Constructions

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

A:B = 5:7

Now,

Draw a ray AX which makes an acute angle ∠BAX, then mark A+B points at equal distance. And, here, A=5 and B=7

Hence, minimum number of these points

= A+B

= 5+7

=12

Given, a line segment AB in the ratio 4:7

A:B = 4:7

Now,

Draw a ray AX which makes an acute angle BAX

Minimum number of points located at equal distances on the ray AX = A+B

= 4+7

= 11

A₁, A₂, A₃, .......... are located at equal distances on the ray AX.

Point B is joined to the last point is A₁₁.

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

A:B = 5:7

Steps of construction:

1. Draw a ray AX, an acute BAX.

2. Draw a ray BY AX,
ABY = BAX.

3. Now, locate the points A₁,A₂,A₃,A₄ and A₅ on AX and B₁,B₂,B₃,B₄,B₅ and B₆ (Because A:B = 5:7)

4. Join A₅B₆.

A₅B₆ intersect AB at a point C.

AC:BC= 5:6

Steps:

1. Locate points B₁, B₂,B₃,B₄,B₅,B₆ and B₇ on BX at equal distance.

2. Join the last points is B₇ to C.

To construct a triangle similar to a triangle, with its sides x/y of the corresponding sides of given triangle, the minimum number of points to be located at an equal distance is equal to the greater of m and n in m/n.

Here, m:n = 8:5 or m/n = 8/5

So, the minimum number of point to be located at equal distance on ray BX is 8.

Construction:

In the given figure:

OP and OQ are the tangents to the circle.

Now,

Sum of opposite angles = 180°

∴ POQ + PRQ=180°

⇒ 60° + θ=180°

⇒ θ=120°

Hence, the angle between them should be 120°.

True

Given,

Ratio=

On simplifying we get,

Required ratio = 3:1

Hence,

Geometrical construction is possible to divide a line segment in the ratio 3:1.

False

Steps of construction

1. Draw a line segment BC.

2. B and C as centers draw two arcs of suitable radius intersecting each other at A.

3. Join BA and CA. ∆ABC is required triangle.

4. From B draw any ray BX downwards making an acute angle CBX.

5. Marked seven points B₁,B₂,B₃,............B₇ on BX (BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇).

6. Join B₃C and from B₇ draw a line B₇C’B₃C intersecting the extended line segment BC at C’.

7. Draw C’ A’CA intersecting the extended line segment BA at A’.

Then, ∆A’BC’ is the required triangle whose sides are 7/3 of the corresponding sides of ∆ABC.

Given that,

segment B₆C’ B₃C. But since our construction is never possible that segment B₆C’ B₃C because the similar triangle A’BC’ has its sides 7/3 of the corresponding sides of triangle ABC.

So, B₇C’ is parallel to B₃C.

False

Let, r = radius of circle and

d = distance of a point from the center.

r = 3.5

And d = 3

r >d

Point P lies in inside the circle, as shown below:

i.e., no tangent is possible.

True

If the angle between the pair of tangents is always > 0 or <180 °,

Then we can construct a pair of tangents to a circle.

Hence, we can draw a pair of tangents to a circle inclined at an angle of 170°, as shown below:

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 A₁, A₂, A₃, A₄, A₅, A₆, A₇, A₈ such that AA₁=A₁A₂=A₂A₃=A₃A₄=A₄A₅=A₅A₆=A₆A₇⁼A₇A₈.

4. Join A₈B.

5. From A₃, draw A₃P ││ A₈B meeting AB at P. (By making an angle equal to ∠BA₈ at A₃).

Then P is the point on AB which divides it in the ratio 3:5. Thus, AP : PB=3:5

Justification:

Let

AA₁=A₁A₂=A₂A₃=A₃A₄=........=A₇A₈=x

In ∆ABA₈, A₃P││A₈B

Hence, AP:PB=3:5

Thinking process:

Here, scale factor, i.e, m<n then the triangle to be constructed is smaller than the given triangle. Use this concept and then construct the required triangle.

Steps of construction:

1. Draw a line segment BC=12 cm.

2. From B draw a line which makes a right angle.

a. Now as point B is the initial point as the centre, draw an arc any radius such that, the arc meets the ray 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 by GBC is 90°. i.e.

∠ GBC = 90°.

