RD Sharma 2021 Solutions for Class 10 Maths Chapter 9 - Constructions

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.

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

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.

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.

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.

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.

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.

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

PQ² + QO² = PQ²

PQ² + (3)² = (5)²

PQ² + 9 = 25

PQ² = 25 − 9

PQ² = 16

PQ = 4 cm

Hence justified.

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).

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).

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).

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).

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).

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 A₁, A₂, A₃, ..., 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 A₁₁.

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 A₁, A₂, A₃, ... are located at equal distances on the ray AX and the point B is joined to A₁₁.

Hence, the correct answer is option (b). 

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 B₁, B₂, B₃, B₄, B₅, B₆, B₇ are marked at equal distance on BX. Then B₇ 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 B₁, B₂, B₃,...., B₇ on BX at equal distances and next step is to join B₇ to C.

Hence, the correct answer is option (c).

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 A₁, A₂, ..., A_m and B₁, B₂, ..., B_n are located at equal distances on rays AX and BY respectively. Then the points A_m and B_n are joined to divide the line segment AB in the ratio m : n.



Here, m = 5 and n = 6

So, the points A₁, A₂, ..., A₅  and B₁, B₂, ..., B₆ are located at equal distances on rays AX and BY respectively. Then the points A₅ and B₆ 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 A₁, A₂, ..., A₅ and B₁, B₂, ..., B₆ are located at equal distances on rays AX and BY respectively. Then the points joined are A₅ and B₆.

Hence, the correct answer is option (a).

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

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

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

Answer:

Steps of construction:

1) Draw a line segment AB = 8 cm.

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

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.

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.

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 AC 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.

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.

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 Q₁, Q₂, Q₃, Q₄, Q₅, Q₆ and Q₇ such that QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₄Q₅ = Q₅Q₆ = Q₆Q₇.

Step 7. Join Q₇R.

Step 8.From Q₅, draw Q₅R' || Q₇R 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.

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.

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.

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$△ABC$.

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.

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.

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.

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.

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.

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.

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 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.

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.

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 R₁, R₂, R₃, R₄ and R₅ such that QR₁ = R₁R₂ = R₂R₃ = R₃R₄ = R₄R₅.

Step 7. Join RR₅.

Step 8. From R₃, draw R₃R' || RR₅ 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.

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 R₁, R₂, R₃, R₄ and R₅ such that QR₁ = R₁R₂ = R₂R₃ = R₃R₄ = R₄R₅ .

Step 7. Join RR₅.

Step 8. From R₃, draw R₃R' || RR₅ 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.

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), B₁, B₂, B₃, B₄, on line segment BX.

Step 4

Join B₄C and draw a line through B₃, parallel to B₄C 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 ΔBB₃C' and ΔBB₄C,

∠B₃BC' = ∠B₄BC (Common)

∠BB₃C' = ∠BB₄C (Corresponding angles)

∴ ΔBB₃C' ∼ ΔBB₄C (AA similarity criterion)

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

⇒

Answer:

Construct a right triangle of sides $\mathrm{AB}=4 \mathrm{cm}, \mathrm{AC}=3 \mathrm{cm} \mathrm{and} \angle \mathrm{A}=90°$ 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.

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 B₁, B₂ and B₃ such that BB₁ = B₁B₂ = B₂B₃.

Step 7. Join B₃C.

Step 8.From B₂, draw B₂C' || B₃C 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.