Mathematics – Class 7 – Chapter 10 – Practical Geometry – Exercise 10.4 – NCERT Exercise Solution

1. Construct ΔABC, given m ∠A =60^O, m ∠B = 30^O and AB = 5.8 cm.

Solution:

Steps:

Step 1. Draw a line segment AB = 5.8 cm.

Step 2. Draw a ray P, at point A to make an angle of 60^O i.e. ∠PAB = 60^O.

Step 3. Draw a ray Q, at point B to make an angle of 30^O i.e. ∠QBA = 30^O.

Step 4. Now the two rays AP and BQ intersect each other at point C.

Hence, ΔABC is the required triangle.

2. Construct ΔPQR if PQ = 5 cm, m∠PQR = 105^o and m∠QRP = 40^o.

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

Solution:

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

Now,

∠PQR + ∠QRP + ∠RPQ = 180^o

105^o+ 40^o+ ∠RPQ = 180^o

145^o + ∠RPQ = 180^o

∠RPQ = 180^o – 145^o

∠RPQ = 35^o

Hence, the measures of ∠RPQ is 35^o.

Steps:

Step 1. Draw a line segment PQ = 5 cm.

Step 2. Draw a ray L, at point P to make an angle of 105^o i.e. ∠LPQ = 35^o.

Step 3. Draw a ray M, at point Q to make an angle of 40^o i.e. ∠MQP = 105^o.

Step 4. Now the two rays PL and QM intersect each other at the point R.

Hence, ΔPQR is the required triangle.

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

Solution:

Given,

EF = 7.2 cm

∠E = 110^o

∠F = 80^o

Now,

According to the question we have to check whether it is possible to construct ΔDEF from the given values.

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^o

∠D = -10^o

Now, we observe that the sum of the two angles is 190^o, which is greater than 180^o.

Hence, it is not possible to construct a triangle.