![Ex 11.4, 10 (ii) - Chapter 11 Class 7 Perimeter and Area - Part 2](https://d1avenlh0i1xmr.cloudfront.net/43e8b3a6-6136-4d57-b80f-49f8d476e1fe/slide61.jpg)
![Ex 11.4, 10 (ii) - Chapter 11 Class 7 Perimeter and Area - Part 3](https://d1avenlh0i1xmr.cloudfront.net/e9494827-a1be-4710-b013-72471e43ba76/slide62.jpg)
![Ex 11.4, 10 (ii) - Chapter 11 Class 7 Perimeter and Area - Part 4](https://d1avenlh0i1xmr.cloudfront.net/8c7e6c5d-0668-40be-806f-8fd42ee1c4ae/slide63.jpg)
![Ex 11.4, 10 (ii) - Chapter 11 Class 7 Perimeter and Area - Part 5](https://d1avenlh0i1xmr.cloudfront.net/bd180773-f615-4659-8c6e-cc2037792f7d/slide64.jpg)
![Ex 11.4, 10 (ii) - Chapter 11 Class 7 Perimeter and Area - Part 6](https://d1avenlh0i1xmr.cloudfront.net/1f8aa80b-aa68-41b7-922d-34bc1907ad37/slide65.jpg)
![Ex 11.4, 10 (ii) - Chapter 11 Class 7 Perimeter and Area - Part 7](https://d1avenlh0i1xmr.cloudfront.net/d8e53205-594b-42d9-8fbb-09877088300a/slide66.jpg)
Area of Path and Cross roads - Worksheet
Area of Path and Cross roads - Worksheet
Last updated at April 16, 2024 by Teachoo
Question 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