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Class IX Chapter 11 – Constructions
Maths
Exercise 11
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Answer:
The below given steps will be followed to construct an angle of 90°
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Draw an arc of some radius taking point P as its centre,
which intersects PQ at R
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(iii) Taking S as centre and with the same radius as before, draw an arc intersecting
the arc at T (see figure)
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(v) Join PU, which is the required ray making 90° with the given ray PQ
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For this, join PS and PT
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In (iii) and (iv) steps of this construction, PU was
drawn as the bisector of
TPS
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Answer:
The below given steps will be followed to construct an angle of 45°
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Draw an arc of some radius taking point P as its centre,
which intersects PQ at R
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(iii) Taking S as centre and with the same radius as before, draw an arc intersecting
the arc at T (see figure)
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(v) Join PU
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(vi) From R and V, draw arcs with radius more than

RV to intersect each other at W
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PW is the required ray making 45° with PQ
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WPQ = 45°
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In (iii) and (iv) steps of this construction, PU was
drawn as the bisector of
TPS
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∠ UPQ

Question 3:
Construct the angles of the following measurements:

(i) 30° (ii)

(iii) 15° Answer:

(i)30°
The below given steps will be followed to construct an angle of 30°
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Taking P as centre and with some radius, draw an arc
of a circle which intersects PQ at R
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Step III: Taking R and S as centre and with radius more than
RS, draw arcs to
intersect each other at T
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(ii)
The below given steps will be followed to construct an angle of


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Draw an arc of some radius, taking point P as its centre,
which intersects PQ at R
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(3) Taking S as centre and with the same radius as before, draw an arc intersecting
the arc at T (see figure)
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(5) Join PU
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(6) From R and V, draw arcs with radius more than

RV to intersect each other at W
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(7) Let it intersect the arc at X
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RX, draw arcs to intersect each other at Y
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(iii) 15°
The below given steps will be followed to construct an angle of 15°
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Taking P as centre and with some radius, draw an arc
Step II: Taking R as centre and with the same radius as before, draw an arc
intersecting the previously drawn arc at point S
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Join PT
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Taking U and R as centre and with radius more

than
ray making 15° with the given ray PQ
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RU, draw an arc to intersect each
Join PV which is the required

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°
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Draw an arc of some radius taking point P as its centre,
which intersects PQ at R
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(3) Taking S as centre and with the same radius as before, draw an arc intersecting
the arc at T (see figure)
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(5) Join PU
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Taking S and V as centre, draw arcs with

radius more than
SV
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Join PW which is the
required ray making 75° with the given ray PQ
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It comes to be
75º
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(1) Take the given ray PQ
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(2) Taking R as centre and with the same radius as before, draw an arc intersecting
the previously drawn arc at S
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(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U
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Let it intersect the arc at V
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Let these arcs intersect each other at W
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The angle so formed can be measured with the help of a protractor
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(iii) 135°
The below given steps will be followed to construct an angle of 135°
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Extend PQ on the opposite side of Q
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(2) Taking R as centre and with the same radius as before, draw an arc intersecting
the previously drawn arc at S
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(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U
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Let it intersect the arc at V
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Join PX, which is the required
ray making 135°with the given line PQ
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It comes to be
135º
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We know that all sides of an equilateral
triangle are equal
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We also
know that each angle of an equilateral triangle is 60º
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Step
I: Draw a line segment AB of 5 cm length
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Let it intersect AB at P
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Join AE
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Join AC and BC
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Justification of Construction:
We can justify the construction by showing ABC as an equilateral triangle i
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, AB = BC
= AC = 5 cm and A = B = C = 60°
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Since AC = AB,
B = C (Angles opposite to equal sides of a triangle)
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°
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(2)
From equations (1) and (2), ∆ABC is an equilateral triangle
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2
Question 1:

Construct a triangle ABC in which BC = 7 cm,

B = 75° and AB + AC = 13 cm
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Step I: Draw a line segment BC of 7 cm
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Step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX
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Step IV: Let CY intersect BX at A
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Question 2:
Construct a triangle ABC in which BC = 8 cm,

B = 45° and AB − AC = 3
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Answer:
The below given steps will be followed to draw the required triangle
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Step II: Cut the line segment BD = 3
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Step III:
Join DC and draw the perpendicular bisector PQ of DC
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Join AC
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Question 3:
Construct a triangle PQR in which QR = 6 cm,

Q = 60° and PR − PQ = 2 cm

Answer:
The below given steps will be followed to construct the required triangle
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At point Q, draw an angle of 60°, say XQR
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(As PR > PQ and PR − PQ = 2 cm)
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Step
III:
Draw perpendicular bisector AB of line segment SR
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Join PQ, PR
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Construct a triangle XYZ in which
Question 4:

Y = 30°,

Z = 90° and XY + YZ + ZX = 11 cm
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Step I: Draw a line segment AB of 11 cm
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Step III: Bisect

PAB and

PAB, of 30° at point A and an angle,

QBA, of 90° at

QBA
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Step IV: Draw perpendicular bisector ST of AX and UV of BX
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Join XY, XZ
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Question 5:
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other
side is 18 cm
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Step I: Draw line segment AB of 12 cm
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Step II:
Cut a line segment AD of 18 cm (as the sum of the other two sides is 18) from ray AX
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Step IV: Let BY intersect AX at C
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∆ABC is the required triangle

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