# 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$\angle$ Z = 60O , $\angle$ X = 30O and ZX = 5.8 cm.

Solution:

Now, constructing a Δ$\Delta$ ZXC where $\angle$ Z = 60O , $\angle$ 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 $\angle$ XZC = 60O  with a compass.

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

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

Thus we have the required triangle ZXC:

Q2:

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

Solution:

Given:

$\angle$ABC  = 105° and $\angle$ BAC = 40°

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

$\angle$ ABC + $\angle$ BCA + $\angle$ CAB = 180°

105. 40 °+ 40°+$\angle$BCA = 180°

145° + $\angle$ BCA = 180°

$\angle$BCA = 180° – 145°

$\angle$BCA = 35°

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

Steps of construction:

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

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

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

(d) QA and BP intersect at point C.

Thus, we have the required triangle ABC.

Q3:

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

Solution:

Given:

In Δ$\Delta$ ABC, $\angle$ A = 120° and m$\angle$ F = 70 °

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

Thus, we have

$\angle$A + $\angle$C+ $\angle$B= 180°

$\angle$A + 120° + 70° = 180°

$\angle$A + 190° = 180°

$\angle$A = 180° − 190° = −10°

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