# Ex 11.4, 10 (ii) - 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 (ii) In the following figures, find the area of the shaded portions: Area of shaded portion = Area of PQRS − Area of ∆PTQ − Area of ∆STU − Area of ∆QRU Area of ∆PQRS Here, QR = 20 cm and SR = SU + UR = 10 + 10 = 20 cm Hence, All side of PQRS are same ∴ PQRS is a square with Side = a = 20 cm Area = a × a = 20 × 20 = 400 cm2 Area of ∆PTQ Here, Δ PTQ is a right angle triangle with Base PQ and height PT Base = b = PQ = SQ = 20 cm Height = h = PT = PS − TS = 20 − 10 = 10 cm Area of ∆PTQ = 1/2 × b × h = 1/2 × 20 × 10 = 10 × 10 = 100 cm2 Area of ∆STU Here, Δ STU is a right angle triangle with Base SU and height ST Base = b = SU = 10 cm Height = h = ST = 10 cm Area of ∆AEF = 1/2 × b × h = 1/2 × 10 × 10 = 5 × 10 = 50 cm2 Area of ∆QRU Here, Δ QRU is a right angle triangle with Base UR and height QR Base = b = UR = 10 cm Height = h = QR = 20 cm Area of ∆QRU = 1/2 × b × h = 1/2 × 10 × 20 = 5 × 20 = 100 cm2 Now, Area of shaded portion = Area of PQRS − Area of ∆PTQ − Area of ∆STU − Area of ∆QRU = 400 − 100 − 50 − 100 = 400 − (100 + 50 + 100) = 400 − 250 = 150 cm2 ∴ Area of shaded portion is 150 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 10 years. He provides courses for Maths and Science at Teachoo.