NCERT Solutions for Class 9 Maths Constructions [Old Book]

NCERT
SOLUTIONS
CLASS - 9th

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Class : 9th
Subject : Maths
Chapter : 11
Chapter Name : CONSTRUCTIONS

Exercise 11.1

Q1 Construct an angle of 900 at the initial point of a given ray and justify the construction.

Answer. The below given steps will be followed to construct an angle of 900.
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which
intersects PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the
previously drawn att at S.
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc
at T (see �gure).
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(v) Join PLJ, which is the required ray making 900 with the given ray PQ.

Justi�cation of Construction:
We can justify the construction, if we can prove ∠UPQ = 90∘
For this, join PS and PT.

We have, ∠SPQ = ∠TPS = 60∘ . In (iii) and (iv) steps of this construction, PU was
drawn as the bisector of ∠TPS .

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

∴ ∠UPS = 12 ∠ TPS = 12 × 60∘ = 30∘
Also, ∠UPQ = ∠SPQ + ∠UPS
= 60∘ + 30∘
= 90∘

Page : 191 , Block Name : Exercise 11.1

Q2 Construct an angle of 450 at the initial point of a given ray and justify the construction.

Answer. The below given steps will be followed to construct an angle of 450.
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which
intersects PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the
previously drawn arc at S.
(iii) Taking S as centre and With the same radius as before, draw an arc intersecting the arc
at T (see �gure).
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(v) join PU. Let it intersect the arc at point V.
(vi) From R and V, draw arcs with radius more than 1/2 RV to intersect each other at W. Join
PW.
PW is the required ray making 450 with PQ.

Justi�cation of Construction:
We can justify the construction, if we can prove ∠WPQ = 45∘
For this, join PS and PT.

We have, ∠SPQ = ∠TPS = 60∘ . In (iii) and (iv) steps of this construction, PU was
drawn as the bisector of ∠TPS .

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

∘
∴ ∠UP S = 12 ∠ TPS = 602 = 30∘
Also, ∠UP Q = ∠SP Q + ∠UP S
= 60∘ + 30∘
= 90∘
In step (vi) of this construction, PW was constructed as the bisector of ∠UPQ.
∘
∴ ∠WPQ = 12 ∠UPQ = 902 = 45∘

Page : 191 , Block Name : Exercise 11.1

Q3 Construct the angles of the following measurements:
∘
(i) 30∘ (ii) 22 12 (iii) 15∘

Answer. (i)300
The below given steps will be followed to construct an angle of 300.
Step I: Draw the given ray PQ. Taking P as centre and with some radius, draw an arc of a
circle which intersects PQ at R.
Step II: Taking R as centre and with the same radius as before, draw an arc intersecting the
previously drawn arc at point S.
Step III: Taking R and S as centre and with radius more than 2 RS, draw arcs to intersect
each other at T. Join PT which is the required ray making 300 with the given ray PQ.

(ii) 22 12 ∘
The below given steps will be followed to construct an angle of 22 12 ∘.
(1) Take the given ray PQ. Draw an arc of some radius, taking point p as its centre, which
intersects PQ at R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the
previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at
T (see �gure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join Let it intersect the arc at point V.
(6) From R and V, draw arcs with radius more than 12 RV to intersect each other at W. Join
PW.
(7) Let it intersect the arc at X. Taking X and R as centre and radius more than 12 RX, draw
arcs to intersect each other at Y.

Joint PY which is the required ray making 22 12 ∘ with the given ray PQ.

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

(iii) 15∘
The below given steps will be followed to construct an angle of 15∘ .
Step I: Draw the given ray PQ. Taking P as centre and with some radius, draw an arc of a
circle which intersects PQ at R.
Step II: Taking R as centre and with the same radius as before, draw an arc intersecting the
previously drawn arc at point S.
Step Ill: Taking R and S as centre and with radius more than 1/2 RS, draw arcs to intersect
each other at T. Join PT.
Step IV: Let it intersect the arc at U. Taking IJ and R as centre and with radius more than 1/2
RU, draw an arc to intersect each other at V. Join PV which is the required ray making 15∘
with the given ray PQ.

Page : 191 , Block Name : Exercise 11.1

Q4 Construct the following angles and verify by measuring them by a protractor:
(i) 75∘ (ii) 105∘ (iii) 135∘

Answer. (i) 75∘
The below given steps will be followed to construct an angle of 75∘ .
(1) Take the given ray PQ. Draw an arc of some radius taking point p as its centre, which
intersects PQ at R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the
previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at
T (see �gure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join PLI. Let it intersect the arc at V. Taking S and V as centre, draw arcs with radius
more than 1/2 SV. Let those intersect each other at W. Join PW which is the required ray
making 75∘ with the given ray PQ.

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

The angle so formed can be measured with the help of a protractor. It comes to be 75∘ .

