 # NCERT Solutions for Class 7 Maths Exercise 10.2 Chapter 10 Practical Geometry

NCERT Solutions for Class 7 Maths Exercise 10.2 Chapter 10 Practical Geometry in simple PDF are given here. This exercise of NCERT Maths Solutions for Class 7 Chapter 10 contains topics related to constructing a triangle when the lengths of its three sides are known (SSS Criterion). By practising these NCERT Solutions for Class 7 Maths, Chapter 10 Practical Geometry, students can score high marks in the exam.

## Download the PDF of NCERT Solutions For Class 7 Maths Chapter 10 Practical Geometry – Exercise 10.2   ### Access Other Exercises of NCERT Solutions For Class 7 Maths Chapter 10 – Practical Geometry

Exercise 10.1 Solutions

Exercise 10.3 Solutions

Exercise 10.4 Solutions

Exercise 10.5 Solutions

### Access Answers to NCERT Class 7 Maths Chapter 10 – Practical Geometry Exercise 10.2

1. Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm

Solution:- Steps of construction:

1. Draw a line segment YZ = 5 cm.

2. With Z as a center and radius 6 cm, draw an arc.

3. With Y as a center and radius 4.5 cm, draw another arc, cutting the previous arc at X.

4. Join XY and XZ.

Then, ΔXYZ is the required triangle.

2. Construct an equilateral triangle of side 5.5 cm.

Solution:- Steps of construction:

1. Draw a line segment AB = 5.5 cm.

2. With A as a center and radius 5.5 cm, draw an arc.

3. With B as a center and radius 5.5 cm, draw another arc, cutting the previous arc at C.

4. Join CA and CB.

Then, ΔABC is the required equilateral triangle.

3. Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?

Solution:- Steps of construction:

1. Draw a line segment QR = 3.5 cm.

2. With Q as a center and radius 4 cm, draw an arc.

3. With R as a center and radius 4 cm, draw another arc, cutting the previous arc at P.

4. Join PQ and PR.

Then, ΔPQR is the required isosceles triangle.

4. Construct ΔABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.

Solution:- 1. Draw a line segment BC = 6 cm.

2. With B as a center and radius 2.5 cm, draw an arc.

3. With C as a center and radius 6.5 cm, draw another arc, cutting the previous arc at A.

4. Join AB and AC.

Then, ΔABC is the required triangle.

5. When we measure the angle B of triangle by protractor, then angle is equal to ∠B = 90o

#### 1 Comment

1. prikshitverma

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