In the given figure ABCD is a trapezium of area 24.5 CM ^2. If AB।। BC <DAB = 90' , AD= 10cm, BC = 4 cm and ABE is a quadrant of a circle. Then find the area of the shaded region.
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Answer:
The area of the shaded region = 14.875 cm².
Step-by-step explanation:
Given :
Area of trapezium ABCD, A = 24.5 cm²
AD || BC, ∠DAB = 90°, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle.
Area of the trapezium, A = ½ (sum of parallel sides) × perpendicualr distance between the parallel sides
A = ½ (AD + BC) × AB
24.5 = ½ (10 + 4) × AB
24.5 × 2 = 14 AB
49 = 14 AB
AB = 49/14
AB = 7/2
AB = 3.5 cm
Radius of the quadrant of the circle ,r = AB = 3.5 cm
Area of the quadrant of the circle = ¼ ×πr²
= (1/4) (22/7 x 3.5 x 3.5)
= 9.625 cm²
Area of the quadrant of the circle,ABE = 9.625 cm²
Area of the shaded region = Area of the trapezium,ABCD - Area of the quadrant of the circle,ABE
= 24.5 - 9.625
Area of the shaded region = 14.875 cm²
Hence, the area of the shaded region = 14.875 cm².
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