# Ex 11.4, 10 (i) - Chapter 11 Class 7 Perimeter and Area

Last updated at Dec. 12, 2018 by Teachoo

Last updated at Dec. 12, 2018 by Teachoo

Transcript

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 9 years. He provides courses for Maths and Science at Teachoo.