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#### Question 1:

Choose the correct answer from the given four options:
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

Given, a line segment AB in the ratio 5 : 7.

Now, draw a ray AX which makes an acute angle ∠BAX, then mark A + B points at equal distance.

Here, A = 5 and B = 7.

Therefore, the minimum number of these points = A + B
= 5 + 7
= 12
Hence, the correct answer is option D.

#### Question 2:

Choose the correct answer from the given four options:
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

Given, a line segment AB in the ratio 4 : 7
$⇒$A : B = 4 : 7
Minimum number of points located at equal distances on the ray AX = A + B
= 4 + 7
= 11
Here, A1, A2, A3, ... are located at equal distances on the ray AX.
Point B is joined to the last point is A11.

Hence, the correct answer is option B.

#### Question 3:

Choose the correct answer from the given four options:
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

Given, a line segment AB in the ratio 5 : 6.
⇒ A : B = 5 : 6

To divide a line segment AB in the ratio 5 : 6, the fifth division of AX and the sixth division of BY are joined. Thus, A5 and B6 are joined.
Hence, the correct answer is option A.

#### Question 4:

Choose the correct answer from the given four options:
To construct a triangle similar to a given $△$ABC with its sides $\frac{3}{7}$ 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

Steps:
1. Locate points B1, B2, B3, B4, B5, B6 and B7 on BX at equal distance. 2. Join the last point B7 to C. Hence, the correct answer is option C.

#### Question 5:

Choose the correct answer from the given four options:
To construct a triangle similar to a given $△$ABC with its sides $\frac{8}{5}$ 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

When a triangle similar to the given triangle with its sides $\frac{m}{n}$ of the corresponding sides of given triangle is to be constructed, the minimum number of points to be located at an equal distance is equal to the greater of m and n.
Here, $\frac{m}{n}=\frac{8}{5}$.

So, the minimum number of point to be located at equal distance on ray BX is 8.
Hence, the correct answer is option C.

#### Question 6:

Choose the correct answer from the given four options:
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°

First, make the figure as per the given information. Here,
OP and OQ are the tangents to the circle. Thus, they intersect the circle at right angles.

Now, the sum of angles of a quadrilateral, here PRQO, is 360°.
∴ ∠POQ + ∠ORP + ∠PRQ + ∠OQR = 360°
⇒ 60° + 90° + θ + 90° = 180°
θ = 120°
The angle between them should be 120°.
Hence, the correct answer is option D.

#### Question 1:

By geometrical construction, it is possible to divide a line segment in the ratio $\sqrt{3}:\frac{1}{\sqrt{3}}$.

The ratio in which the line segment is to be divided is $\sqrt{3}:\frac{1}{\sqrt{3}}$.

So, $\sqrt{3}:\frac{1}{\sqrt{3}}$ can be simplified as 3 : 1 and 3 as well as 1 both are positive integers.
Hence, by geometrical construction is possible to divide a line segment in the ratio 3 : 1.

#### Question 2:

To construct a triangle similar to a given $△$ABC with its sides $\frac{7}{3}$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.

False

To draw a triangle similar to a given $△$ABC with its sides $\frac{7}{3}$ of the corresponding sides of $△$ABC, the points B1, B2, ...., B7 are located at equal distances on BX, B3 is joined to C and then a line segment B7C' is drawn parallel to B3C where C' lies on BC produced.

So, B7C’ is parallel to B3C.

#### Question 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.

False

Let r be radius of circle and d be the distance of the point from the centre.
Here, r = 3.5 cm and d = 3 cm.
⇒ r > d
Point P lies in inside the circle, as shown below: Thus, it is not possible to make tangents to the circle from point P which lies inside the circle.

#### Question 4:

A pair of tangents can be constructed to a circle inclined at an angle of 170°.

True

The angle between the pair of tangents should always be greater than 0° and less than 180 °.

Hence, we can draw a pair of tangents to a circle inclined at an angle of 170°, as shown below: #### Question 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. Draw a line segment AB = 7 cm. 2. Draw a ray AX, making an acute ∠BAX. 3. Along AX, mark 3 + 5 = 8 points namely A1, A2, A3, A4, A5, A6, A7, A8 such that
AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = A7A8. 4. Join A8B. 5. From A3, draw A3P  ∥ A8B meeting AB at P. (By making an angle equal to ∠BA8 at A3).
Then, P is the point on AB which divides it in the ratio 3 : 5. Thus, AP : PB = 3 : 5. Justification:
Let AA1 = A1A2 = A2A3 = A3A4 =........= A7A8 = x
In $△$ABA8, A3P ∥ A8B.
$\therefore \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{{\mathrm{AA}}_{3}}{{\mathrm{A}}_{3}{\mathrm{A}}_{8}}=\frac{3x}{5x}=\frac{3}{5}$
Hence, AP : PB = 3 : 5.

