Area | Racquel Sanderson

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October 13, 2024 · 4:59 am

Tricky Area Ratio Question

A circle of radius $r$ , is inscribed within an isosceles triangle $ABC$ . $CA=CB=5r$ . Given that $\angle{ACB}$ is acute, find the ratio of the area of the circle to that of triangle $ABC$ .

Mathematics Specialist 3AB Question15 page 55

I came across this question while searching for an area of sectors and segments question.

Here’s a diagram

We know $AC, AB and BC$ are tangents to the circle. Because the triangle is isosceles, the distance from $A$ to the circle is the same as the distance from $B$ to the circle.

$CP$ is perpendicular to $AB$ (because the triangle is isosceles).

Because it is proportional, i.e. $r$ and $5r$ , we can let $r=1$

Let $h=CP$

(1) $\begin{equation*}h=\sqrt{25-a^2}\end{equation*}$

but $h$ also equals

(2) $\begin{equation*}h=1+\sqrt{(5-a)^2+1}\end{equation*}$

Set equation $1$ equal to equation $2$

$\begin{equation*}\sqrt{25-a^2}={1+\sqrt{(5-a)^2+1}\end{equation}$

Square both sides

$\begin{equation*}25-a^2=1+(5-a)^2+1+2\sqrt{(5-a)^2+1}\end{equation}$

Expand and simplify

$\begin{equation*}25-a^2=2+25-10a+a^2+2\sqrt{(5-a)^2+1}\end{equation}$

$\begin{equation*}0=2-10a+2a^2+2\sqrt{(5-a)^2+1}\end{equation}$

Divide by $2$

$\begin{equation*}0=1-5a+a^2+\sqrt{(5-a)^2+1}\end{equation}$

$\begin{equation*}-\sqrt{(5-a)^2+1}=a^2-5a+1\end{equation}$

Square both sides

$\begin{equation*}(5-a)^2+1=a^4-5a^3+a^2-5a^3+25a^2-5a+a^2-5a+1\end{equation}$

$\begin{equation*}25-10a+a^2+1=a^4-10a^3+27a^2-10a+1\end{equation}$

$\begin{equation*}0=a^4-10a^3+26a^2-25 \end{equation}$

I solved this using a graphics calculator

$a=-0.8434, a=1.3068, a=4.5367, a=5$

We can reject $a=-0.8434$ , $a=4.5367$ , and $a=5$ . If $a=5$ , there isn’t a triangle, and if $a=4.5367$ $\angle{ACB}$ is not acute.

Hence the area of triangle $ABC=a\times h$

$\begin{equation*}A_{\Delta}=1.3068\times\sqrt{25-1.3068^2}\end{equation}$

(3) $\begin{equation*}A_{\Delta}=6.3069\end{equation*}$

(4) $\begin{equation*}A_{\circ}=\pi\end{equation*}$

Hence, equation $4$ divided by equation $3$ is

$\begin{equation*}\frac{\pi}{6.3069}\approx 0.5\end{equation}$

Perhaps I approached this question in the wrong way. Is there an easier process?

Three Circles – Area Problem

This is question 5 from the UK Maths Trust Senior Challenge October 2023.

I have tackled this in three ways; using non-right trig to find the area, Heron’s Law, and the Shoelace Formula.

Method 1

Use the area of a triangle formula

Use the cosine rule to find cosθ.

Once we have cosθ, we can find sinθ.

Hence the area is,

Method 2

Use Heron’s law.

Heron’s law is a way of calculating area of a triangle from the lengths of the three sides of the triangle.

This is my preferred method – simple and direct.

Method 3

Shoelace formula (Gauss’s Area formula)

We need to allocate each of the vertices a co-ordinate.

The co-ordinates are listed in an anti-clockwise direction.

This is probably a bit over the top, but once you get the hang of it, it’s very easy.

Category Archives: Area

Tricky Area Ratio Question

Three Circles – Area Problem

Method 1

Method 2

Method 3

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