Bihar Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th Maths Solutions Chapter 11 Constructions Ex 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 :

- Draw a ray OA.
- With its initial point O as centre and any radius, draw an arc CDE, cutting OA at C.

- With centre C and same radius (as in step 2), draw an arc, cutting the arc CDE at D.
- With D as centre and the same radius, draw an arc cutting the arc CDE at E.
- With D and E as centres, and any. convenient radius (more than \(\frac { 1 }{ 2 }\) DE ), draw two arcs intersecting at P.
- Join OP. Then ∠AOP = 90°.

Justification :

By construction, OC = CD = OD

∴ ∆ OCD is an equilateral triangle. So, ∠COD = 60°.

Again, OD = DE = EO

∴ ∆ ODE is also an equilateral triangle. So, ∠DOE = 60°.

Since OP bisects ∠DOE, so ∠POD = 30°.

Now, ∠AOP = ∠COD + ∠DOP = 60° + 30° = 90°.

Question 2.

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

Solution:

Steps of Construction

- Draw a ray OA.
- With O as centre and any suitable radius draw an arc cutting OA at B.

- With B as centre and same radius cut the previous drawn arc at C and then with C as centre and same radius cut the arc at D.
- With C as centre and radius more than half CD draw an arc.
- With D as centre and same radius draw another arc to ut the previous arc at E.
- Join OE. Then ∠AOE = 90°.
- Draw the bisector OF of ∠AOE. Then ∠AOF = 45°.

Justification :

By construction, ∠AOE = 90° and OF is the bisector of ∠AOE

∴ ∠AOF = \(\frac { 1 }{ 2 }\)∠AOE = \(\frac { 1 }{ 2 }\) x 90° = 45°.

Question 3.

Construct the angles of the following measurements :

(i) 30°

(ii) 22\(\frac { 1° }{ 2 }\)

(iii) 15°.

Solution:

(i) Steps of Construction :

- Draw a ray OA.
- With its initial point O as centre and anyx radius, draw an arc, cutting OA at C.

- With centre C and same radius (as in step 2). Draw an arc, cutting the arc of step 2 in D.
- With C and D as centres, and any convenient radius (more than \(\frac { 1 }{ 2 }\) CD), draw two arcs intersecting at B.
- Join OB. Then ∠AOB = 30°.

(ii) Steps of Construction :

- Draw an angle AOB = 90°.
- Draw the bisector OC of ∠AOB, then ∠AOC = 45°.

- Bisect ∠AOC, such that ∠AOD = ∠COD = 22.5°.

Thus, ∠AOD = 22.5°.

(iii) Steps of Construction :

- Construct an ∠AOB = 60°.
- Bisect ∠AOB so that ∠AOC = ∠BOC = 30°.

- Bisect ∠AOC, so that ∠AOD = ∠COD = 15°.

Thus, ∠AOD = 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 :

- Draw a ray OA.
- Construct ∠AOB = 60°.
- Construct ∠AOP = 90°.
- Bisect ∠BOP so that

- So, we obtain ∠AOQ = ∠AOP + ∠BOQ = 60° + 15° = 75°.

Verification :

On measuring ∠AOQ, with the protractor, we find ∠AOQ – 75°.

(ii) Steps of Construction :

- Draw a line segment XY.
- Construct ∠XYT = 120° and ∠XYS = 90°, so that ∠SYT = ∠XYT – ∠XYS = 120° – 90° = 30°

- Bisect angle SYT, by drawing its bisector YZ.
- Then ∠XYZ is the required angle of 105°.

(iii) Steps of Construction :

- Draw ∠AOE = 90°. Then, ∠LOE = 90°.

- Draw the bisector OF of ∠LOE.
- Then, ∠AOF = 135°.

Question 5.

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

Solution:

Let us draw an equilateral triangle of side 4.6 cm (say).

Steps of Construction :

- Draw BC = 4.6 cm.
- With B and C as centres and radii equal to BC = 4.6 cm, draw two arcs on the same side of BC, intersecting each- other at A.

- Join AB and AC.
- Then, ABC is the required equilateral triangle.

Justification :

Since by construction :

AB = BC = CA = 4.6 cm.

∴ ∆ ABC is an equilateral triangle.