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

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry
Ex 10.2 are the part of NCERT Solutions for Class 7 Maths. Here you can find
the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2.

**NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1****NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2****NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3****NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4****NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5**

**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 a line segment XY = 4.5 cm.

(ii) With centre Y and radius 5 cm, draw an arc.

(iii) With centre X and radius 6 cm, draw another arc to meet the first arc at Z.

(iv) Join ZY and ZX.

(v) XYZ is the required triangle.

**Ex 10.2 Class 7 Maths Question 2.**

Construct an equilateral triangle of side 5.5 cm.**Solution: **

**Steps of construction:
**(i) Draw a line segment BC = 5.5 cm.

(ii) With centres B and C and same radius of 5.5 cm, draw two arcs to meet each other at point A.

(iii) Join AB and AC.

(iv) ABC is the required equilateral triangle.

**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 a line segment QR = 3.5
cm.

(ii) With centres Q and R and same radius of 4 cm, draw two arcs to meet each other at point P.

(iv) PQR is the required triangle.

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

**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 a line segment BC = 6 cm.

(ii) With centres B and C and radius 2.5 cm and 6.5 cm, respectively, draw two arcs to meet each other at point A.

(iii) Join AB and AC.

(iv) ABC is the required triangle.

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