THE PERIMETER OF A SECTOR OF A CIRCLE WITH CENTRAL ANGLE 90* IS 25CM FIND THE AREA OF THE MINOR - Maths - Areas Related to Circles

Question

THE PERIMETER OF A SECTOR OF A CIRCLE WITH CENTRAL ANGLE
90* IS 25CM FIND THE AREA OF THE MINOR SEGMENT OF THE
CIRCLE

Varun.Rawat · Accepted Answer

Let r be the radius of a sector of a circle with center at O.

Length of the arc ACB = l = θ / 360⁰× 2 π r = 90⁰/ 360⁰× 2 × 22 / 7 × r = 11r / 7 cm

perimeter of the sector AOBCA = 25 cm

r + r + l = 25 ⇒ 2r + 11r / 7 = 25 ⇒ r = 7 cm

area of the minor segment ABCA = Area of the sector AOBCA - area of Δ AOB

$= \frac{π r^{2}}{4} - \frac{1}{2} \times O B \times O A = \frac{22 \times 49}{7 \times 4} - \frac{1}{2} \times 7 \times 7 = \frac{77}{2} - \frac{49}{2} = \frac{28}{2} = 14 {c m}^{2}$

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THE PERIMETER OF A SECTOR OF A CIRCLE WITH CENTRAL ANGLE 90* IS 25CM FIND THE AREA OF THE MINOR SEGMENT OF THE CIRCLE.

**THE PERIMETER OF A SECTOR OF A CIRCLE WITH CENTRAL ANGLE 90* IS 25CM FIND THE AREA OF THE MINOR SEGMENT OF THE CIRCLE.**