NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.3

Question 1.
Construct a ADEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90 .
Answer:
Steps of Construction
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 1
(i) First we draw a rough sketch with measures marked on it.
(ii) Draw a line segment DE = 5 cm.
(iii) At D, construct ∠EDX = 90
(iv) From the ray DX, cut off DF = 3 cm.
(v) Join EF.
Then, ∆DEF is the required triangle.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

Question 2.
Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110.
Answer:
Steps of Construction
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 2
(i) First we draw a rough sketch with measures marked on it.
(ii) Draw a line segment AB = 6.5 cm.
(iii) At B, using protractor construct ∠ABX = 110°.
(iv) From BX, cut off BC = 6.5 cm.
(v) Join AC.
Thus, ∆ABC is the required isosceles triangle.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

Question 3.
Construct ∆ABC with BC = 7.5 cm AC = 5 cm and m∠C – 60
Answer:
Steps of Construction
(i) First we draw a rough sketch with measures marked on it.
(ii) Draw a line segment BC = 7.5 cm.
(iii) At ‘C’ using protractor construct ∠BCX = 60°.
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 3
(iv) From the ray CX, cut off CA = 5 cm.
(v) Join AB.
Thus, ∆ABC is the required triangle.

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