CONSTRUCTIONS

Let use see how this method gives us the required

CONTENTS division.
 To divide a line segment in a given ratio. Since A3C is parallel to A5B, therefore,

 To construct a triangle similar to a AA3 AC


A 3A 5 CB
given triangle as per given scale factor.
(By the Basic Proportionality Theorem)
 To construct a tangent to a circle at a
AA3 3 AC 3
given point on it (using the centre of By construction,  . Therefore,  .
A 3A 5 2 CB 2
the circle).
This shows that C divides AB in the ratio 3 : 2.
 To construct two tangents to a circle Alternative Method
from a point outside the circle (using
Steps of Construction :
the center of the circle).
1. Draw any ray AX making an acute angle with
AB.
TO DIVIDE A LINE SEGMENT IN A GIVEN 2. Draw a ray BY parallel to AX by making

RATIO ABY equal to BAX.

Given a line segment AB, we want to divide it in 3. Locate the points A1, A2, A3 (m = 3) on AX and
the ratio m : n, where both m and n are positive B1, B2 (n = 2) on BY such that
integers. To help you to understand it, we shall take AA1 = A1A2 = A2A3 = BB1 = B1B2.
m = 3 and n = 2.
4. Join A3B2.
Steps of Construction :
Let it in intersect AB at a point C (see figure)
1. Draw any ray AX, making an acute angle with
AB. Y
B2
2. Locate 5(= m + n) points A1, A2, A3, A4 and A5 on B1
AX so that AA1 = A1A2 = A2A3 = A3A4 = A4A5. A C B
3. Join BA5. A1
A2
4. Through the point A3 (m = 3), draw a line A3 X
parallel to A5B (by making an angle equal to
AA5B) at A3 intersecting AB at the point C
Then AC : CB = 3 : 2
(see figure). Then, AC : CB = 3 : 2.
Whey does this method work ? Let us see.
A C
B
Here AA3C is similar to AB2C. (Why ?)
A1
A2 AA3 AC
A3 Then 
A4 BB2 BC
X
A5
 AA3 3 Step VIII : Since we have to construct a
Since by construction,  , triangle each of whose sides is two-third of the
BB2 2
corresponding sides of ABC. So, take two
AC 3 parts out of three equal parts on AX i.e. from
therefore, 
BC 2 point A2, draw
In fact, the methods given above work for dividing A2B' || A3B, meeting AB at B'.
the line segment in any ratio.
C
We now use the idea of the construction above for
constructing a triangle similar to a given triangle C'
whose sides are in a given ratio with the
corresponding sides of the given triangle.
A B
B'
TO CONSTRUCT A TRIANGLE SIMILAR
TO A GIVEN TRIANGLE AS PER GIVEN A1
SCALE FACTOR A2
A3
Scale factor means the ratio of the sides of the
triangle to be constructed with the corresponding X
sides of the given triangle.
Step IX : From B', draw B'C' || BC, meeting
This construction involves two different situations : AC at C'.
(i) The triangle to be constructed is smaller than AB'C' is the required triangle, each of the
the given triangle, here scale factor is less whose sides is two-third of the corresponding
than 1. sides of ABC.
(ii) The triangle to be constructed is bigger than the Justification : Since B'C' || BC.
given triangle, here scale factor is greater than 1.
So, ABC ~ AB'C'
EXAMPLES 
B' C' AC' AB' 2  AB' 2 
        
