NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

# NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

## NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 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 10.4.

### Ex 10.4 Class 7 Maths Question 1.

Construct ∆ABC, given mA = 60°, mB = 30° and AB = 5.8 cm.

Solution:
Steps of construction:
(i) Draw a line segment AB = 5.8 cm.
(ii) Draw
A = 60° and B = 30° to meet each other at C.
(iii) ABC is the required triangle.

### Ex 10.4 Class 7 Maths Question 2.

Construct ∆PQR if PQ = 5 cm, mPQR = 105° and mQRP = 40°.
(Hint: Recall angle-sum property of a triangle)

Solution:
Given: mPQR = 105°, mQRP = 40°
mPQR + mQRP + QPR = 180°   (Angle sum property of a triangle)
105° + 40° +
QPR = 180°
145° +
QPR = 180°
QPR = 180° – 145° = 35°

Steps of construction:
(i) Draw a line segment PQ = 5 cm.
(ii) Draw
QPR = 35° and PQR = 40° to meet each other at R.
(iii) ∆PQR is the required triangle. [Using AAS or ASA criterion]

### Ex 10.4 Class 7 Maths Question 3.

Examine whether you can construct ∆DEF such that EF = 7.2 cm, mE = 110° and mF = 80°. Justify your answer.

Solution:
We cannot construct ∆DEF with the given measurement.
mE + mF = 110° + 80°
= 190° > 180°   [Sum of the angles of a triangle = 180°]
∆DEF is not possible to construct.

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