Find the centre and radius of the circle.
Points $S$ and $T$ lie on the circle such that the origin is the midpoint of $ST$ .
(b) Show that $ST$ has a length of 12.

(a)We need to put the circle equation into completed square form

$\begin{equation*}(x+2)^2-4+(y-3)^2-9=36\end{equation}$

$\begin{equation*}(x+2)^2+(y-3)^2=49\end{equation}$

The centre is $(-2, 3)$ and the radius is $7$ .

(b)Draw a diagram

We know $SO$ and $TO$ are radii of the circle. Hence $\Delta{SOT}$ is isosceles and the line segment from $O$ to the origin is perpendicular to $ST$ .