Ex.1 Construct a ABC in which AB = 4 cm, BC AC AB 3  AB 3 
BC = 5 cm and AC = 6 cm. Now, construct a
  Let ABC be the given triangle and we want to
triangle similar to ABC such that each of its
construct a triangle similar to ABC such that
sides is two-third of the corresponding sides of th
ABC. Also, prove your assertion. m
each of its sides is   of the corresponding
Sol. Steps of construction n
sides of ABC such that m < n. We follow the
Step I : Draw a line segment AB = 4 cm. following steps to construct the same.
Step II : With A as centre and Steps of construction when m > n.
radius = AC = 6 cm, draw an arc.
Step I : construct the given triangle by using
Step III : With B as centre and radius the given data.
= BC = 5 cm, draw another arc, intersecting the
arc drawn in step II at C. Step II : Take any of the three sides of the
given triangle and consider it as the base. Let
Step IV : Join AC and BC to obtain ABC. AB be the base of the given triangle.
Step V : Below AB, make an acute angle Step III : At one end, say A, of base AB
BAX. construct an acute angle BAX below base AB
Step VI : Along AX, mark off three points i.e. on the opposite side of the vertex C.
(greater of 2 and 3 in 2/3) A1, A2, A3 such that Step IV : Along AX, mark-off m (large of m
AA1 = A1A2 = A2A3. and n) points A1, A2,...,Am on AX such that
Step VII : Join A3B. AA1 = A1A2 = ….. = Am–1Am.
 Step V : Join An to B and draw a line through Ex.2 Draw a triangle ABC with side BC = 7 cm, 
Am Parallel to AnB, intersecting the extended   B = 45º, A = 105º. Then construct a triangle
line segment AB at B'. whose sides are (4/3) times the corresponding
Step VI : Draw a line through B' parallel to sides of ABC.
BC intersecting the extended line segment AC Sol. In order to construct ABC, we follow the
at C'. following steps:
C' Step I : Draw BC = 7 cm.
Step II : At B construct CBX = 45º and at C
C construct
BCY = 180º – (45º – 105º) = 30º
Suppose BX and CY intersect at A. ABC so
B B' obtained is the given triangle. To construct a
A
A1 triangle similar to ABC,
A2
we follow the following steps.
An X Step I : Construct an acute angle CBZ at B
on opposite side of vertex A of ABC.
Am
Step II : Mark-off four (greater 4 and 3 in 4/3)
Step VII : AB'C' so obtained is the required points
triangle.
B1, B2, B3, B4 on BZ such that
Justification : For justification of the above
construction consider triangles ABC and AB'C'. BB1 = B1B2 = B2B3 = B3B4.
In these two triangles, we have Step III : Join B3 (the third point) to C and
BAC = B'AC' draw a line through B4 parallel to B3C,
intersecting the extended line segment BC at C'.
ABC = AB'C' [ B'C' || BC]
X
So, by AA similarity criterion, we have
Y
ABC ~ AB'C' A'
A
AB BC AC
   ....(i)
AB' B' C' AC'
45° 30°
B C'
In  A Am B', AnB || Am B'. C
B1
AB AA n
    B2
BB' A n A m
B3
BB' A n A m BB' m  n B4
       Z
AB AA n AB n

AB' AB m  n AB' mn Step IV : Draw a line through C' parallel to CA

     –1= intersecting the extended line segment BA
AB n AB n
at A'.
AB' m Triangle A'BC' so obtained is the required
    ....(ii)
AB n triangle such that
From (i) and (ii), we have A' B BC' A' C' 4
  
AB' B' C' AC' m AB BC AC 3
  
AB BC AC n
Ex.3 Construct a triangle similar to a given triangle Sol. Steps of construction
ABC such that each of its sides is (6/7)th of the
Step I : Draw a line segment AB = 4 cm.
corresponding sides of ABC. It is given that
Step II : Construct ABP = 60º.
AB = 5 cm, AC = 6 cm and BC = 7 cm.
E
Sol. Steps of Construction P
Step I : Draw a line segment BC = 7 cm. G H
Step II : With B as centre and C

radius = AB = 5 cm, draw an arc.

60°
Step III : With C as centre and A 4 cm B D
radius = AC = 6 cm, draw another arc, 6 cm
intersecting the arc drawn in step II at A.
Step III : Draw a line GH || AB at a distance of
A 3 cm, intersecting BP at C.
A´
Step IV : Join CA.
Thus, ABC is obtained.
Step V : Extend AB to D such that AD = 3/2
B C 3 
B1 C´ AB =   4  cm = 6 cm.
B2 2 
B3
B4 Step VI : Draw DE || BC, cutting AC produced
B5
B6 at E.
B7
B8 Then ADE is the required triangle similar to
Step IV : Join AB and AC to obtain the ABC such that each side of ADE is 3/2
triangle ABC. times the corresponding side of ABC.
Step V : Below base BC, construct an acute Proof : Since DE || BC, we have ADE ~ ABC.
angle CBX.
AD DE AE 3
Step VI : Along BX, mark off seven points    
AB BC AC 2
B1, B2, B3, B4, B5, B6, B7 such that
BB1 = B1B2 = ...... = B6B7.  TO CONSTRUCT A TANGENT TO A
CIRCLE AT A GIVEN POINT ON IT
Step VII : Join B7C.
(USING THE CENTRE OF THE CIRCLE)
Step VIII : Since we have to construct a
triangle each of whose sides is (6/7)th of the Steps of Construction
corresponding sides of ABC. So take 6 parts
out of 7 equal parts on BX i.e. from B6, Draw
B6C´ || B7C, intersecting BC at C´.
O
Step IX : From C´, draw C´A´ || CA, meeting
BA at A´.
A´BC´ is the required triangle each of whose
sides is (6/7)th of the corresponding sides of T P T´
ABC.
Ex.4 Construct a ABC in which AB = 4 cm, 
Step I : Take a point O on the plane of the paper
  B = 60º and altitude CL = 3 cm. Construct a and draw a circle of given radius.
ADE similar to ABC such that each side of
ADE is 3/2 times that of the corresponding
side of ABC.
 Step II : Take a point P on the circle.  TO CONSTRUCT TWO TANGENTS TO A
Step III : Join OP. CIRCLE FROM A POINT OUTSIDE THE
CIRCLE (USING CENTRE OF CIRCLE)
Step IV : Construct OPT = 90º.
Steps of construction
Step V : Produce TP to T to get TPT as the
required tangent. Q

