Rd Sharma 2020 Solutions for Class 10 Maths Chapter 9 Constructions are provided here with simple step-by-step explanations. These solutions for Constructions are extremely popular among Class 10 students for Maths Constructions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 Book of Class 10 Maths Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 Solutions. All Rd Sharma 2020 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 9.10:

#### Question 18:

Construct a Δ*ABC** *in which AB = 5 cm. ∠B = 60° altitude *CD* = 3 cm. Construct a ΔAQR similar to Δ*ABC* such that side Δ*AQR* is 1.5 times that of the corresponding sides of Δ*ACB*

#### Answer:

Given that

Construct a trianglein which and then a triangle similar to it whose sides are of the corresponding sides of.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With *B *as centre and draw an angle.

Step: III -From point *A* and *B* construct which cut the line *BS* at point *C*

Step: IV- Join *AC *to obtain.

Step: V- Below *AB, *makes an acute angle.

Step: VI- Along *AX,* mark off five points such that

Step: VII -Join.

Step: VIII -Since we have to construct a triangle each of whose sides is of the corresponding sides of.

So, we draw a line on *AX *from point which is and meeting *AB *at *Q.*

Step: IX- From *Q *point draw and meeting *AC *at *R*

Thus, is the required triangle, each of whose sides is of the corresponding sides of*.*

#### Page No 9.17:

#### Question 1:

Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

#### Answer:

Given that

Construct a circle of radius, and form its centre, construct the pair of tangents to the circle.

Find the length of tangents.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a circle of radius.

Step: II- Make a point *P* at a distance of, and join .

Step: III -Draw a right bisector of, intersecting at *Q .*

Step: IV- Taking *Q *as centre and radius, draw a circle to intersect the given circle at *T* and *T’*.

Step: V- Joins *PT *and *PT’ *to obtain the require tangents.

Thus, are the required tangents.

Find the length of tangents.

As we know that and is right triangle.

Therefore,

In,

Thus, the length of tangents

#### Page No 9.17:

#### Question 2:

Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

#### Answer:

Given that

Construct a circle of radius, and extended diameter each at distance of 7cm from its centre. Construct the pair of tangents to the circle from these two points .

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a circle of radius.

Step: II- Make a line *CD* .

Step: III-Extend the line *CD *in such a way that point* *

Step: IV-* CP* at a distance of, and join draw a right bisector of, intersecting at *R.*

Step* *V:- Similarly,* DQ* at a distance of, and join draw a right bisector of, intersecting at *S.* * *

Step VI: Taking R and S* *as centre and radius, draw a circle to intersect the given circle at *T* and *T’ *

*B* and *B ’*respectively.

Step: VII- Joins *PT *and *PT’ *as well as *QB *and* QB’* to obtain the require tangents.

Thus, are the required tangents.

#### Page No 9.17:

#### Question 3:

Draw a line segment *AB* of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking *B* as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

#### Answer:

Given that

Construct a circle of radius, and extended diameter each at distance of 7cm from its centre. Construct the pair of tangents to the circle from these two points .

We follow the following steps to construct the given

Step of construction

Step: I First of all we draw a line.

Step: II taking *A *as a centre and draw a circle of radius. Similarly, taking *B *as a centre and draw a circle of radius.

Step: III draw the perpendicular bisector of

Step IV : draw the another circle with taking the bisector point as centre and radius = mid point of which cut the point

Step: V joins and respectively*.** *as well as to obtain the require tangents.

Thus, are the required tangents.

#### Page No 9.17:

#### Question 4:

Draw two tangents to a circle of radius 3.5 cm from a point *P* at a distance of 6.2 from its centre. [CBSE 2013]

#### Answer:

**Steps of Construction**

**Step 1**. Draw a circle with O as centre and radius 3.5 cm.

**Step 2**. Mark a point P outside the circle such that OP = 6.2 cm.

**Step 3**. Join OP. Draw the perpendicular bisector XY of OP, cutting OP at Q.

**Step 4**. Draw a circle with Q as centre and radius PQ (or OQ), to intersect the given circle at the points T and T'.

**Step 5**. Join PT and PT'.

Here, PT and PT' are the required tangents.

#### Page No 9.17:

#### Question 5:

Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°. [CBSE 2013]

#### Answer:

**Steps of Construction**

**Step 1**. Draw a circle with centre O and radius 4.5 cm.

**Step 2**. Draw any diameter AOB of the circle.

**Step 3**. Construct $\angle $BOC = 45º such that radius OC cuts the circle at C.

**Step 4**. Draw AM ⊥ AB and CN ⊥ OC. Suppose AM and CN intersect each other at P.

Here, AP and CP are the pair of tangents to the circle inclined to each other at an angle of 45º.

