Since the two tangents drawn to a circle from an external point are equal.
AQ = AR             ...(1)
Similarly, BQ = BP        ...(2)
and CR = CP           ..(3)
Now perimeter of \(\triangle\)ABC
= AB + BC + AC
= AB + (BP + PC) + AC
= AB + (BQ + CR) + AC
= (AB + BQ) + (CR + AC)
= AQ + AR
= AQ + AQ
= 2AQ
\(\left.A Q=\frac{1}{2} \text { (Perimeter of } \triangle A B C\right)\)

Given radius r = 3.6 cm
Length of the tangents PA = PB = 6.2cm

Construction:
Steps
(1) with centre at o, drawn a circle of radius 3.6 cm.
(2) Draw a line OP = 7.2 cm,
(3) Draw a perpendicular bisector of OP, which cuts OP at M.
(4) With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 6.2 cm.

Diameter = 6 cm
Radius =\(\frac { 6 }{ 2 } =3cm\) 

Length of the tangents PA = PB = 4 cm
Construction:
Steps:
(1) With centre O, draw a circle of radius 3cm.
(2) Draw a line OP = 5 cm
(3) Draw a bisector of OP, which cuts OP and M
(4) With M as centre and MO as radius draw a circle which cuts previous circle at A and B
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 4 cm.

Given the radius of the circle = 4 cm

Length of the tangents PA = PB = 10.2 cm
Construction:
Steps:
(1) With centre O, draw a circle of radius 4cm,
(2) Draw a line OP = 11 cm.
(3) Draw a \(\bot '\) bisector of OP, which cuts at M,
(4) With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 10.2 cm.

The distance between the point from the centre is 10 cm.

Length of the tangents PA - PB = 8.7 cm
Construction:
Steps:
(1) With O as centre, draw a circle of radius 5cm.
(2) Draw a line OP = 10 cm.
(3) Draw a perpendicular bisector of OP which cuts OP at M.
(4) With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA and PB = 8.7 cm

Construction:
(1) With O as the centre, draw a circle of radius 4.5 cm.
(2) Taken a point L on the circle through L drawn as chord LM
(3) Taken a point M distinct from L and N on the circle so that L, M, N are anti-clock wise direction. Joined LN and NM.
(4) Through 'L' drawn a tangent TT' such that \(\angle T L M=\angle M N L\)
(5) TT' is the required tangent.

Radius = 3.4 cm
Centre = P
Tangent at any point R

Construction
Steps
(1) Draw a circle with centre O of radius 3.4cm
(2) Take a point P on the circle. Join OP
(3) Drawn perpendicular line TT' to PR which passes through R.
(4) TT^I is the required tangent.

Given, diameter (d) = 6 cm, we find radius \((r)=\cfrac { 6 }{ 2 } =3cm\)

Construction
Step 1: With centre at O, draw a circle of radius 3 cm.
Step 2: Draw a line OP of length 8 cm.
Step 3: Draw a perpendicular bisector of OP, which cuts OP at M.
Step 4: With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
Step5: Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 7.4 cm.
Verification : In the right angle triangle OAP,PA² = OP² - OA² = 64 -9 = 55
\(PA=\sqrt { 55= } 7.4\ cm\) (approximately) .

Given, radius = 4 cm
Construction
Step 1 : With O as the centre, draw a circle of radius 4 cm.
Step 2 : Take a point L on the circle. Through L draw any chord LM.
Step 3 : Take a point M distinct from L and N on the circle, so that L, M and N are in anti clockwise direction. Join LN and NM.
Step 4 : Through L draw a tangent TT' such that \(\angle\)TLM =\(\angle\)MNL
Step 5 : TT' is the required tangent.

Given, radius r = 3 cm
Construction
Step 1: Draw a circle with centre at O of radius 3 cm.
Step 2: Take a point P on the circle. Join OP.
Step 3: Draw perpendicular line to OP which passes through P.
Step 4: TT'  is the required tangent.

Construction:
Steps (1) Draw a line segment PQ = 6.8 cm
Steps (2) At P, draw PE such that \(\angle QPE={ 50 }^{ 0 }\)
Steps (3) At P, draw PF such that \(\angle FPE={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector to PQ, which intersects PF at 'O' and PQ at G.
Steps (5) With O as center and OP as radius drawn a circle.
Steps (6) From P, marked an arc of 5.2 cm on PQ at D
Steps (7) The perpendicular bisector intersects the circle at I. Joined ID
Steps (8) ID produced meets the circle at R now joining PR and QR, we get the required \(\triangle\)PQR.

