Constructions | Exercise 11.1 | NCERT Solution Maths Class 9 | Chapter 11

 Access Answers of NCERT Class 9 Maths Chapter 11 – Constructions

Exercise 11.1 Page: 191

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

Construction Procedure:

To construct an angle 90°, follow the given steps:

1. Draw a ray OA

2. Take O as a centre with any radius, draw an arc DCB is that cuts OA at B.

3. With B as a centre with the same radius, mark a point C on the arc DCB.

4. With C as a centre and the same radius, mark a point D on the arc DCB.

5. Take C and D as centre, draw two arcs which intersect each other with the same radius at P.

6. Finally, the ray OP is joined which makes an angle 90° with OP is formed.

Justification

To prove ∠POA = 90°

In order to prove this, draw a dotted line from the point O to C and O to D and the angles formed are:

From the construction, it is observed that

OB = BC = OC

Therefore, OBC is an equilateral triangle

So that, ∠BOC = 60°.

Similarly,

OD = DC = OC

Therefore, DOC is an equilateral triangle

So that, ∠DOC = 60°.

From SSS triangle congruence rule

△OBC ≅ OCD

So, ∠BOC = ∠DOC [By C.P.C.T]

Therefore, ∠COP = ½ ∠DOC = ½ (60°).

∠COP = 30°

To find the ∠POA = 90°:

∠POA = ∠BOC+∠COP

∠POA = 60°+30°

∠POA = 90°

Hence, justified.

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

Construction Procedure:

1. Draw a ray OA

2. Take O as a centre with any radius, draw an arc DCB that cuts OA at B.

3. With B as a centre with the same radius, mark a point C on the arc DCB.

4. With C as a centre and the same radius, mark a point D on the arc DCB.

5. Take C and D as centre, draw two arcs which intersect each other with the same radius at P.

6. Finally, the ray OP is joined which makes an angle 90° with OP is formed.

7. Take B and Q as centre draw the perpendicular bisector which intersects at the point R

8. Draw a line that joins the point O and R

9. So, the angle formed ∠ROA = 45°

Justification

From the construction,

∠POA = 90°

From the perpendicular bisector from the point B and Q, which divides the ∠POA into two halves. So it becomes

∠ROA = ½ ∠POA

∠ROA = (½)×90° = 45°

Hence, justified

Solution:

(i) 30°

Construction Procedure:

1. Draw a ray OA

2. Take O as a centre with any radius, draw an arc BC which cuts OA at B.

3. With B and C as centres, draw two arcs which intersect each other at the point E and the perpendicular bisector is drawn.

4. Thus, ∠EOA is the required angle making 30° with OA.

Construction Procedure:

1. Draw an angle ∠POA = 90°

2. Take O as a centre with any radius, draw an arc BC which cuts OA at B and OP at Q

3. Now, draw the bisector from the point B and Q where it intersects at the point R such that it makes an angle ∠ROA = 45°.

4. Again, ∠ROA is bisected such that ∠TOA is formed which makes an angle of 22.5° with OA

(iii) 15°

Construction Procedure:

1. An angle ∠DOA = 60° is drawn.

2. Take O as centre with any radius, draw an arc BC which cuts OA at B and OD at C

3. Now, draw the bisector from the point B and C where it intersects at the point E such that it makes an angle ∠EOA = 30°.

4. Again, ∠EOA is bisected such that ∠FOA is formed which makes an angle of 15° with OA.

5. Thus, ∠FOA is the required angle making 15° with OA.

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

(i) 75° (ii) 105° (iii) 135°

Solution:

(i) 75°

Construction Procedure:

1. A ray OA is drawn.

2. With O as centre draw an arc of any radius and intersect at the point B on the ray OA.

3. With B as centre draw an arc C and C as centre draw an arc D.

4. With D and C as centre draw an arc, that intersect at the point P.

5. Join the points O and P

6. The point that arc intersect the ray OP is taken as Q.

7. With Q and C as centre draw an arc, that intersect at the point R.

8. Join the points O and R

9. Thus, ∠AOE is the required angle making 75° with OA.

(ii) 105°

Construction Procedure:

1. A ray OA is drawn.

2. With O as centre draw an arc of any radius and intersect at the point B on the ray OA.

3. With B as centre draw an arc C and C as centre draw an arc D.

4. With D and C as centre draw an arc, that intersect at the point P.

5. Join the points O and P

6. The point that arc intersect the ray OP is taken as Q.

7. With Q and D as centre draw an arc, that intersect at the point R.

8. Join the points O and R

9. Thus, ∠AOR is the required angle making 105° with OA.

(iii) 135°

Construction Procedure:

1. Draw a line AOA‘

2. Draw an arc of any radius that cuts the line AOA‘at the point B and B‘

3. With B as centre, draw an arc of same radius at the point C.

4. With C as centre, draw an arc of same radius at the point D

5. With D and C as centre, draw an arc that intersect at the point P

6. Join OP

7. The point that arc intersect the ray OP is taken as Q and it forms an angle 90°

8. With B‘ and Q as centre, draw an arc that intersects at the point R

9. Thus, ∠AOR is the required angle making 135° with OA.

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

Construction Procedure:

1. Let us draw a line segment AB = 4 cm .

2. With A and B as centres, draw two arcs on the line segment AB and note the point as D and E.

3. With D and E as centres, draw the arcs that cuts the previous arc respectively that forms an angle of 60° each.

4. Now, draw the lines from A and B that are extended to meet each other at the point C.

5. Therefore, ABC is the required triangle.

Justification:

From construction, it is observed that

AB = 4 cm, ∠A = 60° and ∠B = 60°

We know that, the sum of the interior angles of a triangle is equal to 180°

∠A+∠B+∠C = 180°

Substitute the values

⇒ 60°+60°+∠C = 180°

⇒ 120°+∠C = 180°

⇒∠C = 60°

While measuring the sides, we get

BC = CA = 4 cm (Sides opposite to equal angles are equal)

AB = BC = CA = 4 cm

∠A = ∠B = ∠C = 60°

Hence, justified.