Question

A rectangle ABCD with diagonal 14 cm is inscribed in a circle with centre O as shown in the given figure. If the area of the shaded portion is expressed as `a + bsqrt(3)`, find the values of a and b. Also, find the perimeter of the sector OАВО.

Answer 1

Given:

Diagonal of rectangle AC = BD = 14 cm

Since the rectangle is inscribed in the circle, the diagonal is the diameter.

Radius of the circle (r) = `14/2` = 7 cm

Angle ∠BOC = 60°

Finding area of shaded portion:

The shaded portion consists of two segments formed by chords AD and BC.

In ΔOBC, OB = OC = r = 7 cm.

Since ∠BOC = 60°, ΔOBC is an equilateral triangle.

Area of shaded region = 2 × (Area of sector OBC – Area of △OBC)

Area = `2 xx [60/360 xx π xx 7^2 - sqrt(3)/4 xx 7^2]`

Area = `2 xx [1/6 xx 22/7 xx 49 - (49sqrt(3))/4]`

Area = `2 xx [77/3 - (49sqrt(3))/4]`

Area = `154/3 - (49sqrt(3))/2`

Comparing with `a + bsqrt(3)`:

`a = 154/3`

`b = -49/2`

Perimeter of sector OABO:

Angle ∠AOB = 180° – 60° = 120°   ...(Linear pair)

Length of arc AB = `θ/360 xx 2πr`

Length of arc AB = `120/360 xx 2 xx 22/7 xx 7`

Length of arc AB = `1/3 xx 44 = 44/3` cm

Perimeter of sector OABO = OA + OB + arc AB

Perimeter = `7 + 7 + 44/3`

Perimeter = `14 + 44/3`

Perimeter = `(42 + 44)/3` = `86/3` cm