Construction:
Steps (1) Draw a line segment BC = 5.6 cm
Steps (2) At B, draw BE such that \(\angle CBE={ 60 }^{ 0 }\)
Steps (3) At B draw BF such that \(\angle EBF={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector to BC, which intersects BF at O and BC at G.
Steps (5) With O as centre and OB as radius draw a circle
Steps (6) From B, marked an arc of 4 cm on BC at D.
Steps (7) The perpendicular bisector intersects the circle at I. Joined ID.
Steps (8) ID produced meets the circle at A. Now joined AB and AC. Then \(\triangle\)ABC is the required triangle.

Construction:
Step (1) Draw \(\bar { AB } =5.5cm\)
Step (2) Draw \(\angle BAE={ 25 }^{ 0 }\)
Step (3) Draw \(\angle FAE={ 90 }^{ 0 }\)
Step (4) Drawn the perpendicular bisector XY to AB which intersects AF at O and AB at G.
Step (5) With O as center and OA as radius drawn a circle
Step (6) XY intersects AB at G. On XY from G marked an arc at M such that GM = 4 cm
Step (7) Drawn PQ through M which is parallel to AB.
Step (8) PQ meets the circle at C and S.
Step (9) Joined AC and BC. Now \(\triangle\)ABC is the required triangle

Construction:
Steps (1) Draw QR = 6.5 cm.
Steps (2) Draw \(\angle RQE={ 60 }^{ 0 }\)
Steps (3) Draw \(\angle FQE={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector XY to ER which intersects QF at O and ER at G.
Steps (5)  With O as center and OQ as radius drawn a circle
Steps (6) XY intersects QR at G. On XY, from G marked an arc at M, such that GM = 4.5 cm
Steps (7) Drawn AB through M which is parallel to QR
Steps (8)  AB meets the circle at P and S
Steps (9) Joined QP and RP Then \(\triangle\)PQR is the required triangle.
Steps (10) Here \(\triangle\)SQR is also another required triangle.

Construction:
Step (1) Draw a line segment QR = 5 cm.
Step (2) At Q, draw QE such that \(\angle RQE\) = 40°.
Step (3) At Q, draw QF such that \(\angle EQF\) = 90^o
Step (4)Drawn a perpendicular bisector to QR, which intersects QF at 'O' and QR at G.
Step (5) With O as centre and OQ as radius, draw a circle
Step (6) From G marked arcs of radius 4.4 cm on the circle. Marked them as P and S.
Step (7) Joined QP and PR. Now \(\triangle\)PQR is the required triangle
Step (8) From P draw a line PN which is \(\bot \) to LR. LR meets PN at M.
Step (9) The length of the altitude is PM = 2.1cm

Construction:
Step (1) Draw a line segment PQ = 4.5 cm
Step (2) At P, draw PE such that \(\angle QPE={ 35 }^{ 0 }\)
Step (3) At P, draw PF such that \(\angle EPF={ 90 }^{ 0 }\)
Step (4) Draw \(\bot \) bisector to PQ which intersects PF at O.
Step (5) With O centre OP as radius draw a circle.
Step (6) From G, marked arcs of radius 6 cm on the circle marked them as R and S.
Step (7) Joined PR and RQ. Then \(\triangle\)PQR is the required triangle
Step (8) \(\triangle\)PQS is the required triangle

Construction
Step 1 : Draw a line segment BC = 8 cm.
Step 2 : At B, draw BE such that \(\angle\)CBE = 60^o.
Step 3 : At B, draw BF such that \(\angle\)EBF = 90^o .
Step 4 : Draw the perpendicular bisector to BC, which intersects BF at O and BC at G.
Step 5 : With O as centre and OB as radius draw a circle.
Step 6 : From B, mark an arc of 6 cm on BC at D.
Step 7 : The perpendicular bisector intersects the circle at I. Joint ID.
Step 8 : ID produced meets the circle at A. Now join AB and AC. Then \(\triangle\)ABC is the required triangle.

 Given a triangle \(\triangle\)PQR. We have to construct another triangle whose sides are \(\frac { 7 }{ 3 } \) of the corresponding sides of the given \(\triangle\)PQR.