#### Page No 9.18:

#### Question 6:

Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and $\angle $B = 90º. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle. [CBSE 2014]

#### Answer:

**Steps of Construction**

**Step 1**. Draw a line segment AB = 6 cm.

**Step 2**. At B, draw $\angle $ABX = 90º.

**Step 3**. With B as centre and radius 8 cm, draw an arc cutting ray BX at C.

**Step 4**. Join AC. Thus, ∆ABC is the required triangle.

**Step 5**. From B, draw BD ⊥ AC.

**Step 6**. Draw the perpendicular bisector of BC, cutting BC at O.

**Step 7**. With O as centre and radius OB (or OC), draw a circle. This circle passes through B, C and D.

Thus, this is the required circle.

**Step 8**. Join OA.

**Step 9**. Draw the perpendicular bisector of OA, cutting OA at E.

**Step 10**. With E as centre and radius AE (or OE), draw a circle intersecting the circle with centre O at B and F.

**Step 11**. Join AF.

Here, AB and AF are the required tangents.

#### Page No 9.18:

#### Question 7:

Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to smaller circle from a point on the larger circle. Also measure its length.

#### Answer:

Following are the steps to draw tangents on the given circle:

**Step 1**

Draw a circle of 3 cm radius with centre O on the given plane.

**Step 2**

Draw a circle of 5 cm radius, taking O as its centre. Locate a point P on this circle and join OP.

**Step 3**

Bisect OP. Let M be the midpoint of PO.

**Step 4**

Taking M as its centre and MO as its radius, draw a circle. Let it intersect the given circle at points Q and R.

**Step 5**

Join PQ and PR. PQ and PR are the required tangents.

It can be observed that PQ and PR are of length 4 cm each.

In ΔPQO,

Since PQ is a tangent,

∠PQO = 90°

PO = 5 cm

QO = 3 cm

Applying Pythagoras theorem in ΔPQO, we obtain

PQ^{2} + QO^{2} = PQ^{2}

PQ^{2} + (3)^{2} = (5)^{2}

PQ^{2 }+ 9 = 25

PQ^{2 }= 25 − 9

PQ^{2 }= 16

PQ = 4 cm

Hence justified.

#### Page No 9.18:

#### Question 1:

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°

#### Answer:

PA and PB are two tangents drawn from P to circle with centre O.

∠APB = 60°

Now,

∠OAP = 90° (Radius is perpendicular to the tangent at the point of contact)

∠OBP = 90° (Radius is perpendicular to the tangent at the point of contact)

In quadrilateral OAPB,

∠AOB + ∠OAP + ∠OBP + ∠APB = 360º (Angle sum property of quadrilateral)

⇒ ∠AOB + 90º + 90º + 60º = 360º

⇒ ∠AOB = 360º − 240º = 120º

Thus, in order 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 120°.

Hence, the correct answer is option (d).

#### Page No 9.18:

#### Question 2:

To divide a line segment *AB *in the ratio 5:7, first 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

#### Answer:

We know that, in order to divide a line segment AB in the ratio *m : n* (*m, n* are positive integers), first draw a ray AX such that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is *m* + *n*.

Here, *m* = 5 and *n* = 7

∴ *m* + *n* = 5 + 7 = 12

Thus, to divide a line segment AB in the ratio 5 : 7, first 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 12.

Hence, the correct answer is option (d).

#### Page No 9.18:

#### Question 3:

To divide a line segment *AB *in the ratio *m : n* (*m, n* are positive integers), draw a ray *AX* such that ∠*BAX *is an acute angle and then mark points on ray *AX* at equal distances such that the minimum number of these points is

(a) greater than *m* and *n*

(b) *m* + *n*

(c)* m + n – *1

(d) *mn*

#### Answer:

To divide a line segment AB in the ratio *m *:* n* (*m, n* are positive integers), firstly draw a ray AX such that ∠BAX is an acute angle. Then, along AX mark off *m* + *n* points at equal distances.

Here, AP : PB = *m* : *n*

Thus, to divide a line segment AB in the ratio *m : n* (*m, n* are positive integers), draw a ray AX such that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is *m* + *n*.

Hence, the correct answer is option (b).

