Example 2 - Construct triangle PQR, given PQ = 3 cm, QR = 5.5 cm, PQR

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

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Example 2 Construct a triangle PQR, given that PQ = 3 cm, QR = 5.5 cm and ∠PQR = 60°. First we draw a rough sketch We follow these steps Steps of construction 1. Draw a line segment QR of length 5.5 cm 2. Now, we draw 60° from point Q Check Construction 11.3 of Chapter 11 Class 9 NCERT to see how to draw 60° 3. Taking Q as center, 3 cm as radius, we draw an arc Let the point where arc intersects the ray be point P 4. Join PR and label the sides Thus, Δ PQR 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.