Question 1 - Constructing a right triangle when hypotenuse and side is given (RHS) - Practical Geometry

Last updated at Dec. 16, 2024 by

Ex 10.5, 1 - Construct the right angled PQR, where Q = 90, QR = 8 cm

Ex 10.5, 1 - Chapter 10 Class 7 Practical Geometry - Part 2
Ex 10.5, 1 - Chapter 10 Class 7 Practical Geometry - Part 3
Ex 10.5, 1 - Chapter 10 Class 7 Practical Geometry - Part 4
Ex 10.5, 1 - Chapter 10 Class 7 Practical Geometry - Part 5

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Transcript

Question 1 Construct the right angled ∆PQR, where m∠Q = 90°, QR = 8cm and PR = 10 cm. First we draw a rough sketch We follow these steps Steps of construction 1. Draw a line segment QR of length 8 cm 2. Now, we draw 90° from point Q Check Ex 11.1, 1 of Chapter 11 Class 9 NCERT to see how to draw 90° 3. Taking R as center, 10 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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo