Question

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Answer 1

$\large\bf\underline{Question:-}$

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

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$\large\bf\underline{Given:-}$

• OP = 28cm

$\large\bf\underline {To \: find:-}$

• Area of shaded region

$\huge\bf\underline{Solution:-}$

OP = 28cm which is the Radius of the given circle.

We know that,

• Area of circle = πr²

➝ Area of circle = 22/7×(28)²

➝ Area of circle = 22/7×784

➝ Area of circle = 22×112

➝ Area of circle = 2464 cm²

Circle contains a rhombus of diagonals PQ and RS

➝ PQ = PO + QO

➝ PQ = 28 + 28

➝ PQ = 56

➝ PQ = RS (diameter of circle)

• Area of rhombus = 1/2(diagonal1× diagonal 2)

➝ Area of rhombus = 1/2(PQ × RS)

➝ Area of rhombus = 1/2(56×56)

➝ Area of rhombus = 1/2(3136)

➝ Area of rhombus = 1568sq.units

Area of shaded region = Area of circle - area of rhombus /2

➝ Area of shaded region = 2464-1568/2

➝ Area of shaded region = 896/2

➝ Area of shaded region = 448cm²

Hence , Area of shaded region = 448cm²

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Answer 2

$\sf\blue{Question}$

$\sf{In \ the \ figure \ PQ \ and \ RS \ are \ the \ two}$

$\sf{perpendicular \ diameters \ of \ the \ circle \ with}$

$\sf{centre \ 'O'. \ If \ OP=28 \ cm, \ find \ the \ area}$

$\sf{of \ the \ shaded \ region.}$

__________________________________

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{Area \ of \ shaded \ region \ is \ 448 \ cm^{2}}$

$\sf\orange{Given:}$

$\sf{In \ circle \ with \ center \ O,}$

$\sf{\implies{PQ \ and \ RS \ are \ perpendicular \ diameters.}}$

$\sf{\implies{OP=28 \ cm}}$

$\sf\pink{To \ find:}$

$\sf{The \ area \ of \ the \ shaded \ region.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Radius(OP)=28 \ cm...given}$

$\boxed{\sf{Area \ of \ semicircle=\frac{\pi\times \ r^{2}}{2}}}$

$\sf{\therefore{Area \ of \ semicircle=\frac{\frac{22}{7}\times28\times28}{2}}}$

$\sf{Area \ of \ semicircle=22\times2\times28}$

$\sf{Area \ of \ semicircle=1232 \ cm^{2}}$

$\sf{\therefore{Area \ of \ sector \ RPS=1232 \ cm^{2}}}$

$\sf{In \ \triangle \ RPS,}$

$\sf{RS=2\times28=56 \ cm...RS \ is \ diameter}$

$\sf{\angle \ POR=90°...RS \ and \ PQ \ are \ perpendicular}$

$\sf{Area \ of \ \triangle \ RPS=\frac{1}{2}\times \ Base\times \ Height}$

$\sf{A(\triangle \ RPS)=\frac{1}{2}\times \ RS\times \ OP}$

$\sf{A(\triangle \ RPS)=\frac{1}{2}\times56\times28}$

$\sf{A(\triangle \ RPS)=28\times28}$

$\sf{\therefore{A(\triangle \ RPS)=784 \ cm^{2}}}$

$\sf{Area \ of \ shaded \ region}$

$\sf{=Area \ of \ semicircle \ - \ A(\triangle \ RPS)}$

$\sf{=1232-784}$

$\sf{=448 \ cm^{2}}$

$\sf\purple{\tt{\therefore{Area \ of \ shaded \ region \ is \ 448 \ cm^{2}}}}$