NCERT Exemplar: Constructions

## Exercise 10.1

Q.1. To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn so that ∠BAX is an acute angle 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

Let us construct the figure
Here AX is divided into 5 + 7 = 12 parts. Construct a line parallel to BX from line number 5 to AB. Then the line will divide both AB and AX in the ratio 5 : 7So the right option is d. Therefore, the minimum number of points is 12.

Q.2. To divide a line segment AB in the ratio 4 : 7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A1 , A2 , A3 , .... are located at equal distances on the ray AX and the point B is joined to
(a) A12
(b) A11
(c)A10
(d) A
9

It is given that
Line segment AB is divided in the ratio 4 : 7
A : B = 4 : 7
Construct a ray AX which makes an acute angle BAX
So the minimum number of points which are located at equal distances on the ray is
AX = A + B = 4 + 7 = 11
Here A1, A2, A3 …… are located at equal distances on the ray AX and the point B is joined to A11.
Therefore, point B is joined to A11.

Q.3. To divide a line segment AB in the ratio 5 : 6, draw a ray AX such that ∠BAX is an acute angle, then draw a ray BY parallel to AX and the points A1, A2, A3, ... and B1, B2, B3, ... are located at equal distances on ray AX and BY, respectively. Then the points joined are
(a) A5 and B
(b) A6 and B
(c) A4 and B
(d) A5 and B4

It is given that
Line segment AB is divided in the ratio 5 : 6
A : B = 5 : 6
Steps of Construction:
1. Construct a ray AX and an acute angle BAX.
2. Construct a ray BY || AX and ∠ABY = ∠BAX.
3. Let us locate the points A1, A2, A3, A4 and A5 on AX and B1, B2, B3, B4, B5 and B6 as
A: B = 5 : 6
4. Now join A5B6.
5. Here A5 B6 intersects AB at the point C.
AC: BC = 5 : 6
Therefore, the points joined are A5 and B6.

Q.4. To construct a triangle similar to a given ∆ABC with its sides 3/7 of the corresponding sides of ∆ABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B1 , B2 , B, ... on BX at equal distances and next step is to join
(a) B10 to C
(b) B3 to C
(c) B7 to C
(d) B4 to C

In order to construct a triangle similar to a given ∆ABC with its sides 3/7 we have to divide BC in the ratio 3 : 7
BX should have 7 equidistant points on it as 7 is a greater number
Now we have to join B7 to C.
Therefore, the next step is to join B7 to C.

Q.5. To construct a triangle similar to a given ∆ABC with its sides 8/5 of the corresponding sides of ∆ABC draw a ray BX such that ∠CBX is an acute angle 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

We know that
In order to construct a triangle which is similar to a given triangle, with its sides x/y of the corresponding sides of the triangle, the minimum number of points to be located at equal distance is equal to the greater of m and n in m/n
In this question,
m : n = 8 : 5
m/n = 8/5
Therefore, the minimum number of points to be located is 8.

Q.6. To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is 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°

It is given that
O is the centre of a circle to which a pair of tangents PQ and PR from the point P touches the circle at Q and R
∠RPQ = 60°
We know that
∠OQP = 90° = ∠ORP
The angle between a tangent to a circle and the radius of the same circle passing through the point of contact is 90°
Using the angle sum property of quadrilaterals
∠OQP + ∠RPQ + ∠ORP + ∠ROQ = 360°
Substituting the values
90° + 60° + 90° + ∠ROQ = 360°
∠ROQ = 120°
Therefore, the angle between them should be 120°.

## Exercise 10.2

Q.1. By geometrical construction, it is possible to divide a line segment in the ratio √3: 1/√3. Write ‘True’ or ‘False’ and justify your answer.

True
From the question
Ratio = √3 : 1/√3
On further simplification
√3/ (1/√3) = (√3 × √3)/1 = 3 : 1
So the required ratio is 3 : 1.
Therefore, the statement is true.

Q.2. To construct a triangle similar to a given ∆ABC with its sides 7/3 of the corresponding sides of ∆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 B1 , B2 , ...., B7 are located at equal distances on BX, B3 is joined to C and then a line segment B6C' is drawn parallel to B3C where C' lies on BC produced. Finally, line segment A'C' is drawn parallel to AC. Write ‘True’ or ‘False’ and justify your answer.

