Racquel Sanderson

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 $ST$ .

$OT=7$ and the distance from $O$ to the origin is

$\begin{equation*}\sqrt{(-2-0)^2+(3-0)^2}=\sqrt{13}\end{equation}$

We can use Pythagoras to find the distance from the origin to $T$ .

$\begin{equation*}x=\sqrt{7^2-(\sqrt{13})^2}=\sqrt{49-13}=\sqrt{36}=6\end{equation}$

Hence $ST=2\times6=12$

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Filed under Co-ordinate Geometry, Geometry, Pythagoras, Year 11 Mathematical Methods

October 28, 2025 · 5:42 am

Integration Question (much easier with Integration by Parts)

The Year 12 Mathematics Methods course doesn’t cover Integration by Parts, so they end up with questions like the following.

Determine the following:
(a) $\frac{d}{dx} \left (e^{2x}sin(3x) \right )$
(b) $\frac{d}{dx} \left (e^{2x}cos(3x) \right )$
Hence, determine the following integral by considering both parts (a) and (b)
$\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx$

(a) Use the product rule

(1) $\begin{equation*}\frac{d}{dx} \left (e^{2x}sin(3x) \right)= 2e^{2x}sin(3x)+3e^{2x}cos(3x) \end{equation*}$

(b)

(2) $\begin{equation*}\frac{d}{dx} \left (e^{2x}cos(3x) \right )=2e^{2x}cos(3x)-3e^{2x}sin(3x)\end{equation*}$

I need to use equations $1$ and $2$ to find $\int_0^{\frac{\pi}{2}}13e^2xcos(3x) \enspace dx$ .

The $e^{2x}sin(3x)$ terms need to vanish and I need $13$ of the $e^{2x}cos(3x)$ terms.

$3\times \text{equation} (1)+ 2\times \text{equation} (2)$

(3) $\begin{equation*}3\frac{d}{dx} \left (e^{2x}sin(3x) \right)= 6e^{2x}sin(3x)+9e^{2x}cos(3x) \end{equation*}$

(4) $\begin{equation*}2\frac{d}{dx} \left (e^{2x}cos(3x) \right )=4e^{2x}cos(3x)-6e^{2x}sin(3x)\end{equation*}$

Equation $3$ plus equation $4$

(5) $\begin{equation*}3\frac{d}{dx} \left (e^{2x}sin(3x) \right)+2\frac{d}{dx} \left (e^{2x}cos(3x) \right )=13e^{2x}cos(3x)\end{equation*}$

Integrate both sides of the equation

$\int_0^{\frac{\pi}{2}} \left (3\frac{d}{dx} \left (e^{2x}sin(3x) \right)+2\frac{d}{dx} \left (e^{2x}cos(3x) \right ) \right ) dx=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx$

By the fundamental theorem of calculus, we know

$(3e^{2x}sin(3x) \right)+2 \left (e^{2x}cos(3x) \right ]_0^{\frac{\pi}{2}}=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx$

$3e^{\pi}sin(\frac{3\pi}{2}})+2e^{\pi}cos(\frac{3\pi}{2}})-3e^0sin(3(0))-2e^0cos(3(0))=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx$

$\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx=-3e^{\pi}-2$

Integration by Parts

Remember $\int u\enspace dv=uv-\int v \enspace du$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx$

Let $u=cos(3x)$ , then $du=-3sin(3x)$

and $dv=e^{2x}$ , then $v=\frac{e^{2x}}{2}$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx = \left (cos(3x)\left (\frac{e^{2x}}{2}\right ) \right )_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \frac{e^{2x}}{2}(-3sin(3x)) \enspace dx$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=cos(\frac{3\pi}{2})(\frac{e^\pi}{2})-cos(0)(\frac{e^0}{2})+\frac{3}{2}\int_0^{\frac{\pi}{2}}sin(3x)e^{2x} \enspace dx$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}\int_0^{\frac{\pi}{2}}sin(3x)e^{2x} \enspace dx$

Let $u=sin(3x)$ , then $du=3cos(3x)$

and $dv=e^{2x}$ . then $v=\frac{e^{2x}}{2}$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}(\frac{e^{2x}}{2}sin(3x)]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \frac{e^{2x}}{2}(3cos(3x)) \enspace dx)$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}(-\frac{e^\pi}{2}-\frac{3}{2} \int_0^{\frac{\pi}{2}} e^{2x}cos(3x) \enspace dx)$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}-\frac{3e^\pi}{4}-\frac{9}{4} \int_0^{\frac{\pi}{2}} e^{2x}cos(3x) \enspace dx$

