Question

drea on the shaded region.
OR
In the figure PQ and RS are the two perpendicular diameters of the circle
with centre 'O. If OP = 28 cm, find the area of the shaded region.​

Answer 1

Answer:

fig. is in the attachment]

Given: 

Radius of larger circle,(OA) R = 7 cm

Diameter of smaller circle,(OD) = 7 cm

Radius of smaller  circle = 7/2 cm

Height of ΔBCA( OC) = 7 cm

Base of ΔBCA ( AB )= 14 cm

Area of ΔBCA = 1/2 × base × height

Area of ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm²

Area of larger semicircle with radius (OA)7 cm = 1/2 ×π ×7²= 1/2 ×22/7 ×7×7 = 77 cm²

Area of smaller circle with radius 7/2 cm = πr²= 22/7 × 7/2 × 7/2 = 77/2 cm² 

Area of the shaded region =Area of smaller circle with radius 7/2 cm +Area of larger semicircle with radius 7 cm - Area of ΔBCA 

Area of the shaded region = 77/2 + 77 - 49= 77/2 + 28 = (77 +56) /2= 133/2

Area of the shaded region = 66.5 cm²

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Hope this will help you.........

Answer 2

$\underline\mathfrak{Given:-}$

• $\: \: \: \: \: PQ \: \: and \: \: RS \: \: are \: \: perpendicular \: \: diameter$

• $\: \: \: \: \: OP \: \: = \: \: {28cm}$

$\underline\mathfrak{To \: \: Find:-}$

• $\: \: \: \: \: The \: \: area \: \: of \: the \: \: shaded \: \: region.?$

$\underline\mathfrak{Solutions:-}$

• $\: \: \: \: \: \: \: The \: Ratio \: \: of \: \: circle \: \: is \: \: {28cm}$

$\: \: \: \: \: \: \: \underline{Area \: \: of \: \: circle \: \: = \: \: \pi \: {r}^{2}}$

$\: \: \: \: \: \: \: \: \: \leadsto \: \: \dfrac{22}{7} \: \times \: {28}^{2}$

$\: \: \: \: \: \: \: \: \: \leadsto \: \: \dfrac{22}{7} \: \times \: {784}$

$\: \: \: \: \: \: \: \: \: \leadsto \: \: {22} \: \times \: {112}$

$\: \: \: \: \: \: \: \: \: \leadsto \: \: {2464} \: {cm}^{2}$

• $\: \: \: \: \: \underline{A \: \: rhombus \: \: of \: \: diagonal \: \: PQ \: \: and \: \: RS}$

$\: \: \: \: \: \leadsto {PQ} \: = \: \: PO \: + \: QO$

$\: \: \: \: \: \leadsto {PQ} \: = \: \: {28} \: + \: {28}$

$\: \: \: \: \: \leadsto {PQ} \: = \: \: {56}$

$\: \: \: \: \: \leadsto {PQ} \: = \: \: {RS} \: \: \: \: \: \: \: \: \: {(Diagonal \: \: of \: \: circle)}$

• $\: \: \: \: \: \underline{Area \: \: of \: \: rhombus \: \: = \: \: \dfrac{1}{2} \: {(Diagonal \: {1} \: \: \times \: \: Diagonal \: {2})}}$

$\: \: \: \: \: \leadsto \: \: \dfrac{1}{2} \: {({PQ} \: \: \times \: \: {RS})}$

$\: \: \: \: \: \leadsto \: \: \dfrac{1}{2} \: {({56} \: \: \times \: \: {56})}$

$\: \: \: \: \: \leadsto \: \: \dfrac{1}{2} \: {({3136})}$

$\: \: \: \: \: \leadsto \: \: {1568} \: sq. \: unit.$

• $\: \: \: \: \: \underline{The \: \: area \: \: of \: the \: \: shaded \: \: region \: \: = \: \: Area \: \: of \: \: circle - \: \: \dfrac{Area \: \: of \: \: rhombus}{2}}$

$\: \: \: \: \: \leadsto \: \: {2464} \: \: - \: \: \dfrac{1568}{2}$

$\: \: \: \: \: \leadsto \: \: \dfrac{896}{2}$

$\: \: \: \: \: \leadsto \: \: {448} \: {cm}^{2}$

$\: \: \: \: \: Hence,$

$\: \: \: \: \: The \: \: area \: \: of \: the \: \: shaded \: \: region \: \: is \: \: {448} \: {cm}^{2}$

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