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 ,

2

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 ,

2

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

3

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.

7

(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 ,

4

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

x

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

5

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 .

5

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

the radius.

(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

AD||BC

(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 ,

2

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

their corresponding radii are supplementary.

(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

Thus, in quadrilateral YACB

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.