RD Sharma 2020 solution class 9 chapter 16 Construction MCQS

MCQS

Page-16.19

Question 1:

With the help of a ruler and compasses, it is possible to construct an angle1 of
(a) 35°
(b) 40°
(c) $37\frac{1°}{2}$
(d) $47\frac{1°}{2}$

Answer 1:

Using a ruler and compass, construct an angle of 150°.
Bisect the angle to get an angle of 75°.
Again bisect the angle to get an angle of 37.5°.


Hence, the correct option is (c).

Question 2:

With the help of a ruler and compasses it is not possible to construct an angle of
(a) $37\frac{1°}{2}$

(b) 40°

(c) $22\frac{1°}{2}$

(d) $67\frac{1°}{2}$

Answer 2:

(a) Using a ruler and compass, construct an angle of 150°.
Bisect the angle to get an angle of 75°.
Again bisect the angle to get an angle of 37.5°.

(b) It is not possible to construct an angle of 40° with the help of a ruler and compass.

(c) Using a ruler and compass, construct an angle of 90°.
Bisect the angle to get an angle of 45°.
Again bisect the angle to get an angle of 22.5°.

(d) Using a ruler and compass, construct an angle of 270°.
Bisect the angle to get an angle of 135°.
Again bisect the angle to get an angle of 67.5°.
​
Hence, the correct option is (b).

Question 3:

The construction of a triangle ABC in which AB = 4 cm, ∠A = 60° is not possible when difference of BC and AC is equal to
(a) 3.5 cm
(b) 4.5 cm
(c) 3 cm
(d) 2.5 cm

Answer 3:

We know, 
In a triangle, sum of any two sides must be greater than the third side.

$$\begin{gathered} AB+AC>BC \\ \Rightarrow AB>BC-AC \\ \Rightarrow 4>BC-AC \left(\because AB=4 \mathrm{cm}\right) \\ \Rightarrow 4>BC-AC \end{gathered}$$

Thus, difference of BC and AC cannot be greater than 4 cm.
​
Hence, the correct option is (b).

Question 4:

The construction of a triangle ABC, given BC = 6 cm, ∠B = 45° is not possible when difference of AB and AC is equal to
(a) 6.9 cm
(b) 5.2 cm
(c) 5 cm
(d) 4 cm

Answer 4:

We know, 
In a triangle, sum of any two sides must be greater than the third side.

$$\begin{gathered} AB+BC>AC \\ \Rightarrow BC>AC-AB \\ \Rightarrow 6>AC-AB \left(\because BC=6 \mathrm{cm}\right) \\ \Rightarrow 6>AC-AB \end{gathered}$$

Thus, difference of AB and AC cannot be greater than 6 cm.
​
Hence, the correct option is (a).

Question 5:

The construction of a triangle ABC, given that BC = 3 cm, ∠C = 60° is possible when differences of AB and AC is equal to
(a) 3.2 cm
(b) 3.1 cm
(c) 3 cm
(d) 2.8 cm

Answer 5:

We know, 
In a triangle, sum of any two sides must be greater than the third side.

$$\begin{gathered} AB+BC>AC \\ \Rightarrow BC>AC-AB \\ \Rightarrow 3>AC-AB \left(\because BC=3 \mathrm{cm}\right) \\ \Rightarrow 3>AC-AB \end{gathered}$$

Thus, differences of AB and AC can be 2.8 cm.

Hence, the correct option is (d).

Question 6:

The construction of a triangle ABC, given that AB = 5 cm, ∠A = 45° is possible when BC + CA is equal to
(a) 5 cm
(b) 4.5 cm
(c) 8 cm
(d) 4 cm

Answer 6:

We know, 
In a triangle, sum of any two sides must be greater than the third side.

Since, AB = 5 cm
Thus, BC + CA > 5 cm

Hence, the correct option is (c).