APB & AQO are semicircles and AO = OB.if the perimeter of the figure is 40cm,find the area of the shaded region

we assume (π=22/7) Let OA=OB=r 40=(π • r/2) + (π • r+r) 280=40r r=7 shaded area=(1/2 • π • 7/2 • 7/2 + 1/2 • π • 7 • 7)cm² (77 • 5/4)cm² 385/4 cm²
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gggggggggg
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please give solution in proper steps
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urgent expert please help
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please reply expert I have exams and my teacher told this question is going to com pls pls pls
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Pls answer in proper steps. Urgently required
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Let OA=OB =r.
Therefore, 40= 22/7* r/2 + 22/7*r + r
280= 40r
r=7
So, shaded area= (1/2*22/7*7/2*7/2 + 1/2*22/7*7*7)
= 77/4 +77
= 77 (1/4+1)
= 77* 5/4
= 385/4 cm^2
= 96.25 cm^2
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EXPERTS GIVE THE SOLUTION FAST PLEASE
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thanks kanishka..
 
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THUMBS up
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Is this correct. Expert please give the answer
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Check it out

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Here's the proper solution...

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Hope this answer would help.

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Ans= 96.2cm^2
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diagram
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thanks buddy
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Why we add only r ....I think 2r shd be added...plz reply fast
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I thick short method to solve this question is given by me . Hope this helps you. All above are wrong according to me.

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Hope this helps you.

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All the answers are wrong Check oswaal guide pg279 Q10
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Answer is 96.25 cm^2 (or 96 1/4 cm^2), verified from Official CBSE Marking Scheme, so rest assured there is no chance of mistake. Also, this question came in 2015 in Delhi Set 1.

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69.25
 
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Answer

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45
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Answer is 96.25cm^2

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This is the answer

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I think my answer will be very useful😊😊

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Write comments

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Let OA=OB =r. Therefore, 40= 22/7* r/2 + 22/7*r + r 280= 40r r=7 So, shaded area= (1/2*22/7*7/2*7/2 + 1/2*22/7*7*7) = 77/4 +77 = 77 (1/4+1) = 77* 5/4 = 385/4 cm^2 = 96.25 cm^2
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This is the answer

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