Steps of construction :
1. Draw a circle of radius 1·5 cm. Take a point P outside it.
2. Through P draw a secant PAB to meet the circle at A and B.
3. Produce AP to C such that PC = PA. Bisect CBat Q.
4. With CB as diameter and centre as Q, draw a semi circle
5. Draw \(PD\bot CB\) cutting the semi-circle at the point D.
6. With P as centre and PO as radius an arc to cut the circle at T and T
PT and P are the required tangents.

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

1.5 cm, 4.5 cm

Given, a line segment AB and we have to divide it in the ratio 5 : 6.
Steps of Construction
1. Draw a ray AX making an acute \(\angle BAX\)
2. Draw a ray BY parallel to AX by making LABY equal to \(\angle BAX\) 
3. Now, locate the points A₁ , A₂ , A₃ ,A₄ and A₅  (m=5) on AX and B₁, B₂, B₃ , B₄, B₅ and B₆ ( n = 6) on BY such that all the points are at equal distances from each other.
4.Join B6A5' Let it intersect AB at a point C. Then, AC : BC =5 : 6.
Hence, the points joined are A₅ and B₆ .

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.

Given, PQ = 6.4 cm and
 \(\frac { QR }{ PQ } =\frac { 5 }{ 8 } or\quad \frac { QR }{ PR } =\frac { 5 }{ 3 } \)
Steps of Construction
1. Draw a line segment of length PQ = 6.4 cm.
2. Draw any ray QX making an acute \(\angle PQX\) with PQ.
3. Draw a ray parallel to Q)( by making \(\angle QPY=\angle PQX\)
4. Mark five points A₁ , A₂ , A _{3 ,} A₄ and A₅ on QX and three points B₁, B₂ , and, B₃ on PY such that
QA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = PB₁ = B₁B₂ = B₂B₃        
5.Join A₅ to B₃.
Then it intersect PQ at a point R.Thus, R is the point of dividing PQ such that \(\frac { QR }{ PQ } =\frac { 5 }{ 8 } or\quad \frac { QR }{ PR } =\frac { 5 }{ 3 } \) 

The justification will be given by using the basic proportionality theorem and similar triangles.

False

Steps of Construction : 
1. Draw any line l and take any point X on it. 
2. At X, construct a right angle.
3. With X as centre and of radius 4 cm, draw an arc intersecting XY in A.
4. At A and initial line AX, draw angle of 30^o  to each sides and let these lines intersect line I in B and C.
5. Join AB and AC to get the given triangle.
6. At B, construct an acute angle ㄥCBZ (<90^o).
7. Mark three points on BZ such that BB₁= B₁B₂ = B₂B₃.
8. Join B₃C.
9. Through B₂, draw B₂C' || B₃C, intersecting line I in C'.
10. Through C', draw C'A' || CA, intersecting BA in A'. Thus, ΔA'BC' is the required triangle.

Steps of Construction : 
1. Draw a circle with centre O and radius 3.5 cm.
2. Draw the diameter POQ.

3. Construct angle of 90^o at the end points  P and Q.
4. XPY and LQM are the two tangents at P and Q to the circle with centre O. 
XPY || LQM, because ㄥP + ㄥQ = 180^o.

AB'C' is the required triangle.

(i) Draw a circle of radius 3 cm with O as its centre
(ii) Draw AB as diameter of thc circle.
(iii) A point P is taken on outer circle and OP is joined.
(iv) Perpendicular bisector of Op is drawn interesting 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: 
1. Draw a line segment AB = 8 cm. 
2. Taking A as centre draw a circle C of radius 4 cm and taking B as centre draw a circle C' of radius 3 cm. 
3. Draw perpendicular bisector of AB, which intersects AB at point O. 
4. Taking point O as centre draw a circle of radius 4 cm passing through points A and B which intersect circle C at P and S and circle C' at points R and Q. 
5. Join AQ, AR, BP, and BS. These are the required tangents. 
Justification: 
Join AP 
In ΔABP    AP 丄 BP    [Radius 丄 to tangent]
AP = 4 cm and AB=8 cm
AB² = AP² + BP²
⇒ (8)² =(4)²+ BP² ⇒ 64 = 16 + BP²
⇒ 64 - 16 = BP² ⇒ 48 = BP²
⇒ BP= \(\sqrt { 16\times 3 } \)  ⇒ BP = \(4\sqrt { 3 } cm\)
Similarly BS = \(4\sqrt { 3 } cm\)
Join BQ.
In ΔABQ  BQ丄 AQ    [Radius 丄 to tangent]
BQ=3 cm
AB²= BQ² + AQ²⇒ (8)² = (3)² + AQ²
⇒ 64=9+AQ² ⇒ 64-9=AQ²
⇒ 55=AQ² ⇒ AQ=\(\sqrt { 55 } cm\)
Similarly AR=\(\sqrt { 55 } cm\)
A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths. 
∴ AQ = AR and BP = BS.

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.

In Δ ABC, ∠A+∠B+∠C=180^o
⇒ 105^o+ 45^o +∠C = 180^o⇒∠C30^o
Steps of Construction: 
1. Draw a line segment BC = 7 cm. At point B draw ∠B = 45^o and at point C draw ∠C = 30^o and get ΔABC. 
2. Draw an acute ∠CBX on the base BCat point B. Mark the ray BX with B₁,B₂,B₃,B₄, such that BB₁= B₁B₂= B₂B₃= B₃B₄. 
3. Join B₃ to C. Draw B₄C' || B₃C, where C' is point on extended line segment BC. 
4. At C' draw C'A' || AC, where A' is a point on extended line segment BA. 

5. Δ A'BC' is the required triangle.
Justification:
In Δ ABC and Δ A'BC' AC || A'C'
∴ By BPT \(\frac { AB }{ A'B } =\frac { BC }{ BC' } \)    ...(i)
From (i) and (ii) we get
\(\frac { AB }{ A'B } =\frac { 3 }{ 4 } \Rightarrow A'B=\frac { 4 }{ 3 } AB\)
In Δ BB₃C and ΔBB₄C'
By BPT \(\frac { BC }{ BC' } =\frac { { BB }_{ 3 } }{ { BB }_{ 4 } } =\frac { 3 }{ 4 } \)      ...(ii) 
∴ Sides of new triangle formed is \(\frac { 4 }{ 3 } \)  times the corresponding sides of first triangle.