#### Page No 9.18:

#### Question 4:

To draw a pair of tangents to a circle which are inclined to each other at an angle of 35°, it is required to draw tangents at the end points of those two radii of the circle, the angle between which is

(a) 105°

(b) 70°

(c) 140°

(d) 145°

#### Answer:

PA and PB are two tangents drawn from P to circle with centre O.

∠APB = 35°

Now,

∠OAP = 90° (Radius is perpendicular to the tangent at the point of contact)

∠OBP = 90° (Radius is perpendicular to the tangent at the point of contact)

In quadrilateral OAPB,

∠AOB + ∠OAP + ∠OBP + ∠APB = 360º (Angle sum property of quadrilateral)

⇒ ∠AOB + 90º + 90º + 35º = 360º

⇒ ∠AOB = 360º − 215º = 145º

Thus, in order to draw a pair of tangents to a circle which are inclined to each other at an angle of 35°, it is required to draw tangents at the end points of those two radii of the circle, the angle between which is 145º.

Hence, the correct answer is option (d).

#### Page No 9.18:

#### Question 5:

To construct a triangle similar to a given ∆*ABC *with its side 8/5 of the corresponding side 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

#### Answer:

In order to construct a triangle similar to a given triangle with its sides $\frac{m}{n}$ of the corresponding sides of the triangle, the minimum number number of points to be located at equal distances on the ray is *m* or *n*, whichever is greater.

If *m* > *n*, then minimum points to be located at equal distances on the ray is *m*.

If *n* > *m*, then minimum points to be located at equal distances on the ray is *n*.

Here, *m* = 8 and *n* = 5

8 > 5

Thus, in order to construct a triangle similar to a given ∆ABC with its side $\frac{8}{5}$ of the corresponding side 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 8.

Hence, the correct answer is option (b).

#### Page No 9.18:

#### Question 6:

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 *A*_{1}, *A*_{2}, *A*_{3}, ... are located at equal distances on the ray *AX *and the point *B* is joined to

(a) *A*_{12}

(b) *A*_{11}

(c) *A*_{10}

(d) *A*_{9}

#### Answer:

To divide a line segment AB in the ratio *m *:* n* (*m, n* are positive integers), firstly draw a ray AX such that ∠BAX is an acute angle. Now, along AX mark off *m* + *n* points at equal distances.

Suppose these *m* + *n *points marked on ray AX be A_{1}, A_{2}, A_{3}, ..., A* _{m}*, A

_{m }_{+ 1}, ..., A

_{m }_{+ n.}Then,

*join B to A*

_{m }_{+ n}

*.*

Here,

*m*= 4 and

*n*= 7

∴

*m*+

*n*= 4 + 7 = 11

So, B is joined to the point A

_{11}.

Thus, in order 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 A

_{1}, A

_{2}, A

_{3}, ... are located at equal distances on the ray AX and the point B is joined to A

_{11}.

Hence, the correct answer is option (b).

#### Page No 9.19:

#### Question 7:

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 *B*_{1}, *B*_{2}, *B*_{3},.... on *BX *at equal distances and next step is to join

(a) *B*_{10 }to *C*

(b) *B*_{3} to* C*

(c) *B*_{7} to *C*

(d) *B*_{4 }to *C*

#### Answer:

In order to construct a triangle similar to a given triangle with its sides $\frac{m}{n}$ of the corresponding sides of the triangle, the minimum number number of points to be located at equal distances on the ray is *m* or *n*, whichever is greater.

If *m* > *n*, then minimum points to be located at equal distances on the ray is *m*.

If *n* > *m*, then minimum points to be located at equal distances on the ray is *n*.

Here, *m* = 3 and *n* = 7

7 > 3

So, seven points B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}, B_{7} are marked at equal distance on BX. Then B_{7} is joined to C.

Thus, in order 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 B_{1}, B_{2}, B_{3},...., B_{7} on BX at equal distances and next step is to join B_{7} to C.

Hence, the correct answer is option (c).

#### Page No 9.19:

#### Question 8:

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 *A*_{1}, *A*_{2}, .... and *B*_{1}, *B*_{2},..... are located at equal distances on rays *AX *and *BY *respectively. Then the points joined are

(a) *A*_{5} and *B*_{6}

(b) *A*_{6} and *B*_{5}

(c) *A*_{4} and *B*_{5}

(d) *A*_{5} and *B*_{4}

#### Answer:

To divide a line segment AB in the ratio *m* : *n*, draw a ray AX such that ∠BAX is an acute angle. Then draw a ray BY parallel to AX. The points A_{1}, A_{2}, ..., A_{m} and B_{1}, B_{2}, ..., B* _{n}* are located at equal distances on rays AX

*and BY respectively. Then the points A*

_{m}and B

*are joined to divide the line segment AB in the ratio*

_{n}*m*:

*n*.

