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# Facts that Matter, Ex 11.1 NCERT Solutions- Constructions Class 9 Notes | EduRev

## Class 9 : Facts that Matter, Ex 11.1 NCERT Solutions- Constructions Class 9 Notes | EduRev

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.
All you need of Class 9 at this link: Class 9

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

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.
• Divide into 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

Q3. Construct the angles of the following measurements:
(a) 30º
(b) (c) 15º
Ans:

(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 that i.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 of V. 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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