Table of Contents

Toggle## Class 7, Maths, Chapter 10, Exercise 10.4 Solutions

**Q.1. Construct ∆ABC, given m****∠****A= 60°, m****∠****B = 30° and AB = 5.8 cm.**

**Ans:**

** **

**Steps of construction:**

- Draw a line segment AB = 5.8 cm.
- At point A, draw an angle of 60
^{o}. i.e. ∠PAB = 60^{o}. - At point B, draw an angle of 30
^{o}i.e. ∠QBA = 30^{o}. - Now, AP and BQ intersect at the point C.

Then, ΔABC is the required triangle.

**Q.2. Construct ∆PQR if PQ = 5 cm, m****∠****PQR = 105° and m****∠****QRP = 40°. (Hint: Recall angle-sum property of a triangle).**

**Ans:**

We know that the sum of the angles of a triangle is 180^{o}.

∴ m∠PQR + m∠QRP + m∠RPQ = 180^{o}

⟹ 105^{o}+ 40^{o}+ ∠RPQ = 180^{o}

⟹ 145^{o} + ∠RPQ = 180^{o}

⟹ ∠RPQ = 180^{o}– 145^{0}

⟹ ∠RPQ = 35^{o}

**Steps of construction:**

- Draw a line segment PQ = 5 cm.
- At point P, draw an angle of 35
^{o}i.e. ∠XPQ = 35^{o}. - At point Q, draw an angle of 105
^{o}i.e. ∠YQP = 105^{o}. - Now PX and QY intersect at the point R.

Then, ΔPQR is the required triangle.

**Q.3. Examine whether you can construct ∆DEF such that EF = 7.2 cm, m****∠****E = 110° and m****∠****F = 80°. Justify your answer**

**Ans: **Given: in ΔDEF, ∠E = 110^{o} and ∠F = 80^{o}

We know that the sum of the angles of a triangle is 180^{o}.

Then,

⟹ ∠D + ∠E + ∠F = 180^{o}

⟹ ∠D + 110^{o }+ 80^{o}= 180^{o}

⟹ ∠D + 190^{o} = 180^{o}

⟹ ∠D = 180^{o}– 190^{0}

⟹ ∠D = -10^{o}

Here, the sum of two angles is 190^{o} is greater than 180^{o}. So, it is not possible to construct a triangle.

## NCERT Solutions For Class 7 Maths, Chapter 10, Practical Geometry (All Exercises)

**Class 7, Maths, Chapter 9, Practical Geometry **

**Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.1**

**Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.2**

**Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.3**

**Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.4** **← You are here**

**Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.5**