False
Steps of construction:
1. Construct a line segment BC.
2. Taking B and C as centers, construct two arcs of suitable radius which intersects each other at A.
3. Now join BA and CA. ∆ABC is the required triangle.
4. From point B, construct a ray BX downwards which makes an acute angle CBX.
5. On BX, mark 7 points, B1, B2, B3, B4, …….. B7
6. Let us join B3C and from B7 construct a line B7C”|| B3C which intersects the extended line segment BC at C’.
7. Construct C’A’||CA which intersects the extended line segment BA at A’.
So ∆A’B’C’ is the required triangle whose sides are 7/3 of the corresponding sides of ∆ABC.
It is given that
Segment B6C’||B3C
As the construction is not possible that segment B6C’||B3C as the similar triangle A’B’C’ has its sides 7/3 of the corresponding sides of triangle ABC.
Here B7C’||B3C.
Therefore, the statement is false.

Q.3. A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of 3 cm from the centre. Write ‘True’ or ‘False’ and justify your answer.

False
Consider r as the radius of the circle
d as the distance of point from the centre
Here
r = 3.5 cm
d = 3 cm
We know that
r > d
Point P lies inside the circle
So no tangent is possible.
Therefore, the statement is false.

Q.4. A pair of tangents can be constructed to a circle inclined at an angle of 170°. Write ‘True’ or ‘False’ and justify your answer.

True
Construct two tangents from P to the circle with centre at Q and R
Now join OQ and OR
Here OQ and OR are the radii of the circle
∠OQP = ∠ORP = 90° as PQ and PR are the tangents to the circle at Q and R
∠OQP + ∠ORP = 180° …. (1)
From the angle sum property
∠QPR + ∠QOR + (∠OQP + ∠ORP) = 360°
From (1) we get
∠QPR + ∠QOR + 180° = 360°
∠QPR + ∠QOR = 180°
Here ∠QPR and ∠QOR < 180°
The angle given is 170° which is less than 180°
Therefore, the statement is true.

## Exercise 10.3

Q.1. Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5

Steps of Construction
1. Construct a line segment AB = 7 cm
2. Construct a ray AX which makes an acute angle ∠BAX
3. Along the line AX mark 3 + 5 = 8 points
A1, A2, A3, A4, A5, A6, A7, A8 where
AA1 = A1A2 = A2A= A3A4 = A4A5 = A5A6 = A6A7 = A7A8
4. Let us join A8B
5. From the point A3 construct A3C||A8B which meets AB at C
It makes an angle equal to ∠BA8 at A3
6. C is the point on AB which divides it in the ratio 3 : 5
7. So AC: BC = 3 : 5
Verification:
Consider AA1 = A1A2 = A2A3 = A7A8 = x
In triangle ABA8 we know that A3C||A8B
Here
AC/CB = AA3/A3A8 = 3x/5x = 3/5
So AC : CB = 3: 5
Therefore, the point P which divides it in the ratio 3 : 5 is found.

Q.2. Draw a right triangle ABC in which BC = 12 cm, AB = 5 cm and ∠B = 90°. Construct a triangle similar to it and of scale factor 2/3. Is the new triangle also a right triangle.

It is given that
In a right triangle ABC
BC = 12 cm, AB = 5 cm and ∠B = 90°
We have to find if the new triangle is a right triangle
Steps of construction:
1. Construct a line segment AB = 5cm. At the point A draw a right triangle SAB
2. Construct an arc of radius 12 cm with B as its centre which intersects SA at the point C.
3. Now join BC to obtain ABC
4. Construct a ray AX which makes an acute angle with AB opposite to the vertex C
5. Let us locate 3 points A1, A2, A3 on the line segment AX where AA1 = A1A2 = A2A3.
6. Join A3B
7. Construct a line through A2 parallel to A3B which intersects AB at B’.
8. Construct a line parallel to BC which intersects AC at C’ through B’
9. So triangle AB’C’ is the required triangle.
Therefore, the new triangle is also a right triangle.

Q.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.

Steps of Construction
1. Construct a line segment BC = 6cm
2. With S and C as centres, construct two arcs of radii 4 cm and 5 cm which intersect with each other at A.
3. Let us join BA and CA. ∆ABC is the required triangle.
4. From the point B, construct a ray BX downwards which makes an acute angle
5. Now mark 5 points B1, B2, B3, B4, B5 on BX where
BB1 = B1B2 = B2B3 = B3B4 = B4B5
6. Join B3C and from the point B5 construct B5M || B3C which intersects the extended line segment BC at M
7. From the point M construct MN || CA which intersects BA at N
8. Triangle NBM is the required triangle
Therefore, ∆NBM is the required triangle.

Q.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.