(ii) 105∘
The below given steps will be followed to construct an angle of 105∘
(1) Take the given ray PQ. Draw an arc of some radius taking point p as its centre, which
intersects PQ at R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the
previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at
T (see �gure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join PLI. Let it intersect the arc at V. Taking T and V as centre, draw arcs with radius
more than 1/2 TV. Let these arcs intersect each other at W. Join PW which is the required ray
making 105∘ with the given ray PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 105∘ .

(iii) 135∘
The below given steps will be followed to construct an angle of 135∘
(1) Take the given ray PQ. Extend PQ on the opposite side of Q. Draw a semi-circle of some
radius taking point P as its centre, which intersects PQ at R and W.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the
previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at
T (see �gure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

(5) Join PU. Let it intersect the arc at V. Taking V and W as centre and with radius more than
1/2 VW, draw arcs to intersect each other at X. Join PX, which is the required ray making
135∘ with the given line PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 135∘ .

Page : 191 , Block Name : Exercise 11.1

Q5 Construct an equilateral triangle, given its side and justify the construction.

Answer. Let us draw an equilateral triangle of side 5 cm. We know that all sides of an
equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be 5 cm. We
also know that each angle Of an equilateral triangle is 60∘ .
The below given steps Will be followed to draw an equilateral triangle Of 5 cm side.
Step I: Draw a line segment AB of 5 cm length. Draw an arc of some radius, while taking A as
its centre. Let it intersect AB at P.
Step II: Taking P as centre, draw an arc to intersect the previous arc at E. join AE.
Step Ill: Taking A as centre, draw an arc of 5 cm radius, which intersects extended line
segment AE at C. Join AC and BC. ΔABC is the required equilateral triangle of side 5 cm.

Justi�cation of Construction:
We can justify the construction by showing ABC as an equilateral triangle i.e., AB =
BC = AC = 5cm and ∠A = ∠B = ∠C = 60∘ .
In △ABC, we have AC = AB = 5cm and ∠A = 60∘ .
since AC = AB
∠B = ∠C (Angles opposite to equal sides of a triangle)

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

In △ABC,
∠A + ∠B + □C = 180∘ ( Angle sum property of a triangle)
□60∘ + □C + □C = 180∘
□60∘ + 2□C = 180∘
□2□C = 180∘ − 60∘ = 120∘
□□C = 60∘
□□B = □C = 60∘
We have, □A = □B = □C = 60∘ … (1)
□□A = □B and □A = □C
□BC = AC and BC = AB( Sides opposite to equal angles of a triangle)
□AB = BC = AC = 5cm … (2)
from equations (1) and (2), /DeltaABC is an equilateral triangle.

Page : 191 , Block Name : Exercise 11.1

Exercise 11.2

Q1 Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.

Answer. The below given steps will be followed to construct the required triangle.
step I: Draw a line segment BC of 7 cm. At point B, draw an angle of 750, say □XBC.
step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX.
Step III: Join DC and make an angle DCY equal to □BDC.
step IV: Let CY intersect 3X at A. ΔABC is the required triangle.

Page : 195 , Block Name : Exercise 11.2

Q2 Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB – AC = 3.5 cm.

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Answer. The below given steps will be followed to draw the required triangle.
Step I: Draw the line segment BC = 8 cm and at point B, make an angle of 450, say □XBC.
Step II: Cut the line segment BD 3.5 cm (equal to AB - AC) on ray BX.
Step Ill: Join DC and draw the perpendicular bisector PQ of DC.
Step IV: Let it intersect BX at point A. Join AC. ΔABC is the required triangle.

Page : 195 , Block Name : Exercise 11.2

Q3 Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm.

Answer. The below given steps will be followed to construct the required triangle.
Step I: Draw line segment QR of 6 cm. At point Q, draw an angle of 600, say ΔXQR.
Step II: Cut a line segment QS of 2 cm from the line segment QT extended in the opposite
side of line segment XQ (As PR > PQ and PR — PQ 2 cm). Join SR.
Step Ill: Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P.
Join PQ, PR.
ΔPQR is the required triangle.

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Page : 195 , Block Name : Exercise 11.2

Q4 Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

Answer. The below given steps will be followed to construct the required triangle.
Step I: Draw a line segment AB of tl cm.
(As XY + YZ + ZX = 11 cm)
Step II: Construct an angle, □PAB, of 300 at point A and an angle, □QBA, of 900 at point 3.
Step Ill: Bisect □PAB and □QBA. Let these bisectors intersect each other at point X.
Step IV: Draw perpendicular bisector ST of AX and UV of BX.
step V: Let ST intersect Ad at Y and UV intersect AB at Z.
W, XZ.
ΔXYZ is the required triangle.

Page : 195 , Block Name : Exercise 11.2

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Q5 . Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side
is 18 cm.

Answer. The below given steps will be followed to construct the required triangle.
Step I: Draw line segment AB of 12 cm. Draw a ray AX making 900 with AB.
Step II: Cut a line segment AD of 18 cm (as the sum of the other two sides is 18) from ray
AX.
Step III: Join DB and make an angle DBY equal to ADB.
step IV: Let BY intersect AX at C. Join AC, BC.
△ABC is the required triangle.

Page : 195 , Block Name : Exercise 11.2

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