Steps of Construction:
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.

Given: A circle of radius 3 cm with centre O and a point P at a distance of 7 cm from O. Construction: We have to construct the two tangents from P to the circle

Step of construction: 1
1. Draw a line segment PO = 7 cm.
2. From the point 0, draw a circle of radius = 3 cm,
3. Draw a perpendicular bisector of PO. Let M be the mid-point of PO.
4. Taking M as centre and OM as radius a circle.
5. Let this circle intersects the given circle at the point Qand R.
6. Join PQ and PR.
Thus PQ and PR are the required two tangents
Length of the tangent PQ = PR = \(\sqrt { PQ^{ 2 }-OQ^{ 2 } } \)
\(PQ=PR=\sqrt { (7)^{ 2 }-(3)^{ 2 } } \)
\(=\sqrt { 49-9 } =\sqrt { 40 } =6.3\quad cm\) 

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.

Steps of Construction
(i) Draw a line segment AB = 7.6 cm.
(ii) Draw a ray AX, making an acute \(\angle BAX\) with AB.
(iii) Mark 5 + 8 = 13 points, i.e. A₁,A₂,A₃,A₄,...A₁₂,A₁₃ on AX, such that
AA₁=A₁A₂=A₂A₃=...=A₁₂A₁₃
(iv) Join A₁₃B.
(v) From A₅ , draw \({ A }_{ 5 }O\parallel { A }_{ 13 }B\) which intersects AB at O.Then, O is the point on AB which divides it in the ratio 5:8.
So, AO: OB = 5:8.

Justification
In \(\Delta { ABA }_{ 13 }\) , we have
\({ A }_{ 5 }O\parallel { A }_{ 13 }B\)
By construction, \(\frac { AO }{ OB } =\frac { { AA }_{ 5 } }{ { A }_{ 5 }{ A }_{ 13 } } =\frac { 5 }{ 8 } \)
Hence, \(\frac { AO }{ OB } =\frac { 5 }{ 8 } \) or AO: OB=5:8
On measuring, we found that AD = 2.9cm and DB = 4.7 cm

\(2\sqrt { 15 } cm\)

7.75 cm, square

Given, a \(\Delta ABC\) ,in which BC = 6cm,AB = 5 cm and \(\angle ABC=60°\),Here,scale factor= 3/4<1
 

Steps of Construction
1.Draw a line segment AB=5 cm
2.From point B, draw \(\angle ABY=60°\) and cut-ff BC=6 cm from BY.
3.Join AC.Thus, \(\Delta ABC\) is the given triangle.
4.Now, from A, draw any ray AX downwards making an acute \(\angle BAX\)
5.Mark four points B₁ , B₂ , B₃ and B₄ on AX such that AB₁ = B₁B₂ = B₂B₃ =B₃B₄
6.Join B₄B and from B₃,draw B₃M || B₄B intersecting AB at M.
7.From point M, draw MN || BC intersecting AC at N.Then, \(\Delta AMN\) is the required triangle whose sides are 3/4 of the corresponding sides of \(\Delta ABC\) .

In \(\Delta\)PQR 
\(\angle\)Q = \(\angle\)B, PQ = PR = 8 cm.
\(\therefore\)      \(\angle\)R = \(\angle\)Q = \(\angle\)B
\(\therefore\)    Also    \(\angle\)B = \(\angle\)C
\(\Rightarrow\)   \(\angle\)A=\(\angle\)P
In \(\Delta\)ABC and \(\Delta\)PQR
\(\frac { AB }{ AC } =\frac { PQ }{ QR } \) , \(\angle\) A = \(\angle\)P
\(\therefore\)  \(\Delta\)ABC \(\sim \) \(\Delta\) PQR