Ex 11.4, 10 (i) In the following figures, find the area of the shaded portions:
Area of shaded portion = Area of ABCD − Area of ∆AEF − area of ∆BEC
Area of ABCD
ABCD is a rectangle with
Length = l = 18 cm
Breadth = b = 10 cm
Area of ABCD = l × b
= 18 × 10
= 180 cm2
Area of ∆AEF
Since ABCD is a rectangle,
∠ A = 90°
∴ ∆AEF is a right angled triangle
with base as AE & Height as AF
Base = b = 10 cm
Height = h = 6 cm
Area of ∆AEF = 1/2 × b × h
= 1/2 × 10 × 6 = 5 × 6 = 30 cm2
Area of ∆BEC
Since ABCD is a rectangle,
∠ B = 90°
∴ ∆BEC is a right angled triangle
with base as AE & Height as AF
Base = b = 8 cm
Height = h = 10 cm
Area of ∆BEC = 1/2 × b × h
= 1/2 × 8 × 10 = 4 × 10 = 40 cm2
Now,
Area of shaded portion
= Area of ABCD − Area of ∆AEF − Area of ∆BEC
= 180 − 30 − 40
= 150 − 40
= 110 cm2
∴ Area of shaded region is 110 cm2

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.