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.