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

1. Construct ΔABC, given m ∠A =60O, m ∠B = 30O 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 60O i.e. ∠PAB = 60O.

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

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 = 105o and m∠QRP = 40o.

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

Solution:

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

Now,

∠PQR + ∠QRP + ∠RPQ = 180o

105o+ 40o+ ∠RPQ = 180o

145o + ∠RPQ = 180o

∠RPQ = 180o – 145o

∠RPQ = 35o

Hence, the measures of ∠RPQ is 35o.

Steps:

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

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

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

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 = 110o

∠F = 80o

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 180o.

Then,

∠D + ∠E + ∠F = 180o

∠D + 110o+ 80o= 180o

∠D + 190o = 180o

∠D = 180o – 190o

∠D = -10o

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

Hence, it is not possible to construct a triangle.

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