3. From B as centre draw an arc of 5 cm which intersects the line GB at A.

4. Join AC, ABC is the given right triangle.

5. From B draw an acute CBH downwards.

6. On ray BH, mark three point B₁,B₂ and B_3, such that

BB₁=B₁B₂= B₂B₃

7. Join B₃C

8. From point B₂ draw B₂N||B₃C intersect BC at N.

9. From point N draw NM||CA intersect BA at M. ∆MBN is the required triangle. ∆MNB is also a right angled triangle at B.

Justification:

As per the construction, ∆MNB is the required triangle

Let

BB₁=B₁B₂=B₂B₃=x

From triangles ∆BNB₂ and ∆BCB₃, we can say that they are similar

[by AA congruency criteria, ∠ B is common in both triangles and ∠BNB₂ =∠BCB₃ and are corresponding angles of the same transverse as B₃C || B₂N]

---- (i)

Similarly,

∆MBN and ∆ABC are similar.

[by AA congruency criteria, ∠ A is common in both triangles and ∠BAC=∠BMN and are corresponding angles of the same transverse as AC || MN]

Hence

---- (ii)

From (i) and (ii) the constructed triangle ∆MNB is of scale 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.

Thinking process:

Here scale factor i.e., m>n then the triangle to be construct is larger than the given triangle. Use this concept and then constant the required triangle.

Steps of construction:

1. Draw the line segment BC=6cm.

2. Taking B and C as centres, draw two arcs of radii 4 cm and 5cm 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,B_2,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 5/3 of the corresponding sides of the ∆ABC.

Hence, ∆NBM is the required triangle.

Justification:

As per the construction, ∆MNB is the required triangle

Let

BB₁=B₁B₂=B₂B₃= B₃B₄₌ B₄B₅₌x

From triangles ∆BCB₃ and ∆BMB₅, we can say that they are similar

[by AA congruency criteria, ∠ B is common in both triangles and ∠BCB₃ =∠BMB₅ and are corresponding angles of the same transverse as B₅D || B₃C]

---- (i)

Similarly,

∆MBN and ∆ABC are similar.

[by AA congruency criteria, ∠ B is common in both triangles and ∠BAC=∠BNM and are corresponding angles of the same transverse as AC || MN]

Hence

---- (ii)

From (i) and (ii) the constructed triangle ∆MNB is of scale times of the ∆ABC.

Thinking process:

I. Firstly taking the perpendicular bisector of the distance from the centre to the external point. After that taking one half of bisector as radius and draw a circle.

II. Drawing circle intersect the given circle at two points. Now, meet these intersecting points to an external point and get the required tangents.

Given, a point M is at a distance of 6 cm from the centre of a circle of radius 4 cm.

Steps of construction

1. Draw a circle of radius 4 cm. Let centre of this circle is O.

2. Take a point M at 6cm away from the radius.

3. Join OM and bisect it. Now, with M and O as centres and with radius more than half of draw two arcs on the either sides of the line OM. Let the arc meet at A and B,just that, M₁ be mid-point of OM.

4. Taking M₁ as centre and M₁O as radius draw a circle to intersect circle with radius 4 and centre O at two points P and Q.

5. Join PM and QM. PM and QM are the required tangents from M to circle with centre O and radius 4.

Thinking process:

I. Firstly, we find the ratio of AB in which P divides it with the help of the relation A= 3/4 AB.

II. Secondly, we find the ratio of AC in which Q divides it with the help of the relation AQ= 1/4 AC.

III. Now, construct the line segment AB and AC in which P and Q respectively divides it in the ratio from step
(i) and (ii) respectively.

IV. Finally get the points P and Q. After that join PQ and get the required measurement of PQ.

Given that, AB=5cm and AC=7cm

Also, and … (i)

From Eq.(i)

Then, PB=AB-AP

[As P is any point on the AB]

Hence AP: PB = 3:1

i.e. , scale factor of line segment AB is .

Again from Eq. (i).

Then,

[As Q is any point on the AC]

Hence AQ:QC = 1:3

i.e. scale factor of line segment AQ is .

Steps of construction:

1. Draw a line segment AB=5cm.

2. Now draw a ray AZ making an acute ∠BAZ=60°.

a. With A as centre and with any radius, draw another arc cutting the line AB at D.

b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point E.

c. Now join the ray AZ which forms an angle of 60° with the line AB.

