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

- Class 7 Maths Practical Geometry Exercise 10.1
- Class 7 Maths Practical Geometry Exercise 10.2
- Class 7 Maths Practical Geometry Exercise 10.3
- Class 7 Maths Practical Geometry Exercise 10.4
- Class 7 Maths Practical Geometry Exercise 10.5

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

Ex 10.2 Class 7 Maths Question 1.

Construct ∆XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.

Solution:

Steps of construction:

(i) Draw XY = 4.5 cm.

(ii) Draw an arc with centre Y and radius 5 cm.

(iii) Draw another arc with centre X and radius 6 cm to meet the first arc at Z.

(iv) Join ZY and ZX.

(v) XYZ is the required triangle.

Using SSS criterion

Ex 10.2 Class 7 Maths Question 2.

Construct an equilateral triangle of side 5.5 cm. Solution: Steps of construction:

(i) Draw BC = 5.5 cm.

(ii) Draw two arcs with centres B and C and same radius of 5.5 cm to meet each other at A.

(iii) Join AB and AC.

(iv) ABC is the required triangle.

Using SSS criterion

Ex 10.2 Class 7 Maths Question 3.

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

Solution:

Steps of construction:

(i) Draw QR = 3.5 cm.

(ii) Draw two arcs with centre Q and R and same radius of 4 cm to meet each other at P.

(iii) Join PQ and PR.

(iv) PQR is the required triangle.

(v) Since PQ = PR = 4 cm, therefore APQR is an isosceles triangle.

Using SSS Criterion

Ex 10.2 Class 7 Maths Question 4.

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

Solution:

Steps of construction:

(i) Draw BC = 6 cm.

(ii) Draw two arcs with centres B and C and radius 2.5 cm and 6.5 cm respectively to meet each other at A.

(iii) Join AB and AC.

(iv) ABC is the required triangle.

[Using SSS criterion]

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Helpful

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Kabyatara Pradhan says

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Prasanna Choopari says

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Mishti says

In Fourth question where is whole answer.