CBSE Class 10 Answered
Name the point of the full sector where the angle is 90 degrees as O(say)
Name the point where the two semicircles meet as A.
Name the end point on the vertical line segment as B and on the horizontal line segment as C.
From A drop perpendiculars to both semicircles meeting them at M and N (say)
So AM=AN=3.5 cm (1/2 of the diameter)
Area of sector MOA=(90/360)*(22/7)*(3.5)*(3.5)
area triangle MOA=1/2*(3.5)*(3.5)
Subtract the area of the triangle from the area of the sector you wil get the area of half the shaded portion(for the simicircles)=3.5Double that will be the area of the shaded portion common to the semicircles=3.5*2=7 sq cm area of the full big sector=(90/360)(22/7)(7)(7)=77/2=38.5 sq cm
From this if we subtract the area of both the semicircles , we will get the area of the top shaded portion.
now area of the two semicircles = area of both semicircles- area of the common part , the shaded portion.(we need to subtract this as it gets counted twice in the area of the semicircles)
[(22/7)*3.5*3.5)] -7=38.5-7=31.5sq cm
So area of the top shaded part=area of full sector-area of two semi circles=38.5-31.5=7
So area of the total shaded part
=area of top shaded part+area of lower shaded part
=7+7=14 sq cm