 Lesson: Constructions
Exercise 10.1 (6)
Question: 1
To divide a line segment AB in the ratio , first a
5:7
ray AX is drawn so that is an acute angle
BAX
and then at equal distances points are marked on
the ray AX such that the minimum number of these
points is
a.
8
b.
10
c.
11
d.
12
Solution:
(d)
The minimum number of points marked on the ray
5 7 12
Question: 2
To divide a line segment AB in the ratio , a ray
4:7
AX is drawn first such that is an acute angle
BAX and then points .... are located at equal
1
A ,
3
A
distances on the ray AX and the point B is joined to
a.
12
A
b.
11
A
c.
10
A
d.
9
A
Solution:
(b)
On dividing the construction line into
7 4 11
equal parts.
Hence, the part will be joined to B.
th
11
Question: 3
To divide a line segment AB in the ratio , draw a
5:6
ray AX such that is an acute angle, then
BAX
draw a ray BY parallel to AX and the points
1
A ,
,..... and ,.... are located at equal distances
3
A
1
B ,
2
B ,
3
B
on ray AX and BY, respectively. Then the points
joined are
a. and
5
A
6
B
b. and
6
A
5
B
c. and
4
A
5
B
d. and
5
A
4
B Solution:
(a)
In the below figure, segment AB of given length is
divided into two parts of ratio by following
5:6
steps:
(i) Draw a line-segment AB of given length.
(ii) Draw an acute angle BAX as shown in figure
above either up side or down side.
(iii) Draw angle on other side of
ABY BAX
AX, i.e., down side.
(iv) Divide AX into 5 equal parts by using compass.
Divide BX into same distance in 6 equal parts
as AX was divided.
(v) Now, join and which meet AB at P.
5
A
6
B
(vi) P divides AB in ratio .
AP:PB
5:6 Question: 4
To construct a triangle similar to a given
ABC
with its sides of the corresponding sides of
3
7
ABC
, first draw a ray BX such that is an acute
CBX
angle and X lies on the opposite side of A with
respect to BC. Then locate points ,....... on
1
B ,
2
B ,
3
B
BX at equal distances and next step is to join
a. to C
10
B
b. to C
3
B
c. to C
7
B
d. to C
4
B
Solution:
(c)
The ratio is
3
7
1
So, the resulting figure will be smaller than the
original one
Thus, seventh part is to be joined to C such that the
parallel line from third part of BX will meet on BC
without producing further. Question: 5
To construct a triangle similar to a given
ABC
with its sides of the corresponding sides of
8
5
ABC
draw a ray BX such that is an acute angle
CBX
and X is on the opposite side of A with respect to
BC. The minimum number of points to be located at
equal distances on ray BX is
a.
5
b.
8
c.
13
d.
3
Solution:
(b)
In order to construct a triangle which is similar to
with its sides of the corresponding sides of
ABC
8
5
, the minimum number of points to be
ABC
located at equal distances on ray BX is .
8
Question: 6
To draw a pair of tangents to a circle which are
inclined to each other at an angle of , it is
60
required to draw tangents at end points of those two radii of the circle, the angle between them
should be
a.
135
b.
90
c.
60
d.
120
Solution:
(d)
The tangent and radius are perpendicular to each
other at contact point.
Thus, and in quadrilateral QROS formed by
R
S
tangents and radii will be each. So, the sum of
90
Q O 180
Thus,
O 180 60
O 120 Exercise 10.2 (4)
Write True or False and give reasons for your
answer in each of the following:
Question: 1
By geometrical construction, it is possible to divide
a line segment in the ratio
1
3 :
3
Solution:
True
Upon multiplying or dividing a given ratio by a real
number, the ratio always remains the same.
On multiplying the given ratio by , we will get
or
1
3 3 : 3
3
3:1
Thus, the given ratio is possible to divide a
1
3 :
3
line segment in ratio in place of .
3:1
1
3 :
3
Question: 2 To construct a triangle similar to a given
ABC
with its sides of the corresponding sides of
7
3
ABC
, draw a ray BX making acute angle with BC and X
lies on the opposite side of A with respect to BC.
The points are located at equal
1
B ,
2
B ,
7
......,B
distances on BX, is joined to C and then a line
3
B
segment is drawn parallel to where lies
6
B C
3
B C
C
on BC produced. Finally, line segment is
A C
drawn parallel to AC.
