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}