Here,

*m*= 5 and

*n*= 6

So, the points A

_{1}, A

_{2}, ..., A

_{5 }and B

_{1}, B

_{2}, ..., B

_{6 }are located at equal distances on rays AX and BY respectively. Then the points A

_{5}and B

_{6}are joined.

Thus, in order 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 A

_{1}, A

_{2}, ..., A

_{5}and B

_{1}, B

_{2}, ..., B

_{6}are located at equal distances on rays AX and BY respectively. Then the points joined are A

_{5}and B

_{6}.

Hence, the correct answer is option (a).

#### Page No 9.4:

#### Question 1:

Determine a point which divides a line segment of length 12 cm internally in the ratio 2 : 3 Also, justify your construction.

#### Answer:

Given that

Determine a point which divides a line segment of lengthinternally in the ratio of.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- We draw a ray making an acute anglewith.

Step: III- Draw a ray parallel to *AX* by making an acute angle.

Step IV- Mark of two points on and three points on in such a way that.

Step: V- Joins and this line intersects at a point *P*.

Thus, *P* is the point dividing internally in the ratio of

Justification:

In we have

And

So, *AA* similarity criterion, we have

#### Page No 9.4:

#### Question 2:

Divide a line segment of length 9 cm internally in the ratio 4 : 3. Also, give justification of the construction.

#### Answer:

Given that

Determine a point which divides a line segment of lengthinternally in the ratio of.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- We draw a ray making an acute anglewith.

Step: III- Draw a ray parallel to *AX* by making an acute angle.

Step IV- Mark of two points on and three points on in such a way that.

Step: V- Joins and this line intersects at a point *P*.

Thus, *P* is the point dividing internally in the ratio of

Justification:

In we have

And

So, *AA* similarity criterion, we have

#### Page No 9.4:

#### Question 3:

Divide a line segment of length 14 cm internally in the ratio 2 : 5. Also, justify your construction.

#### Answer:

Given that

Determine a point which divides a line segment of lengthinternally in the ratio of.

We follow the following steps to construct the given

Step of construction

Step: I-First of all we draw a line segment.

Step: II- We draw a ray making an acute anglewith.

Step: III- Draw a ray parallel to *AX* by making an acute angle.

Step IV- Mark of two points on and three points on in such a way that.

Step: V- Joins and this line intersects at a point *P*.

Thus, *P* is the point dividing internally in the ratio of

Justification:

In we have

And

So, *AA* similarity criterion, we have

#### Page No 9.4:

#### Question 4:

Draw a line segment of length 8 cm and divide it internally in the ratio 4:5.

#### Answer:

Steps of construction:

1) Draw a line segment AB = 8 cm.

2) Draw a ray AX making an acute angle $\angle BAX=60\xb0$ with AB.

3) Draw a ray BY parallel to AX by making an acute angle $\angle ABY=\angle BAX$.

4) Mark of four points ${A}_{1},{A}_{2},{A}_{3},{A}_{4}$ on AX and five points ${B}_{1},{B}_{2},{B}_{3},{B}_{4},{B}_{5}$ on BY in such a way that $A{A}_{1}={A}_{1}{A}_{2}={A}_{2}{A}_{3}={A}_{3}{A}_{4}$.

5) Joins ${A}_{4}{B}_{5}$ and this line intersects AB at a point *P*.

Thus, *P* is the point dividing AB internally in the ratio of 4 : 5.

#### Page No 9.9:

#### Question 1:

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are (2/3) of the corresponding sides of it.

#### Answer:

Given that

Construct a triangle of sides and then a triangle similar to it whose sides are of the corresponding sides of it.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With *A *as centre and radius, draw an arc.

Step: III- With *B *as centre and radius, draw an arc, intersecting the arc drawn in step II at *C.*

Step: IV- Joins *AC *and *BC *to obtain.

Step: V- Below *AB, *makes an acute angle.

Step: VI- Along *AX,* mark off three points such that

Step: VII- Join.

Step: VIII- Since we have to construct a triangle each of whose sides is two-third of the corresponding sides of.