Steps of Construction
1. Construct a circle of radius 4 cm. Consider O as the centre of the circle.
2. Let us take a point M which is at 6 cm away from the radius.
3. Join OM and bisect it. With M and O as centres and radius more than half of it, construct two arcs on either side of the line OM. Let the arc meet at A and B where M1 is the midpoint of OM.
4. Take M1 as centre and M1O as the radius, construct a circle to intersect circle with radius 4 cm and the centre O at two points P and Q
5. Now join PM and QM. So PM and QM are the required tangents from M to O and radius 4 cm.
Therefore, the figure is drawn.

## Exercise 10.4

Q.1. Two line segments AB and AC include an angle of 60° where AB = 5 cm and AC = 7 cm. Locate points P and Q on AB and AC, respectively such that AP = 3/4 AB and AQ = 1/4 AC. Join P and Q and measure the length PQ.

Steps of Construction
1. Construct a line segment AB = 5 cm
2. Construct ∠BAZ = 60°
3. With A as centre and 7 cm as radius, construct an arc which cuts AZ at the point C
4. Construct a ray AX which makes an angle ∠BAX
5. Let us divide AX into four equal parts AA1 = A1A2 = A2A3 = A3A4
6. Now join A4B
7. Construct A3P||A4B which meets AB at P
8. P is the on AB where AP = 3/4 AB
9. Construct a ray AY which makes an acute angle ∠CAY
10. Now divide AY into four equal parts AB1 = B1B2 = B2B3 = B3B4
11. Join B4C
12. Construct B1Q||B4C which meets AC at Q.
Q is the point on AC where AQ = 1/4 AC
13. Join PQ and measure it
14. Here PQ = 3.25 cm
Therefore, the length of PQ is 3.25 cm.

Q.2. Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and ∠ABC = 60°, divide it into triangles BCD and ABD by the diagonal BD. Construct the triangle BD' C' similar to ∆BDC with scale factor 4/3. Draw the line segment D'A' parallel to DA where A' lies on extended side BA. Is A'BC'D' a parallelogram.

Steps of Construction
1. Construct a line AB = 3 cm
2. Construct a ray BY which makes an acute angle ∠ABY = 60°
3. With B as centre and 5 cm as radius, construct an arc which cuts the point C on BY
4. Construct a ray AZ which makes ∠ZAX’ = 60° as BY || AZ and ∠YBX’ = ∠ZAX’ = 60°
5. With A as centre and 5 cm as radius, construct an arc which cuts the point D on AZ
6. Join CD
7. So we get a parallelogram ABCD
8. Join BD which is the diagonal of parallelogram ABCD
9. Construct a ray BX downwards which makes an acute angle ∠CBX
10. Now locate 4 points B1, B2, B3, B4 on BX where BB= B1B2 = B2B3 = B3B4
11. Join B4C and from B3C construct a line B4C’ || B3C which intersects the extended line segment BC at C’
12. Construct C’D’||CD which intersects the extended line segment BD at D’. ∆D’BC’ is the required triangle whose sides are 4/3 of the corresponding sides of ∆DBC
13. Construct a line segment D’A’||DA where A’ lies on the extended side BA
14. We see that A’BC’D’ is a parallelogram where A’D’ = 6.5 cm, A’B = 4 cm and ∠A’BD’ = 60°
15. Divide it into triangles A’BD’ and BC’D’ by the diagonal BD’.
Therefore, A’BC’D’ is a parallelogram.

Q.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.

Steps of Construction
1. Construct two concentric circles of radii 3 cm and 5 cm
2. Consider a point P on the outer circle. The distance of point P from O is 5 cm
3. Let us bisect the line segment OP and consider it as M
4. Taking M as centre and PM as radius construct a circle. Let this circle intersect the circle with radius 3 cm at the points T and T’
5. Now join PT and PT’. They are the required tangents to the inner circle.
6. 3 cm is the radius of inner circle and 5 cm is the distance of point P from O
7. From the Pythagoras theorem, the length of tangent is 4 cm.
OP = 5 cm
OT = 3 cm
OP² = OT² + PT²
5² = 3² + PT²
PT² = 25 - 9
PT² = 16
So we get
PT = 4 cm
Therefore, the length of the tangent is 4 cm.

