## 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.Answer:**

**Construction steps:**

1. Draw a ray OA.

2. With O as centre and any radius, draw a semicircle that cuts ray OA at B.

3. With B as centre and same radius, draw an arc to cut the semicircle at C.

4. With C as centre and same radius, draw an arc to cut the semicircle at D.

5. With C and D as centres and same radius, draw two arcs which intersect each other at X.

6. Finally, join XO which makes an angle of 90°.

**Justification**

To justify that ∠XOA = 90°, we have to join OC and OD.

We have, ∠COA = ∠DOC = 60°. From the fourth and fifth step of this construction, we know that OX is the bisector of ∠DOC.

∴ ∠XOC = 1/2 of ∠DOC

= ∠XOC = 1/2 x 60° = 30°

Also, ∠XOA = ∠COA + ∠XOC

= 60° + 30° = 90°

Hence justified.

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

Answer:

Answer:

**Construction steps:**

1. Draw a ray OA.

2. With O as centre and any radius, draw a semicircle that cuts OA at B.

3. With B as centre and same radius, draw an arc to cut the semicircle at C.

4. With C as centre and same radius, draw an arc to cut the semicircle at D.

5. With C and D as centres and same radius, draw two arcs which intersect each other at X.

6. Join XO to intersect the semicircle at Y, which makes an angle 90°.

7. Now, with Y and B as centres and radius more than half of YB draw intersecting arcs to meet at E.

8. Join OE. Therefore ∠EOA = 45°.

**Justification**

From the construction,

∠XOA = 90°

As EO is the perpendicular bisector of ∠XOA. So,

∠EOA = 1/2 (∠XOA)

= ∠EOA = 1/2 × 90° = 45°.

Hence justified.

**Question 3: Construct the angles of the following measurements:**

i) 30°

ii) 22½°

iii) 15°

Answer:

i) 30°

ii) 22½°

iii) 15°

Answer:

(i) 30°

**Construction steps:**

1. Draw a ray OA.

2. Taking O as a centre and any radius, draw an arc that cuts ray OA at B.

3. Now with B as centre and same radius draw an arc on the previous arc to cut it at C. This will be 60°.

4. With B and C as centres and radius more than half of BC, draw 2 arcs which intersect each other at E.

5. Join EO. Thus, ∠EOA = 30°.ii) 22½°

**Construction steps:**

1. Draw a ray OA.

2. With O as centre and any radius, draw a arc that cuts OA at B.

3. With B as centre and same radius, draw an arc to cut the semicircle at C.

4. With C as centre and same radius, draw an arc to cut the semicircle at D.

5. With C and D as centres and same radius, draw two arcs which intersect each other at X.

6. Join XO to intersect the semicircle at Y, which makes an angle of 90°.

7. Now, with Y and B as centres and radius more than half of YB draw intersecting arcs to meet at E.

8. Join OE. Therefore ∠EOA = 45°.

9. Finally, with E and B as centres and radius more than the half of BE draw 2 intersecting arcs to meet at F.

10. Join FO. Therefore ∠FOA = 22½°.iii) 15°

**Construction steps:**

1. Draw a ray OA.

2. Taking O as a centre and any radius, draw a semicircle that cuts ray OA at B.

3. Now with B as centre and same radius draw an arc on the semicircle to cut it at C. This will be 60°.

4. With B and C as centres and radius more than half of BC, draw 2 arcs which intersect each other at E.

5. Join EO to intersect the arc at F. Thus, ∠EOA = 30°.

6. With F and B as centres and radius more than the half of FB, draw 2 intersecting arcs to intersect at D. Join DO. Thus, ∠DOA = 15°.

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

i) 75°

ii) 105°

iii) 135°

Answer:

i) 75°

ii) 105°

iii) 135°

Answer:

i) 75°

**Construction steps:**

1. Draw a ray OA.

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

3. With B as centre and same radius draw an arc to cut the semicircle at C.

4. With C as centre and same radius draw an arc to cut the semicircle at D.

5. With D and C as centre and same radius draw 2 arcs to intersect at X. Join OX. Let this line intersect the arc at Q.

6. With Q and C as centre draw 2 arcs to intersect at Y.

7. Join OY. Thus, ∠YOA = 75°.ii) 105°

**Construction steps:**

1. Draw a ray OA.

2. With O as centre and any radius draw a semicircle to cut ray OA at B.

3. With B as centre and same radius draw an arc to cut the semicircle at C.

4. With C as centre and same radius draw an arc on the arc to cut at D.

5. With D and C as centre and same radius draw 2 arcs to intersect at X. Join OX. Let this line intersect the semicircle at Q.

6. With Q and D as centre draw 2 arcs to intersect at Y.

7. Join OY. Thus, ∠YOA = 105°.iii) 135°

**Construction steps:**

1. Draw a line A’OA.

2. With O as centre and any radius draw a semicircle to intersect the line A’OA at B and B’.

3. With B as centre and same radius draw an arc to cut the semicircle at C.

4. With C as centre and same radius draw an arc to cut the semicircle at D.

4. With D and C as centre and same radius draw 2 arcs to intersect at X. Join OX. Let this line intersect the semicircle at Q.

5. With Q and B’ as centre draw 2 arcs to intersect at Y.

6. Join OY. Thus, ∠YOA = 135°.

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

Answer:

Answer:

**Construction steps:**

1. Draw a line segment, AB of any units (say 5cm).

2. With A and B as centres, draw two semicircle on the line segment AB to cut it at D and E respectively.

3. With D and E as centres, draw 2 arcs that cuts the semicircle at G and F respectively to form an angle of 60° each.

4. Now, join AG and BF and extend such that they meet each other at the point C.

5. Therefore, ABC is the required triangle.

**Justification:**

After construction we observe

AB = 5 cm, ∠A = 60° and ∠B = 60°

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

∴ ∠A + ∠B + ∠C = 180°

= 60° + 60° + ∠C = 180° (as ∠A and ∠B are 60° each)

= 120° + ∠C = 180°

= ∠C = 60°

We have one side equal. To prove that other 2 sides are also equal,

∴ BC = CA = 5 cm (Sides opposite to equal angles are equal)

AB = BC = CA = 5 cm

∠A = ∠B = ∠C = 60°

Hence, justified.

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