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

Last updated at August 10, 2026 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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