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

Last updated at April 16, 2024 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

Transcript

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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