EXAMPLES 
P O
Ex.5 Take a point O on the plane of the paper. With M
O as centre draw a circle of radius 3cm. Take a
point P on this circle and draw a tangent at P.
Q
Sol. Steps of Construction
1. Take given circle and a point P outside the
Step I : Take a point O on the plane of the circle. O is centre of the circle
paper and draw a circle of radius 3 cm.
2. Joint OP
Step II : Take a point P on the circle and join OP.
3. Bisect OP and get its mid-point M
T
4. Draw circle with centre M and
radius = PM = MO
5. Circle drawn meets the given circle at Q above
O P PO and at Q below PO.
6. Join PQ and PQ
7. PQ and PQ are the required tangents drawn to
T´ the circle from the point P.
Step III : Construct OPT = 90º We observe that PQ = PQ
Step IV : Produce TP to T to obtain the EXAMPLES 
required tangent TPT.
Ex.6 Draw a circle of radius 4 cm with centre O. Ex.7 Draw a circle of radius 3 cm. Take a point at a
distance of 5.5 cm from the centre of the circle.
Draw a diameter POQ. Through P or Q draw From point P, draw two tangents to the circle.
tangent to the circle.
Sol. Steps of Construction
Sol. Steps of Construction
Step I : Take a point O in the plane of the
Step I : Taking O as centre and radius equal to paper and draw a circle of radius 3 cm.
4 cm draw a circle.
Step II : Mark a point P at a distance of
Step II : Draw diameter POQ. 5.5 cm from the centre O and join OP.
Step III : Construct PQT = 90º Step III : Draw the right bisector of OP,
intersecting OP at Q.
T
T

O P O
P Q Q

T'

T' Step IV : Taking Q as centre and OQ = PQ as

radius, draw a circle to intersect the given circle
Step IV : Produce TQ to Tto obtain the at T and T´.
required tangent TQT. Step V : Join PT and PT´ to get the required
tangents.
Ex.8 Construct a tangent to a circle of radius 4 cm
from a point on the concentric circle of radius P
6 cm and measure its length. Also verify the
measurement by actual calculation. 30
° 60°

Sol. In order to do the desired construction, 30° 60° 60°

B O
30° A
we follow the following steps:
Step I : Take a point O on the plane of the
paper and draw a circle of radius OA = 4 cm.
Also, draw a concentric circle of radius Q
OB = 6 cm.
Step II : Find the mid-point C of OB and draw Justification: In OAP, we have
a circle of radius OC = BC. Suppose this circle
intersects the circle of radius 4 cm at P and Q. OA = OP = 5 cm (= Radius) Also,
AP = 5 cm (= Radius of circle with centre A)
P  OAP is equilateral.
 PAO = 60º BAP = 120º
C O
B A
In BAP, we have
BA = AP and BAP = 120º
Q
 ABP = APB = 30º PBQ = 60º
Ex.10 Draw a circle of radius 3 cm. Draw a pair of
Step III : Join BP and BQ to get the desired tangents to this circle, which are inclined to
tangents from a point B on the circle of radius each other at an angle of 60º.
6 cm.
Sol. Steps of construction
By actual measurement, we find the
Step I : Draw a circle with O as centre and
BP = BQ = 4.5 cm radius = 3 cm.
Justification: In BPO, we have Step II : Draw any diameter AOB of this
OB = 6 cm and OP = 4 cm circle.

