Ncert Solutions For Class 7 Maths Ex 10.4

Ncert Solutions For Class 7 Maths Chapter 10 Ex 10.4

 

Q1:

Construct a triangle ZXC, where Z = 60O , X = 30O and ZX = 5.8 cm.

Solution:

Now, constructing a Δ ZXC where Z = 60O , X = 300 and ZX = 5.8 cm.

Steps for construction:

(a) Make a line segment ZX = 5.8 cm.

(b) At  Z, make an angle XZC = 60O  with a compass.

(c) At X, draw QXZ = 300  with  a compass.

(d) ZP and XQ intersect at the point C.

Thus we have the required triangle ZXC:

4.1

 

Q2:

Construct a triangle ABC if AB = 5 cm, ABC = 105° and BCA = 40°.

Solution:

Given:

ABC  = 105° and BAC = 40°

We know ,sum of the angles of a triangle = 180 .°

ABC + BCA + CAB = 180°

105. 40 °+ 40°+BCA = 180°

145° + BCA = 180°

BCA = 180° – 145°

BCA = 35°

Constructing a triangle ABC  whose A = 35°, B = 105° and AB = 5 cm.

Steps of construction:

(a) Make a line segment PQ = 5 cm.

(b) At  A, make an angle  QAB = 35° with the help of a protractor.

(c) At  B, make an angle ABP = 105° with the help of a protractor.

(d) QA and BP intersect at point C.

Thus, we have the required triangle ABC.

4.2

 

Q3:

Can you construct  a Δ ABC such that AB = 4 cm, B = 120° and C= 70 °. Explain your answer.

Solution:

Given:

In Δ ABC, A = 120° and m F = 70 °

We know the angles of a triangle sum up to 180o

Thus, we have

A + C+ B= 180°

A + 120° + 70° = 180°

A + 190° = 180°

 

A = 180° − 190° = −10°

Since the sum of the angles is coming more than 180o, this triangle construction is not possible.