Solution:
False
The ratio is .
7
3
1
Thus, the resultant triangle will be larger than given
as ||
7
B C
3
B C
BX is equally divided into parts as
7
7 3 (i) Construct the given triangle with the above-
mentioned specifications.
(ii) Draw an acute angle CBX.
(iii) Divide BX into 7 equal parts and mark them
.
1
B ,
2
B ,
7
......,B
(iv) Draw BC and BA as shown in figure above.
(v) Join .
3
B C
(vi) Construct || , is on BC produced.
7
B C
3
B C
C
(vii) Draw . on BA produced is
C A||AC
A
A BC
required triangle i.e.,
A BC 3
ABC 7 Thus, .
7
B C
||
3
B C
Question: 3
A pair of tangents can be constructed from a point
P to a circle of radius cm situated at a distance
3.5
of 3 cm from the centre.
Solution:
False
Any tangent can be drawn on a circle only if the
distance of point to draw tangent is equal to or
more than radius of circle. In the question, the
radius of the circle is 3.5 cm whereas the point is at
3 cm from the centre which is inside the circle.
Thus, drawing a tangent is not possible if the point
is inside the circle.
Question: 4
A pair of tangents can be constructed to a circle
inclined at an angle of .
170
Solution:
True A pair of tangents can be drawn if the angle
between the tangents lies between 0 and .
180
Hence, as the sum of angles between tangents and
radii on tangent are supplementary a pair of
tangents can be constructed to circle inclined at an
angle of .
170
Exercise 10.3 (4)
Question: 1
Draw a line segment of length 7 cm. Find a point P
on it which divides it in the ratio .
3:5
Solution:
(i) Draw a line-segment MN cm.
(ii) Draw by such that and are
ML||NO
M
N
acute angles.
(iii) Divide ML and NO in 3 and 5 parts equally by
compass and mark .
1
M ,
2,
M
3
M ,
1
N ,
2
N ,
3
N ,
5
N (iv) Now, join which intersect MN at P and
3 5
M N
divides .
MP:PN
3:5
Thus, P is the required point on MN which divides
it in .
3:5
Justification:
In and
3
MM P
5
NN P
(By construction)
ML||NO
(Alt. angles)
M N
(Vertically opposite angles)
3 5
M PM N PN
Thus,
3
MM P
||
5
NN P
3
5
MM MP
NN NP
Let us assume each equal part be cm
3x MP
5x NP MP:NP 3:5
Hence proved.
Question: 2
Draw a right triangle ABC in which BC cm, AB
12
cm and . Construct a triangle similar to
B
90
it and of scale factor . Is the new triangle also a
2
3
right triangle?
Solution:
Scale factor given here is which is
2
3
1
Thus, the triangle to be constructed will be smaller
than
ABC (i) Draw BC cm.
12
(ii) Draw .
CBA
90
(iii) Cut BA cm making .
ABC
90
(iv) Join AC.
(v) is the given triangle.
ABC
(vi) Draw an acute so that A and Y are in
CBY
opposite direction with respect to BC.
(vii) Now, divide BY in three equal parts by
marking it as .
1
B ,
2
B ,
3
B
(viii) Join .
3
B C
(ix) Draw by making equal alternate
2 2
B C ||B C
angles at and .
2
B
3
B (x) Now from point , draw by making
C
C A ||CA
equal alternate angles at C and .
C
Hence, is the required triangle of scale
A BC
factor . It is a right triangle.
2
3
Question: 3
Draw a triangle ABC in which BC = 6 cm, CA = 5
cm and AB = 4 cm. Construct a triangle similar to it
and of scale factor .
5
3
Solution:
Scale factor given here is which is .
5
3
1
Thus, the triangle to be constructed will be larger
than .
ABC (i) Draw BC cm.
6
(ii) Draw arc cm from B.
1
BA
4
(iii) Draw arc cm from C.
2
CA
5
(iv) Let, arc and intersect at A.
2
CA
1
BA
(v) Join AB and AC.
(vi) Draw acute angle CBX below BC.
(vii) Cut BX into five equal parts as
1
B ,
2
B ,
3
B ,
4
B ,
5
B
respectively.
(viii) Join .
3
B C
(ix) Draw by making alternate angles.
5 3
B C ||B C
(x) is on BC produced.
