In the given figure, AB=36 cm and M is mid point of . . .

Question : In the given figure, AB=36 cm and M is mid point of AB. Three semicircles are drawn on AB, AM and MB as diameters. A circle with centre C touches all the three circles. Find the area of the shaded region. (Take π=22/7)

Doubt by Saumya S.

Solution :

Radius of the Bigger Circle (R)
= AB/2 = 36/2 = 18 cm
Radius of two smaller circle (r) = AP = BP = 18/2 = 9 cm

Let Radius of the third circle with centre C be a cm

Join PC and QC and Draw CM ⊥ AB

Now, In Rt. ΔCMP

PC = (9+a) cm
PM = 9 cm
CM = R-a
CM = (18-a) cm

By Pythagoras Theorem

PC²=CM²+PM²
(9+a)²=(18-a)²+(9)²

81+a²+18a=324+a²-36a+81
81+18a-324+36a-81=0
54a=324
a=324/54
a=36/6
a=6 cm

Now, Required shaded area = Area of Bigger Semicircle of Radius R - 2×[Area of Semicircle of Radius r] - Area of Third Circle of radius r

=[πR²/2]-2[πr²/2]-πa²
=[πR²/2]-πr²-πa²
=π[R²/2-r²-a²]
=π[(18)²/2-(9)²-(6)²]
=π[324/2-81-36]
=π[162-117]
=π[45]
= (22/7)×45
=990/7
=141.428
≈ 141.43 cm²