RS Aggarwal - Mathematics

Book: RS Aggarwal - Mathematics

Chapter: 18. Area of Circle, Sector and Segment

Subject: Maths - Class 10th

Q. No. 50 of Exercise 18A

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50

From a thin metallic piece in the shape of a trapezium ABCD in which AB CD and BCD = 90°, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of metal sheet.

Given:

AB CD


BCD = 90°


AB = BC = 3.5 cm = EC


DE = 2 cm


DC = DE + EC = 2 + 3.5 = 5.5 cm


Area of Trapezium = × Sum of Parallel Sides × h


= × (AB + DC) × BC


= × (3.5 + 5.5) × 3.5


= × 9 × 3.5


= 15.75 cm2


Area of Quadrant BFEC = × πr2 = × × 3.5 × 3.5


= 9.625 cm2


Thus, Area of remaining part of metal sheet


= Area of Trapezium – Area of Quadrant BFEC


= 15.75 – 9.625 = 6.125 cm2


Hence, the area of the remaining part of metal sheet is 6.125 cm2.


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