NCERT solution class 7 chapter 10 Practical geometry exercise 10.4 mathematics

EXERCISE 10.4

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Q 1, Ex 10.4 - Practical Geometry - Chapter 10 - Maths Class 7th - NCERT

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Page No 202:

Question 1:

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

Answer:

A rough sketch of the required ΔABC is as follows.

The steps of construction are as follows.
(i)Draw a line segment AB of length 5.8 cm.

(ii)At point A, draw a ray AX making 60º angle with AB.

(iii) At point B, draw a ray BY, making 30º angle with AB.

(iv) Point C has to lie on both the rays, AX and BY. Therefore, C is the point of intersection of these two rays.

This is the required triangle ABC.

Q 2, Ex 10.4 - Practical Geometry - Chapter 10 - Maths Class 7th - NCERT

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Question 2:

Construct ΔPQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°.
(Hint: Recall angle sum property of a triangle).

Answer:

A rough sketch of the required ΔPQR is as follows.

In order to construct ΔPQR, the measure of ∠RPQ has to be calculated.
According to the angle sum property of triangles,
∠PQR + ∠PRQ + ∠RPQ = 180º
105º + 40º + ∠RPQ = 180º
145º + ∠RPQ = 180º
∠RPQ = 180° − 145° = 35°
The steps of construction are as follows.
(i) Draw a line segment PQ of length 5 cm.

(ii) At P, draw a ray PX making an angle of 35º with PQ.

(iii) At point Q, draw a ray QY making an angle of 105º with PQ.

(iv)Point R has to lie on both the rays, PX and QY. Therefore, R is the point of intersection of these two rays.

This is the required triangle PQR.

Q 3, Ex 10.4 - Practical Geometry - Chapter 10 - Maths Class 7th - NCERT

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Question 3:

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

Answer:

Given that,
m∠E = 110° and m∠F = 80°
Therefore,
m∠E + m∠F = 110° + 80° = 190°
However, according to the angle sum property of triangles, we should obtain
m∠E + m∠F + m∠D = 180°
Therefore, the angle sum property is not followed by the given triangle. And thus, we cannot construct ΔDEF with the given measurements.

Also, it can be observed that point D should lie on both rays, EX and FY, for constructing the required triangle. However, both rays are not intersecting each other. Therefore, the required triangle cannot be formed.