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Question 3 - Constructing a triangle using SSS congruency criteria - Practical Geometry

Last updated at May 29, 2023 by

Ex 10.2, 3 - Draw PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm

Ex 10.2, 3 - Chapter 10 Class 7 Practical Geometry - Part 2
Ex 10.2, 3 - Chapter 10 Class 7 Practical Geometry - Part 3 Ex 10.2, 3 - Chapter 10 Class 7 Practical Geometry - Part 4 Ex 10.2, 3 - Chapter 10 Class 7 Practical Geometry - Part 5

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Question 3 Draw ∆PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this? First we draw a rough sketch We follow these steps Steps of construction 1. Draw a line segment PQ of length 4 cm 2. Taking 4 cm as radius, and P as center, draw an arc. 3. Taking 3.5 cm as radius, and Q as center, draw another arc. 4.Let R be the point where the two arcs intersect Join PR and QR and label the sides Let R be the point where the two arcs intersect Join PR and QR and label the sides Thus, Δ PQR is the required triangle Since two sides are equal ∴ Δ PQR is an sssssssssss Let R be the point where the two arcs intersect Join PR and QR and label the sides Thus, Δ PQR is the required triangle Since two sides are equal ∴ Δ PQR is an sssssssssss

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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.