So, we take two parts out of three equal parts on *AX *from point draw and meeting *AB *at *C’.*

Step: IX- From *B’ *draw and meeting *AC *at *C’*

Thus, is the required triangle, each of whose sides is two third of the corresponding sides of*.*

#### Page No 9.9:

#### Question 2:

Construct a triangle similar to a given Δ*ABC** *such that each of its sides is (5/7)^{th} of the corresponding sides of Δ* ABC.* It is given that AB - 5 cm, BC = 7 cm and ∠*ABC** *= 50°.

#### Answer:

Given that

Construct a triangle similar to a triangle *ABC *such that each of sides is of the corresponding sides of triangle *ABC.*

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With *B *as centre and draw an angle.

Step: III- With *B *as centre and radius, draw an arc, cut the line* BY *drawn in step II at *C.*

Step: IV- Joins *AC *to obtain.

Step: V- Below *AB, *makes an acute angle.

Step: VI- Along *AX,* mark off seven points such that

Step: VII-Join.

Step: VIII- Since we have to construct a triangle each of whose sides is of the corresponding sides of.

So, we take five parts out of seven equal parts on *AX *from point draw and meeting *AB *at *B’.*

Step: IX- From *B’ *draw and meeting *AC *at *C’*

Thus, is the required triangle, each of whose sides is of the corresponding sides of*.*

#### Page No 9.9:

#### Question 3:

Construct a triangle similar to a given Δ*ABC* such that each of its sides is (2/3)^{rd} of the corresponding sides of Δ*ABC*. It is given that BC = 6 cm, ∠*B* = 50° and ∠*C* = 60°.

#### Answer:

Given that

Construct a triangle of given data, and then a triangle similar to it whose sides are of the corresponding sides of .

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With *B *as centre draw an angle.

Step: III- With *C *as centre draw an anglewhich intersecting the line drawn in step II at *A.*

Step: IV- Joins *AB *and A*C *to obtain.

Step: V -Below *BC, *makes an acute angle.

Step: VI -Along *BX,* mark off three points such that

Step: VII -Join.

Step: VIII -Since we have to construct a triangle each of whose sides is two-third of the corresponding sides of.

So, we take two parts out of three equal parts on *BX *from point draw and meeting *BC *at *C’.*

Step: IX -From *C’ *draw and meeting *AB *at A*’*

Thus, is the required triangle, each of whose sides is two third of the corresponding sides of*.*

#### Page No 9.9:

#### Question 4:

Draw a Δ*ABC* in which *BC* = 6 cm, *AB* = 4 cm and *AC* = 5 cm. Draw a triangle similar to Δ*ABC* with its sides equal to (3/4)^{th} of the corresponding sides of Δ*ABC*.

#### Answer:

Given that

Construct a triangle of sides and then a triangle similar to it whose sides are of the corresponding sides of.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With *A *as centre and radius, draw an arc.

Step: III -With *B *as centre and radius, draw an arc, intersecting the arc drawn in step II at *C.*

Step: IV -Joins *AC *and *BC *to obtain.

Step: V -Below *AB, *makes an acute angle.

Step: VI -Along *AX,* mark off four points such that

Step: VII -Join.

Step: VIII -Since we have to construct a triangle each of whose sides is of the corresponding sides of.

So, we take three parts out of four equal parts on *AX *from point draw and meeting *AB *at B*’.*

Step: IX- From *B’ *draw and meeting *AC *at *C’*

Thus, is the required triangle, each of whose sides is of the corresponding sides of*.*

#### Page No 9.9:

#### Question 5:

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $\frac{5}{7}$ of the corresponding sides of the first triangle.

#### Answer:

**Steps of Construction
Step 1. **Draw a line segment QR = 6 cm.

**Step 2.**With Q as centre and radius 7 cm, draw an arc.

**Step 3.**With R as centre and radius 5 cm, draw an arc cutting the previous arc at P.

**Step 4.**Join PQ and PR. Thus ∆PQR is the required triangle.

**Step 5.**Below QR, draw an acute angle ∠RQX.

**Step 6.**Along QX, mark seven points Q

_{1}, Q

_{2}, Q

_{3}, Q

_{4}, Q

_{5}, Q

_{6}and Q

_{7}such that QQ

_{1}= Q

_{1}Q

_{2}= Q

_{2}Q

_{3}= Q

_{3}Q

_{4}= Q

_{4}Q

_{5}= Q

_{5}Q

_{6}= Q

_{6}Q

_{7}.

