A triangle ABC' is to be constructed such that
\(\frac { CA^{ ' } }{ CA } =\frac { BA^{ ' } }{ BA } =\frac { B^{ ' }C }{ BC } =\frac { 2 }{ 3 } \)
Thi.s means that the ABC 3 triangle ABC is similar to the triangle ABC with scale factor as \(\frac { 2 }{ 3 } \)
Steps of construction: 
Draw a line segment BC = 5 cm
2. With B as centre and radius = AB = 4 cm, draw an arc.
3. With C as centre and radius = AC = 6 cm, draw another arc, meeting the arc drawn in step 2 at the point A.
4. Join AB and AC to obtain  \(\triangle ABC\)
5.. Below BC, make an acute angle \(\angle CBX\)
6.Along BX mark off three points  B₁ ,B_2, , B₃  such that BB₁ = B₁ B₂
7. Join BC
8. From B₂, draw B₂C II B₃C
9. From C, draw CA' II CA, meeting BA at the point A'. Then ABC is the required triangle

The required triangle is \(\triangle\)A' BC'

\(\frac { 3 }{ 5 } \)

1. Draw a circle of centre O and radius 4 cm.
2. Take a point P outside the circle and draw a secant PAB, intersect~ng the circle at A and B.
3. Produce AP to C such that AP = CP.
4. Draw a semi-circle with CB as diameter.
5. Draw \(PD\bot CB\)  intersecting the semi-circle at D.
6. With P as centre and PD as radius, draw arcs to intersect the given circle at T and T'.
7. Join PT and PT'. Thus, PT and PT' are the required tangents.

Given, radius of circle = 5 cm
1.Draw a circle with 0 as centre and radius 5 cm.
2. Draw any chord PQ through the given point P on the circle.
3.Take a point R on the circle and join P and Q to a pointR.
4. Construct \(\angle QPY=\angle PQX\)  on the opposite side of the chord PQ.
5. Produce yP to X to get YPX, as the required tangent.

One and only one 

False

∴ AB'C' is the required triangle.

Steps of Construction:
1. Draw a circle C' with the help ofa bangle, for finding the centre, take three non collinear points A, B and C, lying on the circle. Join AB and BC and draw perpendicular bisector of AB and BC, both intersect at a point O, 'O' is centre of the circle.
2. Take a point P outside the circle. Join OP.
3. Draw perpendicular bisector of OP, which intersects OP at point O'.
4. Take O' as the centre with OO' as radius draw a circle which passes through O and P, intersecting previous circle at points R and Q.
5. Join PQ and PR.
6. PQ and PR are the required pair of tangents.

Justification:

Join OQ and OR.
In ΔOQP and ΔOPR
OQ = OR   [Radii of the circle]
OP = OP   [Common]
ㄥQ = ㄥR = 90^o   [Radius is 丄 to tangent]
ΔOQP ≅ ΔORP   [by RHS]
PQ = PR
A pair Of tangents can be drawn to a circle from an externar point lying outside the circle.
These two tangents are equal in lengths.
∴ PQ = PR

Steps of Construction: 
1. Draw a line segment BC = 6 cm and at point B draw a ㄥABC = 60^o. 
2. Cut AB 5 cm. Join AC. We obtain ABC is triangle. 
3. Draw a ray BX making an acute angle with BC on the side opposite to the vertex A. 
4. Locate 4 points A₁,A₂,A₃ and A₄ on the ray BX so that BA₁=A₁A₂=A₂A₃=A₃A₄.
5. Join A₄ to C. 
6. At A₃ draw A₃C' || A₄C. Where C' is a point on the line segment BC.
7. At C' draw C'A' || CA, where A' is a point on the line segment BA.

Δ A'BC' is the required triangle.
Justification:
In Δ A'BC' and ΔABC A'C' || AC
∴ By BPT \(\frac { A'B }{ AB } =\frac { BC' }{ BC } \)    ...(i)
From (i) and (ii),
\(\frac { A'B' }{ AB } =\frac { 3 }{ 4 } \Rightarrow A'B=\frac { 3 }{ 4 } AB\)
In ΔBA₃C' and ΔBA₄C
\(\frac { BC' }{ BC } =\frac { { BA }_{ 3 } }{ { BA }_{ 4 } } =\frac { 3 }{ 4 } \)      ...(ii) 
∴ Sides of new triangle formed are \(\frac { 3 }{ 4 } \)  times the corresponding sides of first triangle.