Steps of construction:
1. Constructed a PQR with any measurement.
2. Drawn a ray QX making an acute angle with QR on the side opposite to the vertex P.
3. Joined Q₃ to R and drawn a line through Q₇ parallel to Q₃R, intersecting the extended line segment QR at R'
4. Drawn a line through R' parallel to RP intersecting the extended line segment QP at P.
5. Then PQR' is the required triangle each of whose sides is seven-thirds of the corresponding sides of PQR.

Given a triangle ABC, we are required to construct another triangle whose sides are \(\frac { 6 }{ 5 } \) of the corresponding sides of the ABC

Steps of construction:
1. Constructed a  ABC with any measurement
2. Drawn a ray BX making an acute angle with BC on the side opposite to the vertex A.
3. Joined B₅ to C and drawn a line through B₆ parallel to B₅C intersecting the extended line segment BC at C.
4. Drawn a line through C' parallel to CA intersecting the extended BA at A'.
5. Then A'BC' is the required triangle each of whose sides is six-fifths of the corresponding sides of ABC.

Given a triangle LMN, we are required to construct another triangle whose sides are \(\frac { 4 }{ 5 } \) of the corresponding sides of the \(\triangle\)LMN

Steps of construction:
1. Constructed a LMN with any measurement
2. Drawn a ray MX making an acute angle with MN on the side opposite to the vertex L.
3. Located 5 points M₁, M₂, M₃, M₄, M₅ on MX so that
MM₁ = M₁M₂ = M₂M₃ = M₃M₄ = M₄M₅
4. Joined, M₅N and drawn a line through M₄ parallel to M₅N to intersect MN at N.
5. Drawn a line through N, parallel to the line NL to intersect ML at L
Then L'MN' is the required triangle each of whose sides is four-fifth of the corresponding sides of LMN

Given a triangle PQR, we are required to construct another triangle whose sides are \(\frac { 3 }{ 5 } \) of the corresponding sides of the triangle PQR

 
Steps of construction:
1. Constructed a PQR with any measurement.
2. Drawn a ray QX making an acute angle with QR on the side opposite to the vertex P.
Located 3 points Q₁, Q₂ and Q₃ on QX so that Q Q₁ = Q₁ Q₂ = Q₂ Q₃
4. Joined Q₃R and drawn a line through Q₂ parallel to Q₃R to intersect QR at R'.
5. Drawn a line through R' parallel to the line RP to intersect QP at P'. Then PQR is the required triangle each of whose sides is two-thirds of the corresponding sides of PQR.

Given a triangle PQR, we are required to construct another triangle whose sides are \(\frac { 7 }{ 4 } \) of the corresponding sides of the triangle PQR.
Steps of construction
1. Construct a DPQR with any measurement.
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 7 points (the greater of 7 and 4 in \(\frac { 7 }{ 4 } \))
Q_1,Q_2,Q_3,Q_4,Q_5,Q₆ and Q₇ on QX so that 
QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅ = Q₅Q₆ = Q₆Q₇
4. Join Q₄ (the 4th point, 4 being smaller of 4 and 7 in \(\frac { 7 }{ 4 } \)) to R and draw a line through Q₇ parallel to Q₄R, intersecting the extended line segment QR at R'.
5. Draw a line through R' parallel to RP intersecting the extended line segment QP at P'.
Then \(\triangle\)P'QR' is the required triangle each of whose sides is seven-fourths of the corresponding sides of \(\triangle\)PQR.

Given a triangle PQR we are required to construct another triangle whose sides are \(\frac{3}{5}\) of the corresponding sides of the triangle PQR.



Steps of construction
1. Construct a \(\triangle\) PQR with any measurement
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 5 (the greater of 3 and 5 in \(\frac { 3 }{ 5 } \)) points.
Q₁Q_2, Q₃, Q₄ and Q₅ on QX so that QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅
4. Join Q₅R  and draw a line through Q₃ (the third point, 3 being smaller of 3 and 5 in \(\frac { 3 }{ 5 } \)) parallel to Q₅R to intersect QR at R'.
5. Draw line through R' parallel to the line RP to intersect QP at P'.
Then, \(\triangle\)P'QR' is the required triangle each of whose sides is three-fifths of the corresponding sides of \(\triangle\) PQR.