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

## Class 9 Mathematics by Full Circle

Created by: Full Circle

## 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 Full Circle.
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:
(i) A protractor
(ii) A pair of compasses
(iii) A pair of set squares
(iv) A pair of dividers

EXERCISE 11.1
Ques 1: Construct an angle of 90º at the initial point of a given ray and justify the construction.
Sol:
Steps of construction:
I. Draw a ray OA.
II . Taking O as centre and suitable radius, draw a semicircle, which cuts OA at B.
III. Keeping the radius same, divide the semicircle into three equal parts such that IV. Draw V. 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

Ques 2: Construct an angle of 45º at the initial point of a given ray and justify the construction.
Sol: Steps of construction:
I. Draw a ray .
II. Taking O as centre and with a suitable radius, draw a semicircle such that it intersects at B. III. 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.
IV. Divide into two equal parts, such that V. 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]
Since ∠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

Ques 3: Construct the angles of the following measurements:
(i) 30º
(ii) (iii) 15º
Sol: (i) Angle of 30º
Steps of construction:
I . Draw a ray OA.
II . With O as centre and a suitable radius, draw an arc, cutting at B. III. With centre at B and the same radius as above, draw an arc to cut the previous arc at C.
IV. Join and produce, such that ∠BOC = 60º.
V. bisector of ∠BOC, such that Thus, ∠BOD = 30º

(ii) Angle of Steps of construction:
I . Draw a ray II. Draw an angle ∠AOB = 90º
III. Draw OC, the bisector of ∠AOB, such that  IV. Now, draw OD, the bisector of ∠AOC, such that Thus,  (iii) Angle of 15º
Steps of construction:
I. Draw a ray .
II. Construct ∠AOB = 60º.
III. Draw the bisector of ∠AOB, such that i.e. ∠AOC = 30º IV. Draw the angle bisector of ∠AOC such that Thus, ∠AOD = 15º

Ques 4: Construct the following angles and verify by measuring them by a protractor:
(i) 75º
(ii) 105º
(iii) 135º
Sol: (i) Angle of 75º:
Hint: 75º = 60º + 15º
Steps of construction:
I. Draw .
II. With O as centre and having a suitable radius, draw an arc which meets III. With centre B and keeping the radius same, mark a point C on the previous arc.
IV. With centre C and the same radius, mark another point D on the arc of step II. V. Draw the bisector of such that ∠COP VI. Draw , the bisector of ∠COP, such that ∠COQ = 15º
Thus, ∠BOQ = 60º + 15º = 75º
or ∠AOQ = 75º.

(ii) Angle of 105º:
Hint: 105º = 90º + 15º
Steps of construction:
I. Draw II. With centre O and having a suitable radius, draw an arc which meet OA at B.
III. With centre B and keeping the same radius, mark a point C on the arc of step II.
IV. With centre C and keeping the same radius, mark another point D on the arc of step II. V. Draw OP, the bisector of VI. Draw OQ, the bisector of .
Thus, ∠AOQ = 105º

(iii) Angle of 135º:
Hint: 120º + 15º = 135º
Steps of construction:
I. Draw a ray II. With centre O and having a suitable radius draw an arc to meet OP at A.
III. Keeping the same radius and starting from A, mark points Q, R and S on the arc of step II.
IV. Draw the bisector of V. Draw the bisector of  Thus, ∠POM = 135º.

Ques 5: Construct an equilateral triangle, given its side and justify the construction.
Sol:
Let us construct an equilateral triangle, each of whose side = PQ Steps of construction:
I. Draw a ray . II. Taking O as centre and radius equal to PQ, draw an arc to cut OA at B such that OB = PQ
III. Taking B as centre and radius = OB, draw an arc, to intersect the previous arc at C.
IV. 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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