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Chapter 10 Practical Geometry Exercise 10.2


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

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

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

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]

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 Q1
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 Q2
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 Q2.1
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 Q3
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 Q4
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