Rd Sharma 2022 for Class 10 Maths Chapter 9 - Constructions

Answer:

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 ∠ABA₈ 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.
 

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, AB = 5 cm and AC = 7 cm.
Also, $\mathrm{AP}=\frac{3}{4}\mathrm{AB} \mathrm{and} \mathrm{AQ}=\frac{1}{4}\mathrm{AC}$                        .....(1)
$\Rightarrow \mathrm{AP}=\frac{3}{4}\times 5=\frac{15}{4} \mathrm{cm}$
Then, PB = AB − AP
$\Rightarrow \mathrm{PB}=5-\frac{15}{4}=\frac{5}{4} \mathrm{cm}$                   
$\Rightarrow \mathrm{AP}:\mathrm{PB}=\frac{15}{4}:\frac{5}{4}$
Hence, AP : PB = 3 : 1 i.e., scale factor of line segment AB is $\frac{3}{1}$.

Again from (1),
$\mathrm{AQ}=\frac{1}{4}\mathrm{AC}=\frac{1}{4}\times 7=\frac{7}{4} \mathrm{cm}$
Then,
$\mathrm{QC}=\mathrm{AC}-\mathrm{AQ}=7-\frac{7}{4}=\frac{21}{4} \mathrm{cm}$
$\Rightarrow \mathrm{AQ} : \mathrm{QC}=\frac{7}{4}:\frac{21}{4}$
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 B_1, 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.

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:

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).
$$\begin{gathered} \therefore \frac{\mathrm{AM}}{\mathrm{MB}}=\frac{5}{2} \\ \Rightarrow \frac{\mathrm{MB}}{\mathrm{AM}}=\frac{2}{5} \end{gathered}$$

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.
$$\begin{gathered} \therefore △\mathrm{AMN}~△\mathrm{ABC} \\ \Rightarrow \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AC}}=\frac{\mathrm{NM}}{\mathrm{BC}}=\frac{5}{7} \end{gathered}$$

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:

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 BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
6. Join B₃C and from B₅, draw B₅M ∥ 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)

$\Rightarrow \frac{\mathrm{BM}}{\mathrm{BC}}=\frac{{\mathrm{BB}}_5}{{\mathrm{BB}}_3}=\frac{5x}{3x}=\frac{5}{3}$                           .....(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)

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

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

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:

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

 $\Rightarrow \frac{\mathrm{BN}}{\mathrm{BC}}=\frac{{\mathrm{BB}}_2}{{\mathrm{BB}}_3}=\frac{2x}{3x}=\frac{2}{3}$          .....(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)

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

 

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: ∆PQR ~ ∆ABC,  ∆ABC is an isosceles triangle in which AB = AC = 6 cm and BC = 5 cm and in ∆PQR, PQ = 8 cm.
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 such that point Q coincides with point B.

Justification:
Let BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = x.
As per the construction, B₄R ∥ B₃C.
Now,
$\frac{\mathrm{BC}}{\mathrm{CR}}=\frac{{\mathrm{BB}}_3}{{\mathrm{B}}_3{\mathrm{B}}_4}=\frac{3x}{x}=\frac{3}{1}$
$\therefore \frac{\mathrm{BC}}{\mathrm{CR}}=\frac{3}{1}$

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

Also, from the construction, RP ∥ CA.

In $△\mathrm{PBR} \mathrm{and} △\mathrm{ABC}$,
$\angle \mathrm{PBR} = \angle \mathrm{ABC}$
$\angle \mathrm{PRB}=\angle \mathrm{ACB}$                                            [Corresponding Angles]
$\Rightarrow △\mathrm{PBR}~△\mathrm{ABC}$                                        [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.

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.

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:

First draw the rhombus ABCD in which AB = 4 cm and ∠ABC = 60° as given in Figure and join AC.
Construct the triangle AB'C' similar to ∆ABC with scale factor $\frac{2}{3}$.
Finally draw the line segment C'D' parallel to CD.



Now, $\frac{AB\text{'}}{AB}=\frac{2}{3}=\frac{AC\text{'}}{AC}$           (∵ ∆AB'C' ~ ∆ABC )

Also, $\frac{AC\text{'}}{AC}=\frac{C\text{'}D\text{'}}{CD}=\frac{AD\text{'}}{AD}$

Therefore, $AB\text{'}=B\text{'}C\text{'}=C\text{'}D\text{'}=AD\text{'}=\frac{2}{3}AB$

Thus, AB'C'D' is a rhombus.

Answer:

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'.
10. 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.
11. Now draw a line segment D'A' parallel to DA, where A' lies on the extended side BA i.e., a ray BX₁.
12. 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.

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.