3. With A as centre and radius equal to 7 cm draw an arc intersecting the line 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₌A₁A₃₌A₃A₄

6. Join A₄B

7. From A₃ draw A₃P||A₄B meeting AB at P. [by making an angle equal to ∠AA₄ B]

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 point B₁B₂,B₃ and B_4.

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 ratio1:3.

So, AQ:QC=1:3

12. Finally, join PQ and its measurement is 3.25cm.

Thinking process:

I. Firstly we draw a line segment, then either of one end of the line segment with length 5 cm and making an angle 60° with this end. We know that is parallelogram both opposite sides are equal and parallel, then again draw a line with 5 cm making an angle with 60° from other end of line segment. Now, join both parallel line by a line segment whose measurement is 3 cm, we get a parallelogram. After that we draw a diagonal and get a triangle BOC.

II. Now, we construct the triangle BD’C’ similar to ∆BDC with scale factor 4/3.

III. Now, draw the line segment D’A’ parallel to DA.

IV. Finally, we get the required parallelogram A ‘BC’D.

Steps of construction:

1. Draw a line segment AB=3 cm.

2. Now, draw a ray BY making an acute ∠ABY=60°.

To draw the angle 60° at B:

a. With B as centre and with any radius, draw another arc cutting the line AB at E.

b. With E as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point F.

c. Now join the ray BY which forms an angle of 60° with the line AB.

3. With B as centre and radius equal to 5 cm draw an arc cut the point C on BY.

4. Again draw a ray AZ making an acute ∠ZAX₁=60°.(BY││AZ, ∴∠YBX₁=ZAX₁=60°)

Following the similar steps a, b and c as in point 2, we will draw a ray AZ making an angle 60° with AX �₁.

5. With A as centre and radius equal to 5 cm draw an arc cut the point D on AZ.

6. Now, join CD and finally make a parallelogram ABCD.

7. Join BD, which is a diagonal of parallelogram ABCD.

8. From B draw any ray BX downwards making an acute ∠CBX.

9. Locate 4 points B₁,B₂,B₃,B₄ on BX, such that BB₁=B₁B₂=B₂B₃=B₃B₄.

10. Join B₃C and from B₄ draw a line B₄C’||B₃C intersecting the extended line segment BC at C’.

11. From point C’ draw C’D’││CD intersecting the extended line segment BD at D’. Then, ∆D’BC’ is the required triangle whose sides are 4/3 of the corresponding sides of ∆DBC.

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

13. Finally, we observe that A’BC’D’ is a parallelogram in which A’D’=6.5 cm A’B= 4 cm and ∠A’BD’=60° divide it into triangles BC’D’ and A’BD’ by the diagonal BD’.

Justification thatrBCD is similar torBC’D’:

Let

BB₁=B₁B₂=B₂B₃= B₃B₄= x

The triangles ∆BCB₃ and ∆BC’B₄ are similar by

[by AA congruency criteria, ∠ CBB₃ =∠ C’BB₄ as it is common in both triangles and ∠BB₃C=∠BB₄C’ and are corresponding angles of the same transverse as CB₃ || C’B₄]

---- (i)

Similarly, ∆BCD and ∆BC’D’ are similar.

[by AA congruency criteria, ∠ B is common in both triangles and ∠BCD=∠BC’D’ and are corresponding angles of the same transverse as CD || C’D’]

Hence

---- (ii)

From (i) and (ii) we can say that the constructed triangle ∆BC’D’ is of scale times of the ∆BCD.

Justification that A’BC’D’ is a parallelogram:

As per the construction,

CD || C’D’ and BC’ || A’D’ ---- (i)

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° ---- (ii)

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° ---- (iii)

From (i), (ii) and (iii), we can say that A’BC’D’ is a parallelogram.

(As opposite sides and opposite angles are equal).

Given, two concentric circles of radii 3 cm and 5 cm with centre O. We have to draw pair of tangents from point P on outer circle to the other.

Steps of construction:

1. Draw two concentric circles with centre O and radii 3cm and 5cm.

2. Taking any point P on outer circle. Join OP.

3. Bisect OP, let M’ be the mid-point of OP.

To bisect OP:

a. With P as centre and any radius more than half of the length of OP, draw two arcs on either side of OP.

b. Similarly, with O as centre and any radius more than half of the length of OP; draw two arcs on either side of OP which intersect with the previous arcs at M and N.

c. Join MN to meet the line OP at M’, which is the mid-point.