Collect like terms (the integrals are like)

$\frac{13}{4}(\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx)=-\frac{1}{2}-\frac{3e^\pi}{4}$

$(\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx)=\frac{-2-3e^\pi}{13}$

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Filed under Algebra, Calculus, Integration, Integration by Parts, Year 12 Mathematical Methods

Tagged as integration by parts

October 20, 2025 · 11:11 am

Fundamental Theorem of Calculus

(1) $\begin{equation*}\frac{d}{dx} \left (\int_2^{f(x)} g(t) dt \right )=g(f(x))f'(x)\end{equation*}$

My Year 12 Mathematics Methods students are getting ready for their exam, and questions using the above idea have created a bit of consternation. I am going to work through an example, and show why the ‘formula’ works.

Example

Find $\frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right )$ .

$\begin{equation*}=\frac{d}{dx} \left (\frac{4t^3}{3}+\frac{3t^2}{2} \right ]_2^{x^2}\end{equation}$

$\begin{equation*}=\frac{d}{dx} \left ( \frac{4(x^2)^3}{3}+\frac{3(x^2)^2}{2} -(\frac{4(2^3)}{3}+\frac{3(2^2)}{2} \right )\end{equation}$

$\begin{equation*}=\frac{d}{dx} \left (\frac{4x^6}{3}+\frac{3x^4}{2}-\frac{50}{3} \right) \end{equation}$

$\begin{equation*}=\frac{24x^5}{3}+\frac{12x^3}{2}\end{equation}$

(2) $\begin{equation*}=8x^5+6x^3\end{equation*}$

If we used ‘formula’ $1$

$\begin{equation*}\frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right )=(4(x^2)^2+3(x^2))(2x)\end{equation}$

(3) $\begin{equation*}\frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right )=8x^5+6x^3 \end{equation*}$

We can see equation $2$ and $3$ are the same.

More formally

$\begin{equation*}\frac{d}{dx} \left ( \int_c^{f(x)} g(t) dt \right )=\frac{d}{dx} \left (G(f(x))-G(c) \right ) \end{equation}$

Remember $\frac{d}{dx} \left (f(g(x)) \right )=f'(g(x))\times g'(x)$

$\begin{equation*}\frac{d}{dx} \left (G(f(x))-G(c) \right ) =G'(f(x))f'(x)=g(f(x))\times f'(x)\end{equation}$

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Filed under Calculus, Chain Rule, Differentiation, Integration, Year 12 Mathematical Methods

Tagged as differentiation, fundamental theorem of calculus

October 15, 2025 · 6:19 am

Hard Equation Solving Question

Find the value(s) of $k$ such that the equation below has two numerically equal but opposite sign solutions (e.g. $5$ and $-5$ ).

$\begin{equation*}\frac{x^2-2x}{4x-1}=\frac{k-1}{k+1}\end{equation}$

$\begin{equation*}(x^2-2x)(k+1)=(k-1)(4x-1)\end{equation}$

$\begin{equation*}(k+1)x^2-2kx-2x=4kx-k-4x+1\end{equation}$

$\begin{equation*}(k+1)x^2-2kx-4kx-2x+4x-1=0\end{equation}$

$\begin{equation*}(k+1)^2x^2-(6k-2)x-1=0\end{equation}$

For there to be two numerically equal but opposite sign solutions, the $b$ term of the quadratic equation must be $0$ .

$\begin{equation*}6k-2=0\end{equation}$

Hence $k=\frac{1}{3}$ .

When $k=\frac{1}{3}$ the equation becomes

$\begin{equation*}\frac{x^2-2x}{4x-1}=\frac{\frac{-2}{3}}{\frac{4}{3}}\end{equation}$

$\begin{equation*}\frac{x^2-2x}{4x-1}=\frac{-1}{2}\end{equation}$

$\begin{equation*}2x^2-4x=-4x+1\end{equation}$

$\begin{equation*}2x^2-1=0\end{equation}$

$\begin{equation*}x^2=\frac{1}{2}\end{equation}$

$\begin{equation*}x=\pm \frac{1}{\sqrt{2}}\end{equation}$

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Filed under Algebra, Polynomials, Quadratic, Quadratics, Solving, Solving, Solving Equations

Tagged as hard algebra question, quadratic, solving quadratic equations

Geometry Problem

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

Trigonometric Equation

Polynomial Long Division

Geometry Circle Question

Completing the Square

Example 1

Example 2

Equation of a Circle and Geometry Question

Integration Question (much easier with Integration by Parts)

Integration by Parts

Fundamental Theorem of Calculus

Example

More formally

Hard Equation Solving Question

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