Question

An equilateral triangle is inscribed in a circle,which is inscribed in a square.What is the ratio of the areas of the triangle and the square?

Answer 1

Let r be radius of circle
then,it's radius of equilateral triangle.
Hence,r=a/(2×sq.root(3)
see equilateral triangle,,
diagonal of square side=Diameter of circle
=2×r
=a/(root(3)
area of sq.=0.5×d×d

hence,area of sq.=(1/2)×(a^2)/3
required area is (a^2/6)sq.unit

Answer 2

Answer:

3√3 : 16

Step-by-step explanation:

Since, if a triangle having sides a, b and c is inscribed in a circle,

Then the radius of the circle is,

$R=\frac{a\times b\times c}{4\times A}$

Where,

A = area of the triangle,

Suppose an equilateral triangle having side a unit is inscribed in a circle,

∵ Area of the triangle, A = $\frac{\sqrt{3}}{4}\times (side)^2$

$=\frac{\sqrt{3}}{4}a^2$

Hence, the radius of the circle,

$R=\frac{a^3}{4\times \frac{\sqrt{3}}{4}a^2}=\frac{a}{\sqrt{3}}\text{ unit}$

Now, if the circle is inscribed in a square,

Then the side of the square = diameter of the circle = 2R = $\frac{2a}{\sqrt{3}}\text{ unit}$

Finally,

The ratio of the areas of the triangle and the square,

$\frac{\frac{\sqrt{3}}{4}a^2}{\frac{2a}{\sqrt{3}}\times \frac{2a}{\sqrt{3}}}$

$\frac{\frac{\sqrt{3}}{4}a^2}{\frac{4a^2}{3}}$

$\frac{\frac{\sqrt{3}}{4}}{\frac{4}{3}}$

$\frac{3\sqrt{3}}{16}$