4. Taking M’ as centre and OM’ as radius draw a circle dotted which cuts the inner circle at A and B.

5. Join PA and PB. Thus, PA and PB are required tangents.

6. On measuring PA and PB, we find that PA=PB=4 cm.

Actual calculation:

In the right angle ∆OAP,

∠PAO=90°

According to Pythagoras theorem

(hypotenuse)²=(base)² + (perpendicular)²

PA²=(5)²-(3)²=25-9=16

PA=4 cm

Hence, the length of both tangents is 4 cm.

Therefore, PA = PB = 4cm

Thinking process:

I. Here, for making two similar triangles with one vertex is same of base, we assume that,

In ∆ABC and ∆PQR: vertex B=vertex Q

So, we get the required scale factor.

II. Now, construct a ∆ABC and then a ∆PBR, similar to ∆ABC whose sides areof the corresponding sides of the ∆ABC.

Let ∆PQR and ∆ABC are similar triangles, and then its scale factor between the corresponding sides is

Steps of construction:

1. Draw a line segment BC=5 cm.

2. Construct OQ the perpendicular bisector of line segment BC meeting BC at P’.

To bisect BC:

a. With B as centre and any radius more than half of the length of BC, draw two arcs on either side of BC.

b. Similarly, with C as centre and any radius more than half of the length of BC; draw two arcs on either side of BC which intersect with the previous arcs at M and N.

c. Join MN to meet the line BC at P’, which is the mid-point.

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.

5. From B, draw any ray BX making an acute angle, ∠CBX.

6. Locate four points B₁,B₂,B₃ and B₄ on BX such that BB₁=B₁B₂=B₂B₃=B₃B₄

7. Join B₃C and from B₄ draw a line B₄R||B₃C intersecting the extended line segment BC at R.

8. From point R, draw RP||CA meeting the extended line BA at P.

Then, ∆PBR is the required triangles.

Justification:

Let BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = x

As per the construction B₄R││B₃C

Now,

Also, from the construction RP||CA

_{ThereforerABC is congruent torPBR}

_{(by the AA criteria, as∠PBR =∠ABC, also as RP||CA,∠ACB is corresponding to∠PRB so∠ACB =∠PRB)}

and

Hence, the new triangle rPBR is similar to the given isosceles triangle rABC and its sides are times of the corresponding sides of rABC.

Steps of construction:

1. Draw a line segment AB=5 cm.

2. From point B, draw ∠ABY=60°.

To measure angle at B:

a. With B as centre and with any radius, draw another arc cutting the line AB at D.

b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point E.

c. Draw a ray BY passing through E which forms an angle of 60° with the line AB.

3. With B as centre and radius of 6 cm draw an arc intersecting the line BY at C.

4. Join AC, ∆ABC is the required triangle.

5. From A, draw any ray AX downwards making an acute angle.

6. Mark 7 points A₁,A₂,A₃,A₄,A₅,A₆ and A₇ on AX such that AA₁=A₁A₂=A₂A₃=A₃A₄=A₄A₅=A₅A₆=A₆A₇

7. Join A₇B and from A₅ draw A₅M||A₇B intersecting AB at M.

8. From point M draw MN││BC intersecting AC at N. Then, ∆AMN is the required triangle whose sides are equal to of the corresponding sides of ∆ABC.

Justification:

Let AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇ = x

As per the construction A₅M││A₇B

Now,

Also MN || BC,

Therefore rAMC is congruent to rAMN

Thus,

Hence, the new triangle rAMN is similar to the given triangle rABC and its sides are times of the corresponding sides of rABC.

Given,

•Radius of the circle is 4 cm.

•Angle between the tangents is 60°.

In order to draw the pair of tangents, we follow the following steps.

Steps of construction:

1. Take a point O on the plane of the paper and draw a circle of radius OA=4 cm.

2. Extend the line segment OA to B such that OA=AB=4 cm.

3. Taking A as the centre draw a circle of radius AO=AB=4 cm. This circle intersects the first circle drawn in step 1 at P and Q.