Q.4. Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ∆ ABC in which PQ = 8 cm. Also justify the construction

Steps of Construction
1. Construct a line segment BC = 5 cm
2. Draw OQ which is the perpendicular bisector of line segment BC meeting BC at P’.
3. Consider B and C as centres construct two arcs of equal radius 6 cm which intersects each other at the point A
4. Now join BA and CA. ∆ABC is the required triangle.
5. From the point B, construct any ray BX which makes an acute angle ∠CBX
6. Let us locate four points B1, B2, B3 and B4 where BB1 = B1B2 = B2B3 = B3B4
7. Join B3C and from the point B4 construct a line B4R||B3C which intersects the extended line
8. From the point R, construct RP||CA which meets BA produced at P
∆ PBR is the required triangle.
Justification:
Consider BB1 = B1B2 = B2B3 = B3B4 = x
From the construction
B4R||B3C
BC/CR = BB3/B3B4 = 3x/x = 3
BC/CR = 3/1
Here
BR/BC = (BC + CR)/BC = BC/BC + CR/BC
= 1 + 1/3
= 4/3
From construction, RP||CA
So rABC is congruent to rPBR [by AAA criterion]
PB/AB = RP/CA = 4/3
The new triangle PBR is similar to isosceles triangle ABC and its sides are 4/3 times of the corresponding sides of ABC.

Q.5. Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60º . Construct a triangle similar to ∆ ABC with scale factor 5/7. Justify the construction.

Steps of Construction
1. Construct a line segment AB = 5 cm
2. From the point B, construct ∠ABC =60º
In order to find angle B
(i) Taking B as centre and with any radius construct another arc which cuts the line AB at D
(ii) Taking D as centre and with same radius construct the first arc at the point E
(iii) Construct a ray BY which passes through E and forms an angle 60º with the line AB
3. Taking B as centre and radius 6 cm construct an arc which intersects the line BY at C
4. Join AC and triangle ABC is the required triangle.
5. From the point A, construct any ray AX downwards which makes an acute angle
6. Locate 7 points A1, A2, A3, A4, A5, A6 and A7 on AX where AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7
7. Join A7B and from the point A5 construct A5M||A7B which intersects AB at M
8. From M construct MN||BC which intersects AC at N.
∆ AMN is the required triangle whose sides are equal to 5/7 of its corresponding sides of ∆ABC.
Justification:
Consider AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = x
From the construction
A5M || A7B
AM/MB = AA5/A5A7 = 5x/2x = 5/2
AM/MB = 5/2
We know that
AB/AM = (AM + MB)/AM = AM/AM + MB/AM
= 1 + 2/5
= 7/5
Here MN || BC
r AMC is congruent to r AMN
So we get
AN/AC = NM/BC = AM/AB = 5/7
Therefore, a triangle similar to ∆ ABC with scale factor 5/7 is constructed.

Q.6. Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is 60º. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.

Steps of construction:
1. Construct a circle with O as centre and 4 cm radius
2. Construct any diameter AOB
3. Construct an angle ∠AOP = 60º where OP is the radius which intersect the circle at the point P
4. Construct PQ perpendicular to OP and BE perpendicular to OB PQ and BE intersect at the point R
5. RP and RB are the required tangents
6. The measurement of OR is 8 cm
Justification:
PR is the tangent to a circle
∠OPQ = 90º
BR is the tangent to a circle
∠OBR = 90º
So we get
∠POB = 180 - 60 = 120º
In BOPR
∠BRP = 360 - (120 + 90 + 90) = 60º
Therefore, the distance between the centre of the circle and the point of intersection of tangents is 8 cm.

Q.7. Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle similar to ∆ABC with scale factor 3/2. Justify the construction. Are the two triangles congruent? Note that all the three angles and two sides of the two triangles are equal.

Steps of Construction
1. Construct a line segment BC = 6 cm
2. Taking B as centre and 4 cm radius, construct an arc
3. Taking C as centre and 9 cm radius, construct another arc intersecting the previous arc at A
4. Now join BA and CA. ∆ABC is the required triangle
5. Through the point B, construct an acute angle ∠CBX on the side opposite to the vertex A
6. Mark three arcs B1, B2 and B3 on BX where BB1 = B1B2 = B2B3
7. Join B2C
8. Construct B3C’||B2C which intersects the extended line segment BC at the point C’
9. Construct C’A’ || CA which intersects the extended line segment BA at the point A’
∆A’BC’ is the required triangle
Justification:
We know that
B3C’ || B2C
BC/CC’ = 2/1
Here
BC’/BC = (BC + CC’)/BC = 1 + CC’/BC
= 1 + 1/2
= 3/2
Similarly
CC’||CA
∆ABC is similar to ∆A’BC’
A’B/AB = BC’/BC = CC’/AC = 3/2
Therefore, the two triangles are not congruent.

The document NCERT Exemplar: Constructions | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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