#### Question 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 $\frac{2}{3}$. Is the new triangle also a right triangle?

Steps of construction:
1. Draw a line segment BC = 12 cm. 2. From B, draw a line which makes a right angle.

a. Now as point B is the initial point as the centre, draw an arc of any radius such that the arc meets BC at point D. b. With D as centre and with the same radius as before, draw another arc cutting the previous one at point E. c. Now with E as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point F. d. With E and F as centres and with a radius more than half the length of FE, draw two arcs intersecting at point G. e. Join points B and G. The angle formed is a right angle i.e., ∠GBC = 90°. 3. From B as centre, draw an arc of 5 cm which intersects GB at A. 4. Join AC to obtain the right angled triangle ABC. 5. From B, draw an acute ∠CBH downwards. 6. On ray BH, mark three points B1, B2 and B3 such that BB1 = B1B2 = B2B3. 7. Join B3C. 8. From point B2, draw B2N ∥ B3C intersecting BC at N. 9. From point N, draw NM CA intersect BA at M. $△$MBN is the required triangle, right angled at B. As per the construction, $△$MNB is the required triangle
Justification:
Let BB1 = B1B2 = B2B3 = x.

Considering $△$BNB2 and $△$BCB3
∠B = ∠B                                      (common)
∠BNB2 =∠BCB                        (corresponding angles of the same transverse as B3C ∥ B2N)
∴ $△$BNB2 $~$$△$BCB                (by AA criteria)

$⇒\frac{\mathrm{BN}}{\mathrm{BC}}=\frac{{\mathrm{BB}}_{2}}{{\mathrm{BB}}_{3}}=\frac{2x}{3x}=\frac{2}{3}$          .....(1)

Similarly, in $△$MBN and $△$ABC,
∠ A  = ∠ A                                    (common)
∠BAC = ∠BMN                           (corresponding angles of the same transverse as AC ∥ MN)
∴ $△$MBN $~$ $△$ABC                    (by AA criteria)

$⇒\frac{\mathrm{MB}}{\mathrm{AB}}=\frac{\mathrm{MN}}{\mathrm{AC}}=\frac{\mathrm{BN}}{\mathrm{BC}}=\frac{2}{3}$          .....(2)

∴ the constructed triangle $△$MNB is of scale $\frac{2}{3}$ times of the $△$ABC.

Also, as we can clearly see, ∠B = 90° is common in both the triangles $△$ABC and $△$MNB. Hence, the constructed triangle $△$MNB is also a right angle triangle.

#### 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 $\frac{5}{3}$.

Steps of construction:

1. Draw the line segment BC = 6 cm. 2. Taking B and C as centres, draw two arcs of radii 4 cm and 5 cm respectively intersecting each other at A. 3. Join BA and CA. $△$ABC is the required triangle. 4. From B, draw any ray BD downwards making at acute angle. 5. Mark five points B1, B2, B3, B4 and B5 on BD, such that BB= B1B2 = B2B3 = B3B4 = B4B5. 6. Join B3C and from B5, draw B5 B3C intersecting the extended line segment BC at M. 7. From point M, draw MN ∥ CA intersecting the extended line segment BA at N. 8. Then, $△$NBM is the required triangle whose sides are equal to $\frac{5}{3}$ of the corresponding sides of the $△$ABC.

Justification:
As per the construction, ∆MNB is the required triangle
Let BB1 = B1B2 = B2B3 = B3B4 = B4B5 = x.

In triangles $△$BCB3 and $△$BMB5
∠B = ∠B                                   (common)
∠BCB3 = ∠BMB5                    (corresponding angles of the same transverse as B5D ∥ B3C)
∴ $△$BCB3 $~$ $△$BMB5             (by AA congruency criteria)

$⇒\frac{\mathrm{BM}}{\mathrm{BC}}=\frac{{\mathrm{BB}}_{5}}{{\mathrm{BB}}_{3}}=\frac{5x}{3x}=\frac{5}{3}$                           .....(1)

Similarly, in $△$MBN and $△$CBA,
∠B = ∠B                                         (common)
∠BAC = ∠BNM                             (corresponding angles of the same transverse as AC ∥ MN)
∴ $△$MBN $~$ $△$CBA                      (by AA congruency criteria)

$⇒\frac{\mathrm{AB}}{\mathrm{NB}}=\frac{\mathrm{AC}}{\mathrm{NM}}=\frac{\mathrm{BC}}{\mathrm{BM}}=\frac{5}{3}$                           .....(2)

From (1) and (2), the constructed triangle $△$NBM is of scale $\frac{5}{3}$ times of the $△$ABC.

