What is the ratio of the area of the circumscribed circle to the area of the inscribed circle of an equilateral triangle?

Problem: Given an equilateral triangle. What is the ratio of the area of a circle inscribed to the area of a circle circumscribed about the triangle?

Attempt: From what I understood, we have a circle inscribed in an equilateral triangle, and that triangle is inscribed to a circle. So we have two circles, big circle and small circle.

I am asked to find the ratio of the area of the small circle to the big circle. Let "s" be the area of the small circle and "b" for the big circle.

Ratio = Area of small circle / Area of big circle = $\frac{s}{b}$

The radius of the big circle is twice the radius of the small circle, so $R = 2r.$

Substituting the radius, Ratio = $\frac{\pi*r^2}{\pi*R^2}$ = $\frac{\pi*r^2}{\pi*(2r)^2}$ = $\frac{1}{4}$ = $1:4$

Question: Looking at the answer keys, the answer is supposed to be $\frac{1}{2}$ and not $\frac{1}{4}$. Any idea where did I got wrong or am I totally wrong? How to get the $\frac{1}{2}$. Thank you.

Answer

Verified

Hint: We here have been given an equilateral triangle and we have to find the ratio of the area of its circumcircle and incircle. For this, we will first mention that all the centres of an equilateral triangle lie at the same point. Then we will draw its figure and mark the both the radii of the triangle. Then we will use the property of the centroid, i.e. the centroid divides the median into a ratio of 2:1. Thus, this will be the ratio of the circumradius and the inradius. Then we will use the formula for the area of the circles which will be equal to the square of the ratio of the radii. Thus, we will get our answer.

Complete step by step answer:

We have now been given an equilateral triangle with its circumcircle and incircle and we have been asked the ratio of the areas of those circles. Now, we know that the circumcentre and the incentre of an equilateral triangle lie at the same point.Let us assume there to be a $\Delta ABC$ with its circumcircle and incircle at O (as mentioned above, they both will lie on the same point as the triangle is equilateral). If we draw a diagram of it, we’ll get the following figure:

Now, we have assumed that the inradius of the triangle is ‘r’ and the circumradius is ‘R’. We also know that in an equilateral triangle, the centroid of the triangle also lies on the same point as all the other centres. This means that the centroid, the circumcentre and the incentre lies on the same point in the triangle, i.e. O. Now, since AD is a straight line from A to line BC passing through point O, we can say that AD is the median from A to the side BC of the $\Delta ABC$. Now, from the figure, we can see that:$AD=AO+OD$ .....(i)Now, we know that the centroid divides the median in the ratio 2:1. Thus, the point O divides the line AD in the ratio 2:1. Thus, from equation (i) we can say that:$\begin{align} & AO:OD=2:1 \\ & \Rightarrow \dfrac{AO}{OD}=\dfrac{2}{1} \\ \end{align}$ Now, from the figure, we can see that:$\begin{align} & AO=R \\ & OD=r \\ \end{align}$ Thus, we can say that:$\dfrac{R}{r}=\dfrac{2}{1}$ Now, we know that the area of a circle with radius ‘a’ is given as $\pi {{a}^{2}}$. Thus, the area of the circumcentre and incircle is given as:$\begin{align} & ar\left( circumcircle \right)=\pi {{R}^{2}} \\ & ar\left( incircle \right)=\pi {{r}^{2}} \\ \end{align}$ Thus, their ratio is given as:$\dfrac{ar\left( circumcircle \right)}{ar\left( incircle \right)}=\dfrac{\pi {{R}^{2}}}{\pi {{r}^{2}}}$ Solving this, we’ll get: \[\begin{align} & \dfrac{ar\left( circumcircle \right)}{ar\left( incircle \right)}=\dfrac{{{R}^{2}}}{{{r}^{2}}} \\ & \Rightarrow \dfrac{ar\left( circumcircle \right)}{ar\left( incircle \right)}={{\left( \dfrac{R}{r} \right)}^{2}} \\ \end{align}\] Now, we know that $\dfrac{R}{r}=\dfrac{2}{1}$ Thus, we get the required ratio as:$\begin{align} & \dfrac{ar\left( circumcircle \right)}{ar\left( incircle \right)}={{\left( \dfrac{R}{r} \right)}^{2}} \\ &\Rightarrow \dfrac{ar\left( circumcircle \right)}{ar\left( incircle \right)}={{\left( \dfrac{2}{1} \right)}^{2}} \\ &\Rightarrow \dfrac{ar\left( circumcircle \right)}{ar\left( incircle \right)}=\dfrac{4}{1} \\ \end{align}$ Thus the required ratio is 4:1.

So, the correct answer is “Option A”.

Note: Questions related to equilateral triangles are always simple as all the centres of the triangle lie on the same point. Not only that, every median is also an altitude, angle bisector and the perpendicular bisector of the opposite side. This is a very important property which also makes the questions related to equilateral triangles easier.