**Step 7.**Join Q

_{7}R.

**Step 8.**From Q

_{ }_{5}, draw Q

_{5}R' || Q

_{7}R meeting QR at R'.

**Step 9.**From R', draw P'R' || PR meeting PQ in P'.

Here, ∆P'QR' is the required triangle, each of whose sides are $\frac{5}{7}$ of the corresponding sides of ∆PQR.

#### Page No 9.9:

#### Question 6:

Draw a right triangle *ABC* in which *AC* = *AB* = 4.5 cm and ∠A = 90°. Draw a triangle similar to Δ*ABC* with its sides equal to (5/4)th of the corresponding sides of Δ*ABC*.

#### Answer:

Given that

Construct a right triangle of sides and then a triangle similar to it whose sides are of the corresponding sides of.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With *A *as centre and draw an angle.

Step: III- With *A *as centre and radius*.*

Step: IV- Join *BC *to obtain.

Step: V- Below *AB, *makes an acute angle.

Step: VI- Along *AX,* mark off five points such that

Step: VII-Join.

Step: VIII- Since we have to construct a triangle each of whose sides is of the corresponding sides of.

So, we draw a line on *AX *from point which is and meeting *AB *at B*’.*

Step: IX- From *B’ *point draw and meeting *AC *at *C’*

Thus, is the required triangle, each of whose sides is of the corresponding sides of*.*

#### Page No 9.9:

#### Question 7:

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

#### Answer:

Given that

Construct a right triangle of sides and then a triangle similar to it whose sides are of the corresponding sides of.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With *A *as centre and draw an angle.

Step: III- With *A *as centre and radius*.*

Step: IV -Join *BC *to obtain.

Step: V -Below *AB, *makes an acute angle.

Step: VI -Along *AX,* mark off five points such that

Step: VII -Join.

Step: VIII -Since we have to construct a triangle each of whose sides is of the corresponding sides of.

So, we draw a line on *AX *from point which is and meeting *AB *at B*’.*

Step: IX -From *B’ *point draw and meeting *AC *at *C’*

Thus, is the required triangle, each of whose sides is of the corresponding sides of*.*

#### Page No 9.9:

#### Question 8:

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/2 times the corresponding sides of the isosceles triangle.

#### Answer:

Given that

Construct an isosceles triangle ABC in which *AB* = *BC* = 6 cm and altitude = 4 cm then another triangle similar to it whose sides are $\frac{3}{2}$ of the corresponding sides of$\u25b3ABC$.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment *AB* = 6 cm.

Step: II- With *B *as centre and radius = *BC* = 6 cm, draw an arc.

Step: III- From point *A* and *B* construct which cut the line *BS* at point *C*

Step: IV -Join *AC *to obtain.

Step: V- Below *AB, *makes an acute angle.

Step: VI -Along *AX,* mark off five points such that

Step: VII- Join.

So, we draw a line on *AX *from point which is and meeting *AB *at *Q.*

Step: IX -From *Q *point draw and meeting *AC *at *R*

Thus, is the required triangle, each of whose sides is of the corresponding sides of*.*

#### Page No 9.9:

#### Question 9:

Draw a Δ*ABC* with side *BC* = 6 cm. *AB* = 5 cm and ∠ *ABC* = 60°. Then, construct a triangle whose sides are (3/4)^{th} of the corresponding sides of the Δ*ABC*.

#### Answer:

Given that

Construct a of given data, and then a triangle similar to it whose sides are of the corresponding sides of.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With *B *as centre draw an angle.

Step: III- With *B *as centre and radius, draw an arc.

Step: IV- Join *AC *to obtain.

Step: V -Below *AB, *makes an acute angle.

Step: VI -Along *AX,* mark off four points such that

Step: VII -Join.

So, we take three parts out of four equal parts on *AX *from point draw and meeting *AB *at *B’.*

Step: IX- From *B’ *draw and meeting *AC *at *C’*

Thus, is the required triangle, each of whose sides is of the corresponding sides of*.*

#### Page No 9.9:

#### Question 10:

Construct a triangle similar to Δ *ABC* in which *AB* = 4.6 cm, *BC* = 5.1 cm, ∠*A* = 60° with scale factor 4 : 5.

#### Answer:

Given that

Construct a of given data, and then a triangle similar to it whose sides are of the corresponding sides of.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With *A *as centre draw an angle.

