Electronis Discussion

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A circle with centre $\text{O}$ is shown in the figure. A rectangle $\text{PQRS}$ of maximum possible area is inscribed in the circle. If the radius of the circle is $a$, then the area of the shaded portion

is _______.

- $\pi a^{2}-a^{2}$
- $\pi a^{2}-\sqrt{2}a^{2}$
- $\pi a^{2}-2a^{2}$
- $\pi a^{2}-3a^{2}$

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Theorem: “A perpendicular from the Centre of a circle to a chord, bisects the chord.” . (Refer: Proof)

In this case, if a perpendicular is drawn from the center of circle, it will bisect the base side of the rectangle.

The half-length of this chord would be: $a\times cos 45^{\circ}=\frac{a}{\sqrt{2}}$.

Hence the length of the rectangle would be $2\times \frac{a}{\sqrt{2}}=a\sqrt{2}$

Area of circle: $\pi$$a^{2}$ and the area of rectangle : $2a^{2}$

$\therefore$ Area of remaining portion would be: $\pi$$a^{2}-2a^{2}$

Option C is correct.

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