Question

A circle is inscribed in an equilateral triangle and a square is inscribed in the circle. The ratio of the triangle to the area of the square is
A. $\sqrt 3 \,:\,\sqrt 2 $
B. $\sqrt 3 \,:\,1$
C. $3\sqrt 3 \,:\,2$
D. $3\sqrt 2 $

Answer 1

Hint:To solve this problem we should know what a circle inscribed in an equilateral triangle means. It means that a circle is made inside the triangle and circle circumference is touched by the circumference of the triangle.
Area of the equilateral triangle having side dimension $a$ is $ = \dfrac{{\sqrt 3 {a^2}}}{4}$ .
Area of triangle circumscribed a circle of radius $r$ is $ = r\dfrac{{a + b + c}}{2}$

Complete step by step answer:
As, we know that the area of a triangle can be written as $ = r\dfrac{{a + b + c}}{2}$ ,
Here, $a,b$ and $c$ are three sides and $r$ is the circum-radius of the triangle.
As given in question,

For an equilateral triangle, the area of the triangle is $r\dfrac{{3a}}{2}$ .
Also, the area of an equilateral triangle $ = \dfrac{{\sqrt 3 {a^2}}}{4}$ .
Equating the above two equation, we get
 $r\dfrac{{3a}}{2} = \dfrac{{\sqrt 3 {a^2}}}{4}$
By solving it,
 $r = \dfrac{a}{{2\sqrt 3 }}$
So, the in-radius $r = \dfrac{a}{{2\sqrt 3 }}$
Hence, the diameter of the inscribed circle is $2r = 2 \times \dfrac{a}{{2\sqrt 3 }} = \dfrac{a}{{\sqrt 3 }}$ .
The diameter of the inscribed circle is equal to the square diagonal.
Let, each side of the square be $x$ .
As we know, the diagonal of the square will be $\sqrt 2 x$ .
Equating the diameter of the circle with the diagonal of the square we get,
 $\sqrt 2 x = \left( {\dfrac{a}{{\sqrt 3 }}} \right)$
By solving it. We get,
 \[x = \dfrac{a}{{\sqrt 3 .\sqrt 2 }} = \dfrac{a}{{\sqrt 6 }}\]
Side of square will be \[x = \dfrac{a}{{\sqrt 6 }}\] .
Therefore, the area of the square $ = {x^2} = {\left( {\dfrac{a}{{\sqrt 6 }}} \right)^2} = \dfrac{{{a^2}}}{6}$
Then the ratio of area of triangle to area of square $ = \dfrac{{area\,of\,triangle}}{{area\,of\,square}}$
 $\dfrac{{\sqrt 3 {a^2}}}{4}:\dfrac{{{a^2}}}{6} = 3\sqrt 3 :2$
 Hence, the ratio of the triangle to the area of the square is $3\sqrt 3 :2$ .
So, the correct answer is Option C.

Note: In basic mathematics the area of triangle is $base \times height$ . This is applied to any form of triangle. The triangle having all sides of equal length is called an equilateral triangle. The polygon having four sides and all side are equal and parallel to the opposite side is called square.