Monthly Archives: December 2025

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

$\Delta{ABC}=sr$

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