Question

If a square is inscribed in a quadrant of a circle of radius $4\text{cm}$ such that their vertices coincide, then the area of the quadrant covered by the square is

• A. $16\sqrt{2}\text{sq.cm}$
• B. $16\text{sq.cm}$
• C. $8\text{sq.cm}$
• D. $4\text{sq.cm}$

Answer 1

The correct option is C $8\text{sq.cm}$

Let the side of the square $ODCE$ inscribed in quadrant $AOB$ be $a\text{cm}$.
As the diagonal of the square is equal to the radius of the circle.
$\Rightarrow OC=4\text{cm}$
A square of side $a\text{cm}$ will have a diagonal of length $\sqrt{2}a\text{cm}.$

Thus, the square covers an area of 8 cm2 Hence, the correct answer is option (3).
$\therefore \sqrt{2}a=4$
$\Rightarrow 2a^2=16$
$\Rightarrow a^2=8$
$\Rightarrow a=\sqrt{8}$
$\Rightarrow a=2\sqrt{2}\text{cm}$
$\therefore$ Area of square $=a^2=8{\text{cm}}^2$
Thus, the square covers an area of $8{\text{cm}}^2$

Hence, the correct answer is option c.