Ex 6.3
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)
Ex 6.3, 2
Ex 6.3, 3
Ex 6.3, 4 Important
Ex 6.3, 5
Ex 6.3, 6
Ex 6.3, 7
Ex 6.3, 8 Important
Ex 6.3, 9
Ex 6.3, 10
Ex 6.3, 11 Important
Ex 6.3, 12 Important
Ex 6.3, 13 Important
Ex 6.3, 14 Important
Ex 6.3, 15 Important You are here
Ex 6.3, 16
Last updated at April 16, 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