C
(xi) Now, draw which meet BA produced
C A ||CA
at .
A Hence, is the required triangle with the
A BC
scale factor .
5
3
Justification:
(By AA criterion of similarity)
3 5
BCB || BC B
Hence, ( )
3
5
BB BC
BB BC
1 1 2
BB BB .....x
3x BC
5x BC
(By AA criterion of similarity)
ABC|| A BC
Therefore,
A B A C BC
AB AC BC
Hence proved.
Question: 4
Construct a tangent to a circle of radius 4 cm from
a point which is at a distance of 6 cm from its
centre.
Solution:
Tangents can be drawn only if the distance of point
from which tangents are to be drawn is more than (i) Draw a line-segment EF cm.
6
(ii) Taking E as the centre draw a circle of radius 4
cm.
(iii) Draw perpendicular bisector AB of EF which
meets EF at O.
(iv) Now, taking O as centre and OF as radius
draw a circle , which intersect at C and D.
2
R
1
R
(v) Join FC and FD.
Hence, FC and FD are the required tangents.
Exercise 10.4 (7) Question: 1
Two line segments AB and AC include an angle of
where AB cm and AC cm. Locate
60
5
7
points P and Q on AB and AC, respectively such
that AP AB and AQ AC. Join P and Q and
3
4
1
4
measure the length PQ.
Solution:
(i) Draw where AB cm and AC
BAC
60
5
cm.
7
(ii) Draw acute angle CAX and equally mark
1
X ,
respectively.
2
X ,
3
X ,
4
X (iii) Join .
4
X C
(iv) Draw .
1 4
X Q||X C
(v) Draw and equally mark .
BAY
1
Y,
2
Y ,
3
Y ,
4
Y
(vi) Now, join .
4
YB
(vii) Draw .
3 4
YP||YB
(viii) Join P and Q. PQ 3.3 cm.
Question: 2
Draw a parallelogram ABCD in which BC cm,
5
AB cm and , divide it into
3
ABC
60
triangles BCD and ABD by the diagonal BD.
Construct the triangle similar to with
BD C
BDC
scale factor . Draw the line segment parallel
4
3
D A
to DA where lies on extended side BA. Is
A
a parallelogram?
A BC D
Solution:
(i) Draw a line segment AB cm.
3 (i) Draw so that BC cm.
ABC
60
5
(ii) Draw and .
CD||AB
(iii) ABCD is the required parallelogram.
(iv) Join diagonal BD.
(v) Make acute angle CBX on opposite of D with
respect to BC.
(vi) Equally mark respectively.
1
B ,
2
B ,
3
B ,
4
B ,
(vii) Join .
3
B C
(viii) Draw on BC.
3
B C
4
||B C (ix) Now, draw where is on BA. Hence,
D A ||DA
A
parallelogram is similar to
A BC D
parallelogram ABCD with scale factor .
4
3
(x) Lastly, draw where is on BA and
D A ||DA
A
parallelogram is with scale factor .
A B C D
4
3
Question: 3
Draw two concentric circles of radii 3 cm and 5 cm.
Taking a point on outer circle construct the pair of
tangents to the other. Measure the length of a
tangent and verify it by actual calculation.
Solution:
(i) Draw two concentric circles of radii 3
1 2
O ,O
cm and 5 cm taking T as centre. (ii) Draw perpendicular bisector QR of TS. S is any
point on O
2
.
(iii) Draw circle taking radius SC TC with C
3
O
as centre.
(iv) Circle intersect the circle at A and B.
3
O
1
O
(v) Join SA and SB. These are required tangents.
(vi) SA SB cm by measurement.
4.1
By actual calculation:
Join TA. TA and SA are radius and tangent at
contact point A.
Thus, .
SAT
90
Applying Pythagoras theorem in ,
SAT
2 2 2
AS TS TA 2 2
5 3
25 9
16
AS
4
Difference in measurement and by mathematical
calculation AS 4.1 cm 4 cm 0.1 cm.
Question: 4
Draw an isosceles triangle ABC in which AB AC
6 cm and BC 5 cm. Construct a triangle PQR
similar to triangle ABC in which PQ 8 cm. Also,
justify the construction.
Solution:
Draw
PQR|| ABC
PQ 8 cm
Hence, (AB 6 cm)
PQ
AB
8 4
6 3
Thus, PQ QR 8 cm
Draw with scale factor
PQR|| ABC
4
1
3
will be larger than
PQR
ABC (i) Draw BC 5 cm.
