Question

An equilateral triangle circumscribed, a square is inscribed in a circle of radius r. The area of triangle is T and the area of square is S, then \[\dfrac{T}{S}\] is
$
  (a){\text{ }}\dfrac{{3\sqrt 3 }}{2} \\
  (b){\text{ 3}}\sqrt 3 \\
  (c){\text{ 27}} \\
  (d){\text{ }}\dfrac{{9\sqrt 3 }}{2} \\
 $

Answer 1

Hint: Equilateral triangle means that all the sides will be equal and the angles in an equilateral triangle all are equal to ${60^0}$, use this concept along with concept that the length of diagonal of the square will be equal to the diameter of the circle to get the answer.

Complete Step-by-Step solution:

Consider the equilateral triangle ABC with side length a unit.
As we know that all the angles in the equilateral triangle are equal to 60 degrees.
According to given information equilateral triangle is circumscribed as shown in figure and a square is inscribed in the circle as shown in figure.
So the diagonal of the square is passing through the center of the circle so the length of the diagonal is the diameter of the circle.
And it is given that the circle has radius r so the diameter of the circle is twice the radius = 2r.
Let b be the side length of the square.
So apply Pythagoras theorem which is ${\text{hypotenou}}{{\text{s}}^2} = perpendicula{r^2} + {\text{bas}}{{\text{e}}^2}$in triangle DEF we have
$ \Rightarrow {\left( {2r} \right)^2} = {b^2} + {b^2}$
$ \Rightarrow 2{b^2} = 4{r^2}$
$ \Rightarrow {b^2} = 2{r^2}$
Now as we know area (S) of square is square of side.
$ \Rightarrow S = {b^2} = 2{r^2}$ Sq. unit.
Now OG is the perpendicular on the side BC. Therefore it bisect the side BC,
$ \Rightarrow GC = \dfrac{{BC}}{2} = \dfrac{a}{2}$ Unit.
And OG is the radius of the circle therefore OG = r unit
And the angle OCG is half of the angle ACG.
$ \Rightarrow \angle OCG = \dfrac{{\angle ACG}}{2} = \dfrac{{{{60}^0}}}{2} = {30^0}$
So in triangle OCG
$ \Rightarrow \tan {30^0} = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}} = \dfrac{{OG}}{{GC}} = \dfrac{r}{{\dfrac{a}{2}}} = \dfrac{{2r}}{a}$
$ \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{2r}}{a}$
$ \Rightarrow a = 2r\sqrt 3 $ Unit.
Now as we know that the area (T) of equilateral triangle is $\dfrac{{\sqrt 3 }}{4}{a^2}$ where a is the side length of the equilateral triangle.
And we calculate the side length of the triangle.
$ \Rightarrow T = \dfrac{{\sqrt 3 }}{4}{\left( {2r\sqrt 3 } \right)^2} = \dfrac{{\sqrt 3 }}{4}\left( {4{r^2}\left( 3 \right)} \right) = 3\sqrt 3 {r^2}$ Sq. unit.
So the ratio of the area of equilateral triangle to the square is
$ \Rightarrow \dfrac{T}{S} = \dfrac{{3\sqrt 3 {r^2}}}{{2{r^2}}} = \dfrac{{3\sqrt 3 }}{2}$
So this is the required ratio of the area of the equilateral triangle to the square.
Hence option (A) is correct.

Note: Whenever we face such types of problems the key point is to understand the diagrammatic representation for the information provided in the question as it helps in understanding the geometry better. Use of basic properties of equilateral triangles helps understanding the angles and eventually the triangles involved which can contribute in finding the required quantity.