Question

$\huge{\underline{\underline{\bf Question}}}$

ABDC is a quadrant of a circle of radius 28 cm and a semi-circle BEC is drawn with BC as diameter. Find the area of the shaded region.

Class - 10th

chapter - Area related to circles
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STEP BY STEP EXPLANATION ✅

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Answer 1

Answer:

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Step-by-step explanation:

Radius of the quadrant ABC of circle = 28 cm

AB = AC = 28 cm

BC is diameter of semicircle.

ABC is right angled triangle.

By Pythagoras theorem in ΔABC,

BC² = AB² +AC²

⇒ BC² = 28² +28²

⇒ BC = 28√2 cm

Radius of semicircle = 28√2/2 cm = 14√2 cm

Area of ΔABC = ½×AB×AC = ½×28×28 = 392 cm²

Area of quadrant = ¼πr² = (¼)×(22/7)×(28×288) = 616 cm²

Area of the semicircle =½πr² = (½)×(22/7)×14√2×14√2 = 616 cm²

Area of the shaded region =Area of the semicircle + Area of ΔABC – Area of quadrant

= 616 +392- 616 cm²

= 392cm²

Answer 2

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➵In △ BAC, by pythagoras theorem,

BC 2 =AB 2 +AC 2

➵BC 2 =784+784 BC=28 2 cm2

➵BC =142 cm

➵Now,

➵A=ar(BDCEB)

➵A=ar(BCEB)−ar(BCDB)

➵A=ar(BCEB)−[ar(BACDB)−ar(BAC)]

➵A=[ 21( 722×(14 2 ) 2 )−( 41×722×28 2− 21

➵[(×28×28)]A=[ 21× 722 ×196×2-41×722

➵[28×28+21×28×28]A=616−616+392=392 cm 2