#### 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.

Steps of construction:

1. Draw a circle of radius 4 cm. Let centre of this circle be O. 2. Take a point M at 6 cm away from the radius. 3. Join OM and bisect it. Now, with M and O as centres and with radius more than half of  OM, draw two arcs on the either sides of the line. Let the arcs meet at A and B and M1 be mid-point of OM. 4. Taking M1 as centre and M1O as radius, draw a circle to intersect the previous circle with radius 4 and centre O at two points P and Q. 5. Join PM and MQ. PM and MQ are the required tangents from the point M to circle with centre O and radius 4 cm. Hence, obtain the tangents PM and QM to the circle with centre O and radius 4 cm.

#### Question 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 . Join P and Q and measure the length PQ.

Thinking process:
I. Firstly, we find the ratio of AB in which P divides it with the help of the relation AP = $\frac{3}{4}$AB.
II. Secondly, we find the ratio of AC in which Q divides it with the help of the relation AQ = $\frac{1}{4}$AC.
III. Now, construct the line segment AB and AC in which P and Q respectively divides it in the ratio from step I and II respectively.
IV. Finally, get the points P and Q. After that, join PQ and get the required measurement of PQ.

Given that, AB = 5 cm and AC = 7 cm.
Also,                         .....(1)

Then, PB = AB − AP

$⇒\mathrm{AP}:\mathrm{PB}=\frac{15}{4}:\frac{5}{4}$
Hence, AP : PB = 3 : 1 i.e., scale factor of line segment AB is $\frac{3}{1}$.

Again from (1),

Then,

Hence, AQ : QC = 1 : 3 i.e., scale factor of line segment AQ is $\frac{1}{3}$.

Steps of construction:
1. Draw a line segment AB = 5 cm. 2. Now, draw a ray AZ making an acute ∠BAZ = 60°. 3. With A as centre and radius 7 cm, draw an arc intersecting AZ at C. 4. Draw a ray AX making an acute ∠BAX. 5. Along AX, mark 1 + 3 = 4 points A1, A2, A3 and A4 such that A1A2 = A2A3 = A3A4. 6. Join A4B. 7. From A3, draw A3P ∥ A4B meeting AB at P. [by making an angle equal to ∠AA4B]
Then, P is the point on AB which divides it in the ratio 3 : 1.
So, AP : PB = 3 : 1 8. Draw a ray AY making an acute ∠CAY. 9. Along AY, mark 3 + 1 = 4 points B1, B2, B3 and B4 such that AB1 = B1B2 = B2B3 = B3B4. 10. Join B4C. 11. From B1, draw B1Q ∥ B4C meeting AC at Q.   [by making an angle equal to ∠AB4C]
Then, Q is the point on AC which divides it in the ratio 1 : 3.
So, AQ : QC = 1 : 3. 12. Finally, join PQ and its measurement is 3.25 cm. #### Question 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 $\frac{4}{3}$. Draw the line segment D'A' parallel to DA where A' lies on extended side BA. Is A'BC'D' a parallelogram?

Thinking process:
I. Firstly we draw a line segment, then either of one end of the line segment with length 5 cm and making an angle 60° with this end. We know that is parallelogram both opposite sides are equal and parallel, then again draw a line with 5 cm making an angle with 60° from other end of line segment. Now, join both parallel line by a line segment whose measurement is 3 cm, we get a parallelogram. After that we draw a diagonal and get a triangle $△$BOC.
II. Now, we construct $△$BD'C' similar to $△$BDC with scale factor $\frac{4}{3}$.
III. Now, draw the line segment D'A' parallel to DA.
IV. Finally, we get the required parallelogram A'BC'D'.

