Question

ABCD is a square with one side equal to 7cm. It consists of two quadrants ABC and BCD. Their point of intersection is P. Find the area of the shaded region.

Answer:

Area of the shaded region = 2.1266 sq cm.

Area of the shaded region = Area of square - (Area of the quadrant ABC + Area of quadrant BCD) + (Area of segment BQP + Area of segment CRP + Area of triangle CBP) ... (1)
Note: We are subtracting the middle area twice and that is why we have added that area in the end.

As we know ABC and BCD are quadrants, B is the center of the quadrant ABC and C is the center of the quadrant BCD. Therefore, radius of quadrant ABC = BC = 7cm and radius of quadrant BCD = CB = 7cm. Therefore, the two quadrants are congruent.
Join BP and CP. Since BP and CP are the radii of the two quadrants, BP = CP = BC = 7cm. Therefore, triangle BPC is an equilateral triangle with side = 7cm and so all of its interior angles are equal to 60 degrees, i.e., angle PBC = angle PCB = 60
Area of segment BQP:
= Area of sector CBP - Area of equilateral triangle CBP
= π*7*7*60/360 - 7*7√3/4
= (49/2)[π/3 - √3/2] sq cm
Similarly we get area of segment CRP = (49/2)[π/3 - √3/2] sq cm = Area of segment BQP
Sum of areas of the two segments = 2*(49/2)[π/3 - √3/2] = 49[π/3 - √3/2] sq cm
Area of triangle CBP = 7*7√3/4 = 49√3/4 sq cm
Area of quadrant ABC = Area of quadrant BCD = π*7*7*90/360 = π*7*7/4 = 49*π/4 sq cm
Sum of areas of the two quadrants = 2*49*π/4 = 49*π/2 sq cm
Area of square = 7*7 = 49 sq cm Therefore, area of the shaded region: [From (1)]
= 49 - (49*π/2) + (49[π/3 - √3/2] + 49*sqrt(3)/4
= 49[1 - π/2 + π/3 - √3/2 + sqrt(3)/4]
= 49[1 - π/6 - √3/4]
= 49[1 - 0.5236 - 0.433]
= 49[0.0434]
= 2.1266 sq cm.


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