# 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
• 17
gggggggggg
• -5
please give solution in proper steps
• 2
• -5
please reply expert I have exams and my teacher told this question is going to com pls pls pls
• -3
Pls answer in proper steps. Urgently required
• -11
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
• 7
EXPERTS GIVE THE SOLUTION FAST PLEASE
• -6
thanks kanishka..

• 0
THUMBS up
• -6
• 2
Check it out

• -14
Here's the proper solution...

• 55

• 13
Ans= 96.2cm^2
• -6
diagram
• 0
thanks buddy
• 2
• 9
I thick short method to solve this question is given by me . Hope this helps you. All above are wrong according to me.

• 7
Hope this helps you.

• -11
All the answers are wrong Check oswaal guide pg279 Q10
• 2
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.

• 28
69.25

• -4

• 8
45
• 1

• 4

• 0
I think my answer will be very useful😊😊

• 4