Haryana State Board HBSE 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.4 Textbook Exercise Questions and Answers.

## Haryana Board 7th Class Maths Solutions Chapter 10 Practical Geometry Exercise 10.4

Question 1.

Construct ΔABC, given m∠A = 60°, m∠B = 30° and AB = 5 cm.

Solution:

Draw a rough sketch of ΔABC and indicate the measures of the two angles and length of the included side.

Steps of Construction :

1. Draw a ray BP.

2. From ray BP, cut off a line-segment BA = 5 cm.

3. At M, construct ∠XBA of measure 30°. (Fig.)

4. At A construct ∠YAB of measure 60°. Let the rays BX and AY intersect at C. Then ΔABC is the required triangle. (Fig.)

Question 2.

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

Solution:

Angle sum property of a triangle

∠P + ∠Q + ∠R = 180°

∠P + 105° + 40° = 180°

or, ∠P = 180°-145°

∴ ∠P = 35°

Steps of Construction :

1. Construct PQ = 5 cm.

2. Using protractor, draw ∠Q = 105° and ∠P = 35°.

3. Let their new arms meet at R. Now ΔPQR is the required triangle.

Question 3.

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

Solution:

In ΔDEF, EF = 7.2 cm and

∠E + ∠F + ∠D = 180°

But, 110+ 80 +∠D ≠ 180°

or, 190° + ∠D ≠ 180°

Which is not possible as the sum of three angles of a triangle is 180°, So this triangle cannot be constructed.