Last updated at Dec. 13, 2024 by Teachoo
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
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About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo