Question

The area of the smallest circle containing a square of side a is

• A. $\frac{\pi a}{\sqrt{2}}$
• B. $2\pi a^2$
• C. $\frac{\pi a^2}{2}$
• D. $\frac{\pi }{2}a$

Answer 1

The correct option is C $\frac{\pi a^2}{2}$

Let radius of circle $=r$

then Diameter of circle $=2r$

In square ABCD inscribed in a circle; $\angle ABC={90}^o$

According to Central Angle Theorem; $\angle AOC={180}^o$

It means AC is diameter of circle and it is also hypotenuse of right triangle ABC.

Using Pythagoras Theorem in right triangle ABC;

$A{C}^2=A{B}^2+B{C}^2$

$(2r{)}^2=a^2+a^2$

$4r^2=2a^2$

$r^2=\frac{a^2}{2}$

Area of circle $=\pi r^2=\pi \left(\frac{a^2}{2}\right)=\frac{{\pi a}^2}{2}$

The area of the smallest circle containing a square of side a is $\frac{{\pi a}^2}{2}$