Example 4 - Practical Geometry

Last updated at Dec. 16, 2024 by

Example 4 - Construct LMN, right-angled at M, given LN = 5 cm, MN = 3

Example 4 - Chapter 10 Class 7 Practical Geometry - Part 2
Example 4 - Chapter 10 Class 7 Practical Geometry - Part 3
Example 4 - Chapter 10 Class 7 Practical Geometry - Part 4
Example 4 - Chapter 10 Class 7 Practical Geometry - Part 5

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Transcript

Example 4 Construct ∆LMN, right-angled at M, given that LN = 5 cm and MN = 3 cm. First we draw a rough sketch We follow these steps Steps of construction 1. Draw a line segment MN of length 3 cm 2. Now, we draw 90° from point M Check Ex 11.1, 1 of Chapter 11 Class 9 NCERT to see how to draw 90° 3. Taking N as center, 5 cm as radius, we draw an arc Let the point where arc intersects the ray be point L 4. Join LN and label the sides Thus, Δ LMN is the required triangle

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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