Tag Archives: tangents to circles

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?

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