Last updated at June 23, 2017 by Teachoo

Transcript

Ex 6.3, 15 A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. Given: Height of pole = AB = 6m Length of pole of shadow = BC = 4 m Length of shadow of tower = EF = 28 To Find : Height of tower i.e ED Solution:- In ∆ 𝐴𝐵𝐶 and ∆ 𝐷𝐸𝐹 ∠ B = ∠ E = 90° ∠𝐶=∠𝐹 ∴ ∆ 𝐴𝐵𝐶 ∼ ∆ 𝐷𝐸𝐹 ∆ 𝐴𝐵𝐶 ∼ ∆ 𝐷𝐸𝐹 We know that if two triangles are similar, ratio of their sides are in proportion So, 𝐴𝐵/𝐷𝐸=𝐵𝐶/𝐸𝐹 6/𝐷𝐸=4/28 6 ×28=𝐷𝐸×4 (6 × 28)/4= 𝐷𝐸 6 ×7=𝐷𝐸 𝐷𝐸 = 42 Hence, the height of the tower is 42 metres

Ex 6.3

Ex 6.3, 1 (i)

Ex 6.3, 1 (ii)

Ex 6.3, 1 (iii)

Ex 6.3, 1 (iv)

Ex 6.3, 1 (v) Important

Ex 6.3, 1 (vi)

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Ex 6.3, 8 Important

Ex 6.3, 9

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Ex 6.3, 11 Important

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Ex 6.3, 15 Important You are here

Ex 6.3, 16

About the Author

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.