OB2= BP2 + OP2 [Using Pythagoras theorem] Step III : Construct BOC = 60º such that
radius OC meets the circle at C.
BP = OB 2  OP 2 Step IV : Draw AM  AB and CN OC.
 BP = 36  16  20 cm = 4.47 cm ~ 4.5 cm Let AM and CN intersect each other at P.
Then, PA and PC are the desired tangents to the
Similarly, BQ = 4.47cm ~
 4.5 cm given circle, inclined at an angle of 60º
Ex.9 Draw a pair of tangents to a circle of radius
5 cm which are inclined to each other an angle
of 60º. A
O
Sol. In order to draw the pair of tangents, we follow
the following steps. 60° B
Step I : Take a point O on the plane of the
paper and draw a circle of radius OA = 5 cm. N 60°
Step II : Produce OA to B such that P
OA = AB = 5 cm. M
Step III : Taking A as the centre draw a circle Proof : AOC = (180º – 60º) = 120º
of radius AO = AB = 5 cm. Suppose it cuts the
circle drawn in step I at P and Q. In quad. OAPC, we have
Step IV : Join BP and BQ to get the desired OAP = 90º, AOC = 120º, OCP = 90º.
tangents. APC = [360º – (90º + 120º + 90º)] = 60º.
 EXERCISE
Q.1 Draw a line segment of length 7 cm and divide Q.9 Draw a triangle with side BC = 6 cm,
it in the ratio 2 : 3. Measure the two parts. B = 45º and A = 105º. Then, construct
4
Q.2 Draw a line segment of length 7.8 cm and similar triangle whose sides are times the
3
divide it in the ratio 5 : 8. Measure the two
corresponding sides of ABC.
parts.
Q.10 Draw right triangle in which the sides (other
Q.3 Construct a triangle with sides 5 cm, 6 cm and
than hypotenuse) are of lengths 3 cm and 4 cm.
7 cm and then construct another triangle similar
Then construct another similar triangle whose
2
to it whose sides are of the corresponding 5
3 sides are times the corresponding sides of
3
sides of the first triangle.
given triangle.

Q.4 Construct a triangle with sides 5 cm, 6.5 cm

Q.11 Draw a circle of radius 4.6 cm. Take a point P
and 7.6 cm and then construct another triangle
on it. Construct a tangent to the circle at the
7 point P. Also write the steps of construction.
similar to it whose sides are of the
5
corresponding sides of the first triangle. Q.12 Draw a circle of radius 3.5 cm. Construct two
tangents to it inclined at an angle of 60º to each
Q.5 Construct a triangle ABC whose sides are 5 cm,
other.
12 cm and 13 cm. Construct another triangle
3 Q.13 Draw a circle of radius 4 cm. Mark its centre as
similar to ABC and with sides th of the
5 O. Mark a point P such that OP = 5 cm. Using
corresponding sides of the given triangle. ruler and compasses only, construct two
tangents from P to the circle. Measure the
Q.6 Construct a triangle similar to a given triangle length of one of them.
with sides 6 cm, 7 cm and 8 cm and whose
sides are 1.4 times the corresponding sides of Q.14 Draw a circle of diameter 8 cm. From a point P,
the given triangle. 7 cm away from its centre, construct a pair of
tangents to the circle. Measure the lengths of
Q.7 Construct an isosceles triangle whose base is the tangent segments.
7 cm and 4 cm and then construct another
similar triangle whose sides are 1 12 times the Q.15 Draw a circle of radius 3 cm. Take two points P
corresponding sides of the isosceles triangle. and Q on one of its extended diameter each at a
distance of 7 cm from its centre. Draw tangents
Q.8 Draw a triangle ABC with side BC = 8 cm, to the circle from these two points P and Q.
AB = 6 cm and ABC = 60º. Construct
another triangle similar to ABC whose sides Q.16 Draw a circle with the help of a bangle. Take a
point outside the circle. Construct the pair of
3
are of the corresponding sides of the triangle tangents from this point to the circle.
4
ABC.
Q.17 Draw a circle of diameter 6 cm with centre O. Q.19 Draw a circle of radius 3 cm. From a point
Draw a diameter AOB. Through A or B draw 5 cm away from the centre of the circle, draw
tangent to the circle. two tangents to the circle. Find the length of the
tangents.
Q.18 Construct a ABC in which AB = 5cm,
B = 60º and altitude CD = 3 cm. Construct
AQR similar to ABC such that each side of
AQR is 1.5 times that of the corresponding
side of ABC.

ANSWER KEY
1. 2.8 cm, 4.2 cm 2. 3 cm, 4.8 cm 13. 3 cm
14. 5.7 cm (approx) 19. 4 cm each