Step: III- With *B *as centre and radius, draw an arc, intersecting the arc drawn in step II at *C.*

Step: IV- Joins *BC *to obtain.

Step: V- Below *AB, *makes an acute angle.

Step: VI- Along *AX,* mark off five points such that

Step: VII- Join.

Step: VIII- Since we have to construct a triangle each of whose sides is of the corresponding sides of.

So, we take four parts out of five equal parts on *AX *from point draw and meeting *AB *at *B’.*

Step: IX- From *B’ *draw and meeting *AC *at *C’*

Thus, is the required triangle, each of whose sides is of the corresponding sides of*.*

#### Page No 9.9:

#### Question 11:

Construct a triangle similar to a given Δ*XYZ* with its sides equal to (3/4)^{th} of the corresponding sides of Δ*XYZ*. Write the steps of construction.

#### Answer:

Given that

Construct a of given data, and then a triangle similar to it whose sides are of the corresponding sides of.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With *Y *as centre draw an angle.

Step: III- With *Y *as centre and radius, draw an arc.

Step: IV- Join *XZ *to obtain.

Step: V- Below *XY, *makes an acute angle.

Step: VI -Along *XP,* mark off four points such that

Step: VII- Join.

So, we take three parts out of four equal parts on *XP *from point draw and meeting *XY *at *Y’.*

Step: IX- From *Y’ *draw and meeting *XZ *at Z*’*

Thus, is the required triangle, each of whose sides is of the corresponding sides of*.*

#### Page No 9.9:

#### Question 12:

Draw a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of the first triangle.

#### Answer:

Given that

Construct a right triangle of sides and then a triangle similar to it whose sides are of the corresponding sides of.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With *A *as centre and draw an angle.

Step: III- With *A *as centre and radius*.*

Step: IV-Join *BC *to obtain right .

Step: V- Below *AB, *makes an acute angle.

Step: VI- Along *AX,* mark off five points such that

Step: VII- Join.

Step: VIII -Since we have to construct a triangle each of whose sides is of the corresponding sides of right .

So, we draw a line on *AX *from point which is and meeting *AB *at B*’.*

Step: IX- From *B’ *point draw and meeting *AC *at *C’*

Thus, is the required triangle, each of whose sides is of the corresponding sides of*.*

#### Page No 9.9:

#### Question 13:

Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are $\frac{3}{5}$ times the corresponding sides of the given triangle. [CBSE 2014]

#### Answer:

**Steps of Construction**

**Step 1**. Draw a line segment QR = 5.5 cm.

**Step 2**. With Q as centre and radius 5 cm, draw an arc.

**Step 3**. With R as centre and radius 6.5 cm, draw an arc cutting the previous arc at P.

**Step 4**. Join PQ and PR. Thus, ∆PQR is the required triangle.

**Step 5**. Below QR, draw an acute angle $\angle $RQX.

**Step 6**. Along QX, mark five points R_{1}, R_{2}, R_{3}, R_{4 }and R_{5 }such that QR_{1} = R_{1}R_{2} = R_{2}R_{3} = R_{3}R_{4 }= R_{4}R_{5}.

**Step 7**. Join RR_{5}.

**Step 8**. From R_{3}, draw R_{3}R' || RR_{5} meeting QR at R'.

**Step 9**. From R', draw P'R' || PR meeting PQ in P'.

Here, ∆P'QR' is the required triangle, each of whose sides are $\frac{3}{5}$ times the corresponding sides of ∆PQR.

#### Page No 9.9:

#### Question 14:

Construct a triangle PQR with sides QR = 7 cm, PQ = 6 cm and $\angle $PQR = 60º. Then construct another triangle whose sides are $\frac{3}{5}$ of the corresponiding sides of ∆PQR. [CBSE 2014, 2015]

#### Answer:

**Steps of Construction**

**Step 1**. Draw a line segment QR = 7 cm.

**Step 2**. At B, draw $\angle $XQR = 60º.

**Step 3**. With Q as centre and radius 6 cm, draw an arc cutting the ray QX at P.

**Step 4**. Join PR. Thus, ∆PQR is the required triangle.

**Step 5**. Below QR, draw an acute angle $\angle $YQR.

**Step 6**. Along QY, mark five points R_{1}, R_{2}, R_{3}, R_{4 }and R_{5 }such that QR_{1} = R_{1}R_{2} = R_{2}R_{3} = R_{3}R_{4} = R_{4}R_{5} .

