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

Example 5
Important

Example 8 Important

Example 10 Important

Example 14 Important

Theorem 6.1 - Basic Proportionality Theorem Important

Theorem 6.7 Important

Ex 6.2, 4 Important

Ex 6.2, 5 Important

Ex 6.2, 6 Important

Ex 6.2, 9 Important

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 6.4, 1 Important

Ex 6.4, 3 Important

Ex 6.4, 5 Important

Ex 6.5, 2 Important

Ex 6.5, 3 Important

Ex 6.5, 8 Important

Ex 6.5, 11 Important

Ex 6.5, 12 Important

Ex 6.5, 15 Important

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.