4. Join BP and BQ to get desired tangents.

Justification:

In ∆OAP, we have

OA=OP=4 cm

(Radius)

Also, AP=4 cm

(Radius of circle with centre A)

∴ ∆OAP is equilateral

So, ∠PAO=60°

Now,

∠BAP + ∠PAO = 180° (linear angle)

∠BAP + 60° = 180°

∠ BAP = 60°

In ∆BAP, we have

BA=AP = 4 cm (radii of the circle with centre A)

∠BAP=120°

As two sides BA and AP are equal rABP is isosceles.

So, ∠ABP=∠APB

Let ∠ABP=∠APB = α

As the sum of angles is a triangle is 180°

∠ABP + ∠APB + ∠BAP = 180°

α + α + 120° = 180°

2α = 60°

α = 30°

Therefore ∠ABP=∠APB=30°

Hence ∠PBQ=60°

Alternate Method:

Steps of construction:

1. Take a point O the plane of the paper and draw a circle with centre O and radius OA=4 cm.

2. At O construct radii OA and OB such that to ∠AOB = 120° i.e., supplement of the angle between the tangents.

To draw 120° angle at O:

a. Now with point O as the centre, draw an arc any radius such that, the arc meets the ray OA at point C.

b. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

c. Now with D as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point E.

d. With E and D as centres, and with a radius more than half the length of DE, draw two arcs intersecting at point F.

e. Join points O and F. The angle formed by FOA is 90°. i.e. ∠ FOA = 90°.

f. Now as point O is the initial point as the centre, draw an arc any radius such that, the arc meets the ray OF at point G.

g. With G as centre and with the same radius as before, draw another arc cutting the previous one (drawn in point f) at point H.

h. With G and H are centres and with same radius (which is more than half of the length of GH), draw two arcs intersecting at Y.

i. Draw the ray OY which intersects the circle with radius OA = 4 cm at B. This line OB forms an angle of 120° at point O with the line segment OA.

3. Draw perpendicular to OA and OB at A and B respectively.

The perpendicular lines to OA ad OB will make an angle of 90° at the point of intersection i.e. a point A and B respectively. (So, we can measure 90° or construct 90° angles by taking A and B as the points of construction like in point 2.)

4. Let us suppose these perpendiculars intersect at P. Then, PA and PB are required tangents.

5. Now if we measure the ∠APB = 60° and the distance between PO = 8 cm.

Justification:

From the quadrilateral OAPB, we have

∠OAP=∠OBP=90°

(As PA and PB are perpendicular to OA and OB)

∠AOB=120°

(As per construction)

So

∠OAP+∠OBP+∠AOB+∠APB=360°

90°+90°+120°+∠APB=360°

∠APB=360°-(90°+90+120°)

=360°-300°=60°

Therefore ∠APB = 60°.

Thinking process:

I. Triangles are congruent when all corresponding sides and interior angles are congruent. The triangles will have the same shape and size, but one may be a mirror image of the other.

II. So, first we construct a triangle a triangle similar to ∆ABC with scale factor 3/2 and use the above concept to check the triangles are congruent or not.

Steps of construction:

1. Draw a line segment BC=6 cm.

2. Taking B and C as centres, draw two arcs of radii 4 cm and 9 cm intersecting each other at A.

3. Join BA and CA, ∆ABC is the required triangle.

4. From B, draw any ray BX downwards making an acute angle.

5. Mark three points B₁,B₂,B₃ on BX, such that BB₁=B₁B₂=B₂B₃.

6. Join B₂C and from B₃ draw B₃M││B₂C intersecting the extended line segment BC at M.

7. From point M, draw MN││CA intersecting the extended line segment BA to N.

8. Then, ∆NBM is the required triangle whose sides are equal to the corresponding sides of the ∆ABC.

Justification:

Let BB₁= B₁B₂ = B₂B₃ = x

As per the construction B₃M││B₂C

Now,

Also from the construction MN||CA

_{ThereforerABC is congruent torNBM}

_{(by the AA criteria, as∠NBM =∠ABC , also as MN||CA,∠ACB is corresponding to∠NMB so∠ACB =∠NMB)}

and

Hence, the new triangle rNBM is similar to the given triangle rABC and its sides are times of the corresponding sides of rABC.

The two triangles are not congruent because, if two triangles are congruent, then they have same shape and same size. Here, all the three angles are same but three sides are not same i.e., one side is different.