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Q.

A right angled isosceles triangle is inscribed in a circle of radius $r$ . What is the area of the remaining portion of the circle?

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a

\frac{π r^{2}}{2}

b

(π - \frac{1}{2}) r^{2}

c

(π - 1) r^{2}

d

(π - 2) r^{2}

answer is C.

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Detailed Solution

Given, a right angled isosceles triangle is inscribed in a circle of radius

r

.

Question Image

Let r be the radius of the circle and a be the length of the sides that contain the right angle. The diameter of the circle is the hypotenuse of a right-angled triangle.
Pythagoras theorem states that the square of the hypotenuse side in a right-angled triangle equals the sum of the squares of the other two sides.
So, by applying pythagoras theorem,

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

.

\Rightarrow 2 a^{2} = 4 r^{2}

\Rightarrow a = \sqrt{2} r

∴

Remaining area = Area of circle

-

Area of triangle.
We know that area of a circle is

π r^{2}

and area of the triangle=

\frac{1}{2} (Base) (Height)

.

\Rightarrow

Remaining area

= π r^{2} - \frac{1}{2} (r \sqrt{2}) (r \sqrt{2})

.

\Rightarrow

Remaining area

= π r^{2} - r^{2}

\Rightarrow

Remaining area

= r^{2} (π - 1)

Hence, option 3 is correct.

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