NCERT Exemplar: Constructions | Mathematics for SSS 1 PDF Download

Exercise 10.1

Correct Answer is option (d)
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) A₉

_{Correct Answer is option (b)}
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 B₆ 
(b) A₆ and B₅ 
(c) A₄ and B₅ 
(d) A₅ and B₄

_{Correct Answer is option (a)}
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 , B₃ , ... 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

_{Correct Answer is option (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

Correct Answer is option (b)
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°

Correct Answer is option (d)
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

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

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 = A₂A₃ = 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

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 BB₁ = 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.