Question

in the fig.find the area if minor segment AQBP...
plzz..help me

Answer 1

$area = \frac{ \alpha }{360} \times \pi {r}^{2} - \frac{1}{2} {r}^{2} \sin( \alpha )$

$area = \frac{1}{4} \times100 \pi - \frac{1}{2} 100 \times \sin(90)$
$area = 25\pi - 50$
$area = 25(\pi - 2)$
$area = 25(3.14 - 2)$
$area = 25 \times 1.14$
$area = 28.5 {cm}^{2}$
HOPE IT HELPS!!
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Answer 2

Answer:

28.5 cm²

Step-by-step explanation:

Given, radius r = 10 cm and θ = 90°{right-angle}

(i) Area of OAPB:

= θ/360 * πr²

= 90/360 * (3.14) * (10)²

= (1/4) * 314

= 78.5 cm²

(ii) Area of ΔAOB:

= (1/2) * OA * OB

= (1/2) * 10 * 10

= 50 cm²

∴ Area of the minor segment AQBP:

Area of sector OAPB - Area of ΔAOB

= 78.5 - 50

= 28.5 cm²

Hope it helps!