Question

A circle is inscribed in an equilateral triangle of side a. The area of any square inside the triangle is $\dfrac{{{a}^{2}}}{k}$. Find value of k.

Answer 1

Hint: Now we know that the radius of the in-circle of the equilateral triangle is given by $\dfrac{side}{2\sqrt{3}}$ . Hence we can also find the diameter of the in-circle. Now the diameter is nothing but the diagonal of the square. Hence using the Pythagoras theorem in square we can find the value of ${{\left( side \right)}^{2}}$ . Now the area of the square is also given by ${{\left( side \right)}^{2}}$ hence we have the area of the square.

Complete step by step answer:
Now we have that a circle is inscribed in an equilateral triangle and the square is inscribed in the triangle. Consider the figure.

Now since the circle with center O is inscribed in the triangle we can say that the circle is in-circle of the triangle ABC.
Now ABC is an equilateral triangle with side a.
Now we know that the radius of the in-circle for an equilateral triangle is given by $\dfrac{side}{2\sqrt{3}}$ .
Hence the radius of the circle with center O is given by $\dfrac{a}{2\sqrt{3}}$ .
Now we know that the diameter is twice the radius.
Hence the diameter of the circle with radius $\dfrac{a}{2\sqrt{3}}$ is $\dfrac{a}{\sqrt{3}}$ .
Now from the figure we know that PR is the diameter of the circle.
Hence we have $PR=\dfrac{a}{\sqrt{3}}...........................\left( 1 \right)$
Now we know that PQRS is a square.
Hence $\angle PQR={{90}^{\circ }}$. This means $\Delta PQR$ is a right angle triangle.
Hence by Pythagoras theorem $P{{Q}^{2}}+Q{{R}^{2}}=P{{R}^{2}}$
Now since PQRS is a square we have PQ = QR.
Hence $2P{{Q}^{2}}=P{{R}^{2}}$
Now substituting the value of PR we get, $2P{{Q}^{2}}={{\left( \dfrac{a}{\sqrt{3}} \right)}^{2}}$
Hence we get $P{{Q}^{2}}=\dfrac{a}{6}$ .
Now we know that the area of square is given by ${{\left( side \right)}^{2}}$
Hence the area of the square inscribed is $\dfrac{a}{6}$ .

So, the correct answer is “k = 6”.

Note: Now note that the circle inscribed inside the triangle is called in-circle and if the triangle is inscribed inside the circle then the circle is called circumcircle. Now the radius of circumcircle of equilateral triangle is given by $\dfrac{a}{\sqrt{3}}$ while the radius of in-circle is given by $\dfrac{a}{2\sqrt{3}}$ Hence do not make a mistake by taking the radius of the in-circle as $\dfrac{a}{\sqrt{3}}$ .