(ii) Draw two arcs of 6 cm each from B and C in
same direction.
(iii) Join AB and AC.
(iv) Draw acute and equally mark B,
CBX
.
1 2 3 4
B ,B ,B ,B
(v) Join .
3
B C
(vi) Draw . R is on BC.
4 3
B R||B C
(vii) Draw . P is on BA.
RP||CA
Hence, with PQ QR 8 cm with
PQR|| ABC
scale factor .
4
3 Question: 5
Draw a triangle ABC in which AB 5 cm, BC 6
cm and ABC . Construct a triangle similar to
60
ABC with scale factor . Justify the construction.
5
7
Solution:
(i) Draw AB 5 cm.
(ii) Draw .
ABC
60
(iii) Cut BC 6 cm and join AC.
(iv) Draw acute and equally mark
BAX
1
A ,
.
7
......,A
(v) Join and draw . is on segment
7
A B
5 7
A B A BA
B
AB.
(vi) Draw . is on AC.
B C ||BC
C
with scale factor .
AB C || ABC
5
7 Justification:
In and
5
AA B
7
AA B
7 5
A B||A B
Hence, (corresponding angles)
5 7
A A
5 7
BAA BAA
Hence, (By AA criterion of
5
AA B
||
7
AA B
similarity)
5
7
AB AA 5x 5
AB AA 7x 7
where
1 1 2 6 7
x AA A A ....A A
Similarly, (By AA criterion of
AB C || ABC
similarity)
AB AC B C
AB AC BC 5 AC B C
7 AC BC
Therefore, with scale factor .
AB C || ABC
5
7
Question: 6
Draw a circle of radius 4 cm. Construct a pair of
tangents to it, the angle between which is . Also
60
justify the construction. Measure the distance
between the centre of the circle and the point of
intersection of tangents.
Solution:
Angle between tangents is . Thus, the angles
60
between their radii will be calculated as
. The angles between tangents and
180 60 120
(i) Draw a circle of radius 4 cm. (ii) Draw any diameter ACX.
(iii) Draw CB making .
ACB
120
(iv) Draw tangent at A by drawing .
CAY
90
(v) Draw so a tangent can be drawn.
CBY
90
(vi) Both AY, BY tangents intersect at Y making an
angle of .
60
Thus, the two tangents on circle are inclined at .
60
Justification:
As the radius CA and tangent AY at contact point
makes angle and .
YAC
90
YBC
90
Y A C B
360
Y 90 120 90 360
( )
C
120
Y
360 300 60
.
Y
60
Hence proved.
Question: 7
Draw a triangle ABC in which AB 4 cm, BC 6
cm and AC 9 cm. Construct a triangle similar to with scale factor . Justify the construction.
ABC
3
2
Are the two triangles congruent? Note that all the
three angles and two sides of the two triangles are
equal.
Solution:
Given scale factor is .
3
2
1
Thus, the resultant figure will be larger than .
ABC
(i) Draw line segment BC 6 cm.
(ii) Draw an arc of 6 cm from B as centre.
1
A
(iii) Draw another arc of 9 cm from C as centre.
2
A
(iv) Arcs and intersect at A.
1
A
2
A
(v) Join A to B and C.
(vi) Draw acute angle on other side of A.
CBX
(vii) Mark equally spaced .
1
B ,
2
B ,
3
B
(viii) Join .
2
B C
(ix) Draw . is on BC.
3 2
B C ||B C
C
(x) Draw . is on BA.
CA||C A
A
Hence, with scale factor of .
A BC || ABC
3
2 w
Justification:
In and
3
BB C
2
BB C
(Common)w
B B
(By construction)
3 2
B C ||B C
(Corresponding angles)
2 3
BB C BB C
Therefore, (By AA criterion of
3 2
BB C || BB C
similarity)
3
2
BC BB
CB BB
3x 3
2x 2
1 1 2 2 3
BB BB B B x
BC 3
BC 2
In and ,
ABC
A BC
(Common)
B B
Thus,
A C ||AC (Corresponding angles)
A C B ACB
(By AA criterion of similarity)
ABC|| A BC
A C A B C B
= =
AC AB BC
A C A B 3
AC AB 2
Hence proved.