Ex 10.3, 1 - Construct DEF such that DE = 5 cm, DF = 3 cm, EDF = 90

Ex 10.3, 1 - Chapter 10 Class 7 Practical Geometry - Part 2
Ex 10.3, 1 - Chapter 10 Class 7 Practical Geometry - Part 3 Ex 10.3, 1 - Chapter 10 Class 7 Practical Geometry - Part 4 Ex 10.3, 1 - Chapter 10 Class 7 Practical Geometry - Part 5

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Question 1 Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°. First we draw a rough sketch To construct it, We follow these steps Steps of construction 1. Draw a line segment DF of length 3 cm 2. Now, we draw 90° from point D Check Ex 11.1, 1 of Chapter 11 Class 9 NCERT to see how to draw 90° 3. Taking D as center, 5 cm as radius, we draw an arc Let the point where arc intersects the ray be point E 4. Join EF and label the sides Thus, Δ DEF is the required triangle

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