Steps of Construction:
1. Draw a line segment AB of 7 cm
2. Taking A and B as centre draw two circle of 3 cm and 2 cm radius

3. Bisect the line AB. Let mid-point of AB is C.
4. Taking C as centre draw a circle of radius AC which will intersect the circle at point P, Q, R and S.
5. Join BP, BQ, AS and AR. These are the required tangents.

Steps of Construction

(i) Draw a line segment AB = 6.5 cm and a ray AX making an acute angle with the line segment AB.
(ii) Draw another ray \(BY\parallel AX\) such that \(\angle ABY=\angle BAX\)
(iii) Mark off 3 points A₁ ,A₂,A₃ on AX and 5 points B₁,B₂,B₃,B₄,B₅ on BY such that AA₁ = A₁A₂ = A₂A₃= BB₁= B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅
(iv) Join A₃B₅ which intersect AB at point C. Thus, C divides AB in the ratio 3: 5, i.e. AC:CB = 3:5. 
Justification:
In \(\Delta AC{ A }_{ 3 }\) and \(\Delta BC{ B }_{ 5 }\)
\(\angle CB{ B }_{ 5 }=\angle CA{ A }_{ 3 }\)
\(\angle AC{ A }_{ 3 }=\angle BC{ B }_{ 5 }\)
\(\Delta AC{ A }_{ 3 }\sim \Delta BC{ B }_{ 5 }\)
\(\Rightarrow \frac { A{ A }_{ 3 } }{ B{ B }_{ 5 } } =\frac { AC }{ BC } \)
By construction, \(\frac { A{ A }_{ 3 } }{ B{ B }_{ 5 } } =\frac { 3 }{ 5 } \)
On equating Eqs.(i) and (ii), we get \(\frac { AC }{ BC } =\frac { 3 }{ 5 } \)
This shows that C divides the line segment AB internally in the ratio 3:5.

The tangents can be constructed on the given circles as follows.
Step 1
Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius.
Step 2
Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect the circles at points P, Q, R, and S. Join BP, BQ, AS, and AR. These are the required tangents.

Justification
The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B and radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). For this, join AP, AQ, BS, and BR.

∠ASB is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.
∴ ∠ASB = 90°
⇒ BS ⊥ AS
Since BS is the radius of the circle, AS has to be a tangent of the circle. Similarly, AR, BP, and BQ are the tangents.

Steps of Construction:
(i) With O as centre a circle of radius 3 cm is drawn.
(ii) With same centre O another circle of radius 5 cm is drawn.
(iii) A point P is taken on outer circle and OP is joined.
(iv) Perpendicular bisector of OP is drawn intersecting OP at Q.
(v) With Q as centre and OQ as radius a circle is drawn intersecting the smaller circle at A and B.
(vi)  PA and PB is joined.
(vii) PA and PB are the required tangents.
 Length of tangent = 4 cm.

Steps of construction:
Draw a line segment BC = 7 cm.

1. Draw a line segment BC of length 5 cm.
2. At B, draw LMBC = 60° and produced line BM.
3. From point C draw a line making an angle of 30°.
4. Both the lines intersect at A.
5.MBC is the given triangle.
6. Draw a ray BXmaking an acute angle.
7. Locate three points B1, Bz, and B3on line segment BX.
8. Join BC
9. Draw a parallel line through B3to B3C intersecting extended line BCat C.
10. Through C' draw a line parallel to AC intersecting extended line segment BA at A'. A'BC is the required triangle.

Steps of construction:
1. Draw a circle of radius 4 cm with 0 as centre.
2. Take a point A on the circumference of the circle and join OA. Draw perpendicular to OA at point A.
3. Draw a radius OB, making an angle of 1200 with OA.
4. Draw the perpendicular to OB at point B. Let both the perpendiculars intersect at point P.
5. Join OP. PA and PB are required tangents, which make an angle of 600 to each other.

No, two triangles are not congruent.

yes

\(\frac { AP }{ AB } =\frac { 3 }{ 4 } ;\frac { AQ }{ AC } =\frac { 1 }{ 4 }\)
\( \\ PQ\cong 3.25\quad cm\)