The document Facts that Matter, Ex 11.1 NCERT Solutions- Constructions Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by VP Classes.

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**Basic Constructions**

A geometrical construction means to draw geometrical figures, such as an angle, a circle, a triangle, a quadrilateral, and a polygon, etc.**We normally use all or some of the following instruments for drawing geometrical figures:**

- A protractor
- A pair of compasses
- A pair of set squares
- A pair of dividers
- A graduated scale

**Exercise 11.1**

**Q1. Construct an angle of 90º at the initial point of a given ray and justify the construction.****Ans:** **Steps of construction:**

- Draw a ray OA.
- Taking O as the centre and suitable radius, draw a semicircle, which cuts OA at B.
- Keeping the radius the same, divide the semicircle into three equal parts such that
- Draw
- Draw, the bisector of ∠COD.

Thus, ∠AOF = 90º.

**Justification:**

∵ O is the centre of the semicircle and it is divided into 3 equal parts.

∴

∠BOC = ∠COD = ∠DOE**∵ Equal chords subtend equal angles at the centre**

∴ ∠BOC + ∠COD + ∠DOE = 180º

∠BOC + ∠BOC + ∠BOC = 180º

3∠BOC = 180°

∴ ∠BOC = 60º

Similarly, ∠COD = 60º and ∠DOE = 60º

∵ OF is the bisector of ∠COD.

∴ = 30º

Now, ∠BOC + ∠COF = 60º + 30º

∠BOF = 90º or ∠AOF = 90º**Q2. Construct an angle of 45º at the initial point of a given ray and justify the construction.****Ans:** **Steps of construction:**

- Draw a ray.
- Taking O as the centre and with a suitable radius, draw a semicircle such that it intersects at B.
- Taking B as centre and keeping the same radius, cut the semicircle at C. Similarly, cut the semicircle at D and E, such that . Join OC and produce.
- Divideinto two equal parts, such that
- Draw OG, the angle bisector of ∠FOC.

Thus, ∠BOG = 45º or ∠AOG = 45º**Justification:**∵

∴ ∠BOC = ∠COD = ∠DOE

∵ Equal chords subtend equal angles at the centre

∴ ∠BOC + ∠COD + ∠DOE = 180º

∠BOC = 60º

∵ is the bisector of ∠BOC/

∴ .. (1)

Also, is the bisector of ∠COF.

∴ ... (2)

Adding (1) and (2), we get∠COF + ∠FOG = 30º + 15º = 45º

∠BOF + ∠FOG = 45º [∵ ∠COF = ∠BOF]

∠BOG = 45º

**(a) **Angle of 30º**Steps of construction:**

- Draw a ray OA.
- With O as the centre and a suitable radius, draw an arc, cutting at B.
- With centre at B and the same radius as above, draw an arc to cut the previous arc at C.
- Join and produce, such that ∠BOC = 60º.
- bisector of ∠BOC, such that

Thus, ∠BOD = 30º**(b)** Angle of **Steps of construction:**

- Draw a ray
- Draw an angle ∠AOB = 90º
- Draw OC, the bisector of ∠AOB, such that
- Now, draw OD, the bisector of ∠AOC, such that

Thus,

**(c)** Angle of 15º**Steps of construction:**

- Draw a ray.
- Construct ∠AOB = 60º.
- Draw the bisector of ∠AOB, such thati.e. ∠AOC = 30º
- Draw the angle bisector of ∠AOC such that

Thus, ∠AOD = 15º**Q4. Construct the following angles and verify by measuring them by a protractor:****(a) 75º****(b) 105º****(c) 135º****Ans:**

**(a)** Angle of 75º (**Hint:** 75º = 60º + 15º)**Steps of construction:**

- Draw .
- With O as centre and having a suitable radius, draw an arc which meets
- With centre B and keeping the radius same, mark a point C on the previous arc.
- With centre C and the same radius, mark another point D on the arc of step 2.
- Draw the bisector of such that ∠COP
- Draw , the bisector of ∠COP, such that ∠COQ = 15º

Thus, ∠BOQ = 60º + 15º = 75ºor ∠AOQ = 75º.

**(b)** Angle of 105º (**Hint:** 105º = 90º + 15º)

**Steps of construction:**

- Draw
- With centre O and having a suitable radius, draw an arc that meets OA at B.
- With centre B and keeping the same radius, mark a point C on the arc of step 2.
- With centre C and keeping the same radius, mark another point D on the arc of step 2.
- Draw OP, the bisector of
- Draw OQ, the bisector of.

Thus, ∠AOQ = 105º

**(c)** Angle of 135º (**Hint**: 120º + 15º = 135º)**Steps of construction:**

- Draw a ray
- With centre O and having a suitable radius draw an arc to meet OP at A.
- Keeping the same radius and starting from A, mark points Q, R and S on the arc of step 2.
- Draw the bisector ofV. Draw the bisector of

Thus, ∠POM = 135º.**Q5. Construct an equilateral triangle, given its side and justify the construction.****Ans:** Let us construct an equilateral triangle, each of whose side = PQ**Steps of construction:**

- Draw a ray.
- Taking O as centre and radius equal to PQ, draw an arc to cut OA at B such that OB = PQ
- Taking B as centre and radius = OB, draw an arc, to intersect the previous arc at C.
- Join OC and OB.

Thus, ΔOBC is the required equilateral triangle.**Justification:**∵ The are drawn with the same radius.

∴ ⇒

∵ Chords corresponding to equal arcs are equal.

∵ OC = OB = BC

∴ ΔOBC is an equilateral triangle.

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