Steps of construction:
1. Draw a line segment AB = 3 cm. 2. Now, draw a ray BY making an acute ∠ABY = 60°. 3. With B as centre and radius equal to 5 cm, draw an arc cutting BY at point C. 4. Again draw a ray AZ making an acute ∠ZAX1 = 60°. 5. With A as centre and radius equal to 5 cm, draw an arc on AZ cutting it at point D. 6. Now, join CD and finally make a parallelogram ABCD. 7. Join BD, which is a diagonal of parallelogram ABCD and from B draw any ray BX making an acute angle ∠CBX. 8. Locate 4 points B1, B2, B3, B4 on BX, such that BB1 = B1B2 = B2B3 = B3B4. 9. Join B3C and from B4, draw a line B4C' ∥ B3C intersecting the extended line segment BC at C'. 11. From point C', draw C'D' ∥ CD intersecting the extended line segment BD at D'. Then, $△$D'BC' is the required triangle whose sides are $\frac{4}{3}$ of the corresponding sides of $△$DBC. 12. Now draw a line segment D'A' parallel to DA, where A' lies on the extended side BA i.e., a ray BX1. 13. Finally, we observe that A'BC'D' is a parallelogram in which A'D' = 6.5 cm, A'B = 4 cm and ∠A'BD' = 60° divide it into triangles BC'D' and A'BD' by the diagonal BD'.

Justification that A'BC'D' is a parallelogram:
As per the construction,
CD ∥ C'D' and BC' ∥ A'D'                                                                           .....(1)

Also,
As per the given data, ∠ ABC = 60°.

Consider the parallelogram ABCD, the sum of complementary angles is 180°.
So, ∠ABC + ∠BCD = 180°
⇒ ∠BCD = 180° − ∠ ABC
= 180° − 60°
= 120°
Therefore, ∠BCD = 120°.

Now, as CD ∥ C'D',
⇒ ∠BCD = ∠BC'D'                                          (as corresponding angles)
Hence, ∠BC'D' = 120°                                                                               .....(2)

As BC ∥ AD and AD ∥ A'D', hence BC ∥ A'D',
∠ABC = ∠D'A'X1                                          (corresponding angles)
So, ∠D'A'X1 = ∠ABC = 60°
Now, consider ∠D'A'X1 + ∠D'A'B = 180°           (linear angle)
60° + ∠D’A’B = 180°
⇒ ∠D’A’B = 180° − 60° = 120°                                                               .....(3)

From (1), (2) and (3), we can say that A’BC'D' is a parallelogram as opposite sides and opposite angles are equal.

#### 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.

Given, two concentric circles of radii 3 cm and 5 cm with centre O. We have to draw pair of tangents from point P on outer circle to the other.
Steps of construction:
1. Draw two concentric circles with centre O and radii 3cm and 5cm. 2. Taking any point P on outer circle. Join OP. 3. Bisect OP, let M’ be the mid-point of OP. 4. Taking M’ as centre and OM’ as radius, draw a dotted circle cutting the inner circle at points A and B. 5. Join PA and PB. Thus, PA and PB are required tangents. 6. On measuring PA and PB, we find that PA = PB = 4 cm.

Actual calculation:
In the right angle $△$OAP,
∠PAO = 90°

According to Pythagoras theorem,
(hypotenuse)2 = (base)2 + (perpendicular)2
⇒ PA2 = (5)2 $-$ (3)2
= 25 $-$ 9
= 16 cm
⇒ PA = 4 cm         (∵ side cannot be negative)
Hence, the length of both tangents is 4 cm.
Therefore, PA = PB = 4 cm.

#### Question 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. Draw a line segment BC = 5 cm. 2. Construct MN, the perpendicular bisector of line segment BC meeting BC at P'. 3. Taking B and C as centres, draw two arcs of equal radius 6 cm intersecting each other at A. 4. Join BA and CA. So, $△$ABC is the required isosceles triangle and from B, draw any ray BX making an acute angle, ∠CBX. 5. Locate four points B1, B2, B3 and B4 on BX such that BB1 = B1B2 = B2B3 = B3B4. 6. Join B3C and from B4, draw a line B4R ∥ B3C intersecting the extended line segment BC at R. 7. From point R, draw RP ∥ CA meeting the extended line BA at P. Then, $△$PBR is the required triangle.

Justification:
Let BB1 = B1B2 = B2B3 = B3B4 = x.
As per the construction, B4R ∥ B3C.
Now,
$\frac{\mathrm{BC}}{\mathrm{CR}}=\frac{{\mathrm{BB}}_{3}}{{\mathrm{B}}_{3}{\mathrm{B}}_{4}}=\frac{3x}{x}=\frac{3}{1}$
$\therefore \frac{\mathrm{BC}}{\mathrm{CR}}=\frac{3}{1}$

$⇒\frac{\mathrm{BR}}{\mathrm{BC}}=\frac{\mathrm{BC}+\mathrm{CR}}{\mathrm{BC}}=\frac{\mathrm{BC}}{\mathrm{BC}}+\frac{\mathrm{CR}}{\mathrm{BC}}=1+\frac{1}{3}=\frac{4}{3}$

Also, from the construction, RP ∥ CA.

