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Question

In a right-angled triangle, the lengths of the sides containing the right angle are a and b. With the midpoint of each side as centre, three semicircular areas are drawn outside the triangle. The total area enclosed is

A
π2(a2+b2)+(a+b)2
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B
π2(a2+b2)+12ab
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C
π2(a2+b2)+18(a+b)2
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D
π4(a2+b2)+ab
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Solution

The correct option is B π2(a2+b2)+12ab
sol:
(I) Area of semicircle (^ODA)D=OA=a
=OA2=a2 [radius]
=πr22=π2(a2)2=πa28
(II) Area of semicricle(^OEB)=πr22=π2(b2)2
(III) Area of semicircle (^OEA)=π2(a2+b2)
=π(a2+b2)2
(IV) Area of triangle (ΔOAB)=12×a×b
=ab2
Adding all areas=I+II+III+IV
=πa28+πb28+π(a2+b2)2+ab2
=π4(a2+b2+a2+b2)+ab2.

1129369_97815_ans_e6bc9537e48c42b6932ac8a66df3197d.jpg

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