NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

 

Question 1. 

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

Solution:-

Steps of construction:
1. Draw a line segment AB = 5.8 cm.
2. At point A, draw a ray P to making an angle of 60^o i.e. ∠PAB = 60^o.
3. At point B, draw a ray Q to making an angle of 30^o i.e. ∠QBA = 30^o.
4. Now the two rays AP and BQ intersect at the point C.
Then, ΔABC is the required triangle.

Question 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.
∴ ∠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 of construction:

1. Draw a line segment PQ = 5 cm.
2. At point P, draw a ray L to making an angle of 105^o i.e. ∠LPQ = 35^o.
3. At point Q, draw a ray M to making an angle of 40^o i.e. ∠MQP = 105^o.
4. Now the two rays PL and QM intersect at the point R.
Then, 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:-

From the question it is given that,
EF = 7.2 cm
∠E = 110^o
∠F = 80^o
Now 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
We may observe that the sum of two angles is 190^o is greater than 180^o. So, it is not possible to construct a triangle.