**Step 7**. Join RR_{5}.

**Step 8**. From R_{3}, draw R_{3}R' || RR_{5} meeting QR at R'.

**Step 9**. From R', draw P'R' || PR meeting PQ in P'.

Here, ∆P'QR' is the required triangle whose sides are $\frac{3}{5}$ of the corresponding sides of ∆PQR.

#### Page No 9.9:

#### Question 15:

Draw a ∆ABC in which base BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are $\frac{3}{4}$ of the corresponding sides of ∆ABC.

#### Answer:

ΔA'BC' whose sides are of the corresponding sides of ΔABC can be drawn as follows.

**Step 1**

Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

**Step 2**

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

**Step 3**

Locate 4 points (as 4 is greater in 3 and 4), B_{1}, B_{2}, B_{3}, B_{4}, on line segment BX.

**Step 4**

Join B_{4}C and draw a line through B_{3}, parallel to B_{4}C intersecting BC at C'.

**Step 5**

Draw a line through C' parallel to AC intersecting AB at A'. ΔA'BC' is the required triangle.

**Justification**

The construction can be justified by proving

In ΔA'BC' and ΔABC,

∠A'C'B = ∠ACB (Corresponding angles)

∠A'BC' = ∠ABC (Common)

∴ ΔA'BC' ∼ ΔABC (AA similarity criterion)

… (1)

In ΔBB_{3}C' and ΔBB_{4}C,

∠B_{3}BC' = ∠B_{4}BC (Common)

∠BB_{3}C' = ∠BB_{4}C (Corresponding angles)

∴ ΔBB_{3}C' ∼ ΔBB_{4}C (AA similarity criterion)

From equations (1) and (2), we obtain

⇒

#### Page No 9.9:

#### Question 16:

Draw a right triangle in which the sides (other than the hypotenuse) are of lengths 4 cm and 3 cm. Now construct another triangle whose sides are $\frac{3}{5}$ times the corresponding sides of the given triangle.

#### Answer:

Construct a right triangle of sides $\mathrm{AB}=4\mathrm{cm},\mathrm{AC}=3\mathrm{cm}\mathrm{and}\angle \mathrm{A}=90\xb0$ and then a triangle similar to it whose sides are ${\left(\frac{3}{5}\right)}^{\mathrm{th}}$of the corresponding sides of.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment AB = 4 cm.

Step: II- With *A *as centre and draw an angle.

Step: III- With *A *as centre and radius AC = 3 cm*.*

Step: IV-Join *BC *to obtain right .

Step: V- Below *AB, *makes an acute angle $\angle BAX$.

Step: VI- Along *AX,* mark off five points ${A}_{1},{A}_{2},{A}_{3},{A}_{4}\mathrm{and}{A}_{5}$ such that $={A}_{4}{A}_{5}$.

Step: VII- Join ${A}_{5}B$.

Step: VIII -Since we have to construct a triangle each of whose sides is ${\left(\frac{3}{5}\right)}^{\mathrm{th}}$ of the corresponding sides of right .

So, we draw a line on *AX *from point which is ${A}_{3}B\parallel {A}_{5}B$ and meeting *AB *at B*’.*

Step: IX- From *B’ *point draw and meeting *AC *at *C’*

Thus, is the required triangle, each of whose sides is ${\left(\frac{3}{5}\right)}^{\mathrm{th}}$ of the corresponding sides of*.*

#### Page No 9.9:

#### Question 17:

Construct an equilateral triangle with each side 5 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of ∆*ABC*.

#### Answer:

**Steps of Construction
Step 1. **Draw a line segment BC = 5 cm.

**Step 2.**With B as centre and radius 5 cm, draw an arc.

**Step 3.**With C as centre and radius 5 cm, draw an arc cutting the previous arc at A.

**Step 4.**Join AB and AC. Thus, ∆ABC is the required equilateral triangle.

**Step 5.**Below BC, draw an acute angle ∠CBX.

**Step 6.**Along BX, mark three points B

_{1}, B

_{2}and B

_{3}such that BB

_{1}= B

_{1}B

_{2}= B

_{2}B

_{3}.

**Step 7.**Join B

_{3}C.

**Step 8.**From B

_{ }_{2}, draw B

_{2}C' || B

_{3}C meeting BC at C'.

**Step 9.**From C', draw A'C' || AC meeting AB in A'.

Here, ∆A'BC' is the required triangle, each of whose sides are $\frac{2}{3}$ times the corresponding sides of ∆ABC.

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