In ,

$\angle \mathrm{PRB}=\angle \mathrm{ACB}$                                            [Corresponding Angles]
$⇒△\mathrm{PBR}~△\mathrm{ABC}$                                        [By AA criteria]
and $\frac{\mathrm{PB}}{\mathrm{AB}}=\frac{\mathrm{RP}}{\mathrm{CA}}=\frac{\mathrm{BR}}{\mathrm{BC}}=\frac{4}{3}$

Hence, the new triangle is similar to the given triangle whose sides are $\frac{4}{3}$ times of the corresponding sides of the isosceles $△$ABC.

#### Question 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 $\frac{5}{7}$. Justify the construction.

Steps of construction:
1. Draw a line segment AB = 5 cm.
2. From point B, draw ∠ABY = 60° and mark an arc on it such that BC = 6 cm.
3. Join AC, $△$ABC is the required triangle.
4. From A, draw any ray AX downwards making an acute angle ∠BAX.
5. Mark 7 points B1, B2, B3, B4, B5, B6 and B7 on AX, such that AB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
6. Join B7B and from B5, draw B5M ∥ B7B intersecting AB at point M.
7. From point M, draw MN ∥ BC intersecting AC at N. Then, $△$AMN is the required triangle whose sides are equal to $\frac{5}{7}$ of the corresponding sides of the $△$ABC. Justification:
Here, B5M ∥ B7B (by construction).
$\therefore \frac{\mathrm{AM}}{\mathrm{MB}}=\frac{5}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{MB}}{\mathrm{AM}}=\frac{2}{5}$

Now,
$\begin{array}{rcl}\frac{\mathrm{AB}}{\mathrm{AM}}& =& \frac{\mathrm{AM}+\mathrm{MB}}{\mathrm{AM}}\\ & =& 1+\frac{\mathrm{MB}}{\mathrm{AM}}\\ & =& 1+\frac{2}{5}\\ & =& \frac{7}{5}\end{array}$

Also, MN ∥ BC.
$\therefore △\mathrm{AMN}~△\mathrm{ABC}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AC}}=\frac{\mathrm{NM}}{\mathrm{BC}}=\frac{5}{7}$

#### Question 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:
Step 1:  A circle with center O and radius 4 cm is drawn.
Step 2:  OQ is joined when Q is any point on the circle.
Step 3: ∠OQR = 30$°$ is drawn and QR intersects the circle at R.
Step 4: Centering Q and R and with radius QR two arcs are drawn, they intersect at P.
Step 5: PQ and PR are joined.
PQ and PR are the required tangents. Justification:
PQ = QR = RP (By construction)

Thus, PQ is a tangent at Q.
Again in $△$OQR, we have
$\angle \mathrm{OQR}=\angle \mathrm{ORQ}=30°$

Thus, PR is a tangent to circle at R.
Distance between center and intersection of tangent i.e., OP = 8 cm.

#### Question 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 $\frac{3}{2}$. 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:
Step 1:
Construct a triangle ABC with AB = 4 cm, BC = 6 cm and AC = 9 cm.
Step 2: Draw a ray AX making an acute angle with the base AC and mark 3 points A1​, A2​, A3​ such that AA1​ = A1​A2 ​= A2​A3.
Step 3: Join A2C and draw a line A3​C′ such that A2C is parallel to A3​C′ where C′ lies on the produced AC.
Step 4: Now draw another line parallel to BC at C′ such that it meets the produced AB at B′.

Here, $△$AB′C′ is the required triangle similar to $△$ABC with scale factor $\frac{3}{2}$. Justification:
Let AA1 = A1A2 = A2A3 = x.
As per the construction, A3​C′ ∥ A2C.
$⇒\frac{\mathrm{AC}}{\mathrm{CC}\text{'}}=\frac{2}{1}$

$\begin{array}{rcl}\frac{\mathrm{AC}\text{'}}{\mathrm{AC}}& =& \frac{\mathrm{AC}+\mathrm{CC}\text{'}}{\mathrm{AC}}\\ & =& 1+\frac{\mathrm{CC}\text{'}}{\mathrm{AC}}\\ & =& 1+\frac{1}{2}\\ & =& \frac{3}{2}\end{array}$
Again, BC ∥ B'C'.
$\therefore \frac{\mathrm{AB}\text{'}}{\mathrm{AB}}=\frac{\mathrm{B}\text{'}\mathrm{C}\text{'}}{\mathrm{BC}}=\frac{\mathrm{AC}\text{'}}{\mathrm{AC}}=\frac{3}{2}$
Hence, $△\mathrm{AB}\text{'}\mathrm{C}\text{'}~△\mathrm{ABC}$.

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