Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each. [Use Π = 22/7]
The designed area is the common region between two sectors BAEC and DAFC.
Area of sector BAEC = `90^@/360^@ xx 22/7xx(8)^2`
`=1/4xx22/7xx64`
`=(22xx16)/7 cm^2`
`= 352/7 cm^2`
Area of ΔBAC = `1/2xxBAxxBC`
`= 1/2xx8xx7 = 32 cm^2`
Area of the designed portion = 2 × (Area of segment AEC)
= 2 × (Area of sector BAEC − Area of ΔBAC)
`= 2xx(352/7 - 32) = 2((352-224)/4)`
`= (2xx128)/7`
`= 256/7 cm^2`
Concept: Areas of Combinations of Plane Figures
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Solution:
We use the formula for the area of sectors of the circle and the area of the square to solve the problem.
In a circle with radius r and the angle at the centre with degree measure θ,
Area of a sector = θ/360° × πr2
Area of a quadrant = 90°/360° × πr2 = 1/4 πr2
Area of plain or unshaded part of the square = Area of square - Area of quadrant
= side2 - (1/4 × πr2)
= (8 cm × 8 cm) - (1/4 × 22/7 × 8 cm × 8 cm)
= 64 cm2 - 352/7 cm2
= 96/7 cm2
Area of the designed region = Area of the quadrant - Area of plain or unshaded part of the square
= (1/4 × 22/7 × 8 cm × 8 cm) - 96/7 cm2
= 352/7 cm2 - 96/7 cm2
= 256/7 cm2
☛ Check: NCERT Solutions for Class 10 Maths Chapter 12
Video Solution:
NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 16
Summary:
The area of the designed region common between the two quadrants of circles of radius 8 cm each is 256/7 cm2.
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