Steps of Construction: 
(i) A line segment BC = 6 cm is drawn. 
(ii) An arc is drawn from B of radius 5 cm. 
(iii) An arc is drawn from C of radius 7 cm, cutting the first arc at A. AB and AC are joined to get ΔABC. 
(iv) An acute angle CBX is drawn below BC. 
(v) On BX, points B₁,B₂,B₃,B₄,B₅ are taken such that BB1=B₁B₂=B₂B₃=B₃B₄=B₄B_5.
(vi) B₅ and C are joined. 
(vii) B₃C' is drawn parallel to B₅C meeting BC at C'. 
(viii) C'A' is drawn parallel to CA, meeting BA at A'.
(ix) Then ΔA'BC' is the required triangle similar to ΔABC, where sides are \(\frac { 3 }{ 5 } \)  corresponding sides of ΔABC.

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.

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) Draw a circle of radius 4 cm with O as its centre.
(ii) Draw AB as diameter of the circle.
(iii) Take P and Q as two points on extended diameter AB such that OP = OQ = 6 cm.
(iv) Draw perpendicular bisector of OP and OQ intersecting OP and OQ at M and N respectively.
(v) With M as centre and OM as radius draw a circle intersecting the 1^st circle at T and S.
(vi) With N as centre and QN as radius intersecting the 1^st circle at D and E.
(vii) Join PT, PS, QD and QE.
viii) PT, PS, QD and QE are required tangents.

 
Given, a circle of radius 4 cm whose centre is O and a point P, 8 cm away from its centre.
Steps of Construction
1. Draw a circle with O as centre and radius is equal to 4cm.
2. Take a point P such that QP = 8 cm and bisect it. Let M be the mid-point of QP.
3. Taking M as centre and MO as radius, draw a dotted circle. Let this circle cuts the given circle at Qand R.
4. Join PQ and PR. Thus, PQ and PR are the required tangents. By measurement (using scale), PQ = PR = 7 cm

yes