These NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 11 Constructions Exercise 11.1

Question 1.

Construct an angle of 90° at the initial point of a given ray and justify the construction.

Solution:

Steps of construction:

1. Draw a straight line AB.

2. Taking A as centre, draw an arc, which intersects AB at a point D.

3. Now, taking D as a centre, the same radius AD, intersect at E and F on the succession arc. (here AD = \(\widehat{\text { DE }}\) = \(\widehat{\text { EF }}\))

4. From point E take a radius (which should be greater than \(\frac {1}{2}\) \(\widehat{\text { EF }}\)) then with the same radius from F intersect the arc at point C.

5. Now, join C to A.

6. The required angle ∠ABC = 90°.

Question 2.

Construct an angle of 45° at the initial point of a given ray and justify the construction.

Solution:

Steps of Constructions:

1. Follow the instruction of the previous question upto ∠ABC = 90°.

2. Take an arc from the points H and G each which Intersect at I.

3. Here ∠FBC is half of ∠ABC, FB is the angle bisector.

4. So, the required ∠FBC = 45°

Question 3.

Construct the angles of the following measurements:

(i) 30°

(ii) 22½°

(iii) 15°

Solution:

(i) Steps of Construction:

1. Take a straight line AB.

2. Draw an arc, taking A as centre which intersects AB at C.

3. From C take another arc \(\widehat{\mathrm{CD}}\) such that AB = \(\widehat{\mathrm{CD}}\)

4. From C and D take similar arc which intersects at E.

5. The required ∠FAB = 30°.

(ii) Steps of construction:

1. Follow the instruction of question (2) i.e. ∠FAB = 45°.

2. Take the same arc from points C and K, intersect at G.

3. The required ∠HAB = 22½°

(iii) Step of construction:

1. Follow the instruction of 3:

(i) ∠EAB = 30°.

2. Take a small arc from the points G and C that intersects at F.

3. The required angle ∠FAB = 15°.

Question 4.

Construct the following angles and verify by measuring them by a protractor:

(i) 75°

(ii) 105°

(iii) 135°.

Solution:

(i) Steps of construction:

1. Follow as in question 1. i.e. ∠EAC = 90°.

2. Take the same arc from points D and E which intersect at G.

3. The required ∠KAB = 75°.

(ii) Steps of construction:

1. Follow the instruction as in the previous question upto ∠EAB = 90°.

2. Take the same arc from points D and E intersect at F.

3. Join F to A.

4. The required ∠GAB = 105°.

(iii) Steps of construction:

1. Follow the instruction as ∠EAB = 90°.

2. Take the same arc from the points E and F, Intersect at G.

3. Join G to A.

4. The required ∠HAB = 135°.

Question 5.

Construct an equilateral triangle, given its side, and justify the construction.

Solution:

Steps of construction:

1. Take a straight line XY.

2. By measuring 6 cm on the scale, cut one point C on XY line the reverse the same to get B point. Here, BC = 6 cm

3. From B and C points draw an arc of the same length intersect at A.

4. We find here AB = BC = CA = 6 cm. Thus ABC is an equilateral triangle.