Constructions

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NCERT Solutions for Chapter 11 Constructions Class 9 Maths

Book Solutions

1

Construct an angle of 90º at the initial point of a given ray and justify the construction.

Answer

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.

Now, ∠BOC + ∠COF = 60º + 30º
⇒ ∠BOF = 90º
or ∠AOF = 90º

Exercise 11.1 Page Number 191

2

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

Answer

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. Divideinto two equal parts, such that
V. Draw OG, the angle bisector of ∠FOC.
Thus, ∠BOG = 45º
or ∠AOG = 45º

 

Justification:

Adding (1) and (2), we get
∠COF + ∠FOG = 30º + 15º = 45º
∠BOF + ∠FOG = 45º   [∵ ∠COF = ∠BOF]
⇒ ∠BOG = 45º

Exercise 11.1 Page Number 191

3

Construct the angles of the following measurements: 
(i) 30º 

(iii) 15º

Answer

(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. Draw   bisector of ∠BOC, such that

Steps of construction:
I . Draw a ray
II. Draw an angle ∠AOB = 90º
III. Draw OC, the bisector of ∠AOB, such that

(iii) Angle of 15º
Steps of construction:
I. Draw a ray .
II. Construct ∠AOB = 60º.
III. Draw  the bisector of ∠AOB, such that
Exercise 11.1 Page Number 191

4

Construct the following angles and verify by measuring them by a protractor: 
(i) 75º 
(ii) 105º 
(iii) 135º

Answer

(i) Angle of 75º:
Hint: 75º = 60º + 15º

Steps of constructions:

I. A ray OY is drawn.
II. An arc BAE is drawn with O as a centre.
III. With E as a centre, two arcs are A and C are made on the arc BAE.
IV. With A and B as centres, arcs are made to intersect at X and ∠XOY = 90° is made.
V. With A and C as centres, arcs are made to intersect at D 
VI OD is joined and and ∠DOY = 75° is constructed.
Thus, ∠DOY is the required angle making 75° with OY.
 
(ii) Angle of 105º:
Steps of construction:
I. Draw .
II. With O as centre and having a suitable radius, draw an arc which meets at B
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.

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.

Exercise 11.1 Page Number 191

5

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

Answer

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:
Exercise 11.1 Page Number 191

1

Construct a triangle ABC in which BC = 7 cm, ∠B = 75º and AB + AC = 13 cm.

Answer

Steps of construction:
I . Draw a ray BX.

V.  Join D and C.
VI. Bisect DC such that the bisector of DC meets BD at A.
VII. Join AC.
Thus, ΔABC is the required triangle.

Exercise 11.2 Page Number 195

2

Construct a triangle ABC in which BC = 8 cm, ∠ B = 45º and AB ∠ AC = 3.5 cm.

Answer

Steps of construction:
I . Draw a ray
II. From    , cut off BC = 8 cm.
III. Construct ∠CBY = 45º.

V.  Join D and C.
VI. Draw PQ, perpendicular bisector of DC, which intersects at A.
VII. Join AC.
Thus, ABC is the required triangle.
Exercise 11.2 Page Number 195

3

Construct a triangle PQR in which QR = 6 cm, ∠Q = 60º and PR ∠ PQ = 2cm.

Answer

Steps of construction:
I . Draw a ray .
II. From , cut off QR = 6 cm.
III. Construct a line YQY' such that ∠RQY = 60º.

IV. Cut off QS = 2 cm (from QY').
V. Join ‘S’ and ‘R’.
VI. Draw MN, perpendicular bisector of SR, which intersects QY at P.
VII. Join P and R.
Thus, PQR is the required triangle.

Exercise 11.2 Page Number 195

4

Construct a triangle XYZ in which ∠ Y = 30º , ∠ Z = 90º and XY + YZ + ZX = 11 cm.

Answer

Steps of construction:

I . Draw a line segment AB = 11 cm = (XY + YZ + ZX)
II. Construct ∠BAP = 30º = ∠Y
III. Construct ∠ABQ = 90º = ∠Z
IV. Draw the bisector of ∠BAP.
V. Draw  , the bisector of ∠ABQ,
such that   and  intersect each other at X.
VI. Draw perpendicular bisector of AX, which intersects AB at Y.
VII. Draw perpendicular bisector of XB, which intersects AB at Z.
VIII. Join XY and XZ.
Thus, XYZ is the required triangle.

Exercise 11.2 Page Number 195

5

Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.

Answer

Steps of construction:
I . Draw  = 12 cm.
II. Construct ∠CBY = 90º.

III. From  cut off BX = 18 cm.
IV. Join CX.
V. Draw PQ, the perpendicular bisector of CX, such that PQ meets BX at A.
VI. Join AC.
Thus, ABC is the required triangle.
Exercise 11.2 Page Number 195

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