Tag Archives: 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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