Question

In fig. 5 is a chord AB of a circle, with centre O and radius 10 cm, that subtends a right angle at the centre of the circle. Find the area of the minor segment AQBP. Hence find the area of major segment ALBQA. (use π = 3.14)

Answer 1

Given: Radius of the circle, r = 10 cm

Area of the circle  = \[\pi \left( 10 \right)^2 = 3 . 14 \times 100 = 314 {cm}^2\]

Area of the sector OAPB 

\[= \frac{90^o}{360^o} \times \pi \left( 10 \right)^2 \]
\[ = \frac{1}{4} \times 314 \left[ From \left( 1 \right) \right]\]
\[ = 78 . 5 {cm}^2\]

Area of ∆BOA

\[= \frac{1}{2} \times BO \times OA\]
\[ = \frac{1}{2} \times 10 \times 10\]
\[ = 50 {cm}^2\]

Area of the minor segment AQBP = Area of the sector OAPB − Area of the triangle BOA 
                                                = (78.5 − 50) cm² 
                                                = 28.5 cm²

^{Area of the major segment ALBQ = Area of the circle − Area of the minor segment AQBP
                                                      = (314 − 28.5) cm2
                                                      = 285.5 cm2}