These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.

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

Answer:

Steps of Construction

(i) Draw a rough diagram with measures marked on it.

(ii) Draw a line segment AB = 5.8 cm.

(iii) Construct ∠BAX = 60° at A.

(iv) At B, construct ∠ABY = 30° .

(v) Let the rays AX and BY intersect at C. Thus, ∆ABC is the required triangle.

Question 2.

Construct ∆PQR if PQ = 5 cm, m∠PQR = 105° and m ∠QRP = 40°.

(Hint : Recall angle-sum property of a triangle)

Answer:

Here the line segment PQ is given. We can construct the triangle, if the measure of ∠QPR is known.

∠QPR = 180° – [∠PQR + ∠QRP]

= 180°-(105°+ 40°)

= 180° – 145°

∠QPR = 35°

Steps of Construction

(i) Draw a rough diagram with measures marked on it.

(ii) Draw a line, segment PQ = 5 cm.

(iii) At P, construct ∠QPX = 35 .

(iv) At Q, construct ∠PQY = 105 .

(v) Let the rays \(\overline{\mathrm{PX}}\) and \(\overline{\mathrm{QY}}\) intersect at R.

Thus, ∆PQR is the required triangle.

Question 3.

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

Answer:

Here m∠E =110 and m∠F = 80°

i. e. m∠E + m∠F= 110°+ 80°= 190°>180°

Since, sum of the three angles of a triangle is 180°.

∆DEF is not possible, having sum of two angles > 180°

Thus, ∆DEF cannot be constructed.