Yes. we can draw two tangents perpendicular to each other.

Yes. we can draw two tangents parallel to each other on a circle in two different points on the circle.

No, AII the three sides of a right angled triangle cannot be odd numbers
If \(\triangle\)ABC is right angled triangle with \(\angle\)B = 90^o, then AC² = AB² + BC²
If AB and BC are odd, then their squares AB2 and BC2 are odd numbers.
Their sum AB² + BC² is an even number.
Square root of AB² + BC² is also an even number.
AC is even
So three measures cannot be odd.

(3,4,5), (9, 12, l5), (12,16,20), (8, 15,17) (15, 20, 25)

Give two different examples of pair of
(i) similar figures   
(ii) non-similar figures
(i) Two examples of pair of similar figures.
Two equilateral triangles of sides 2 cm and 6 cm.

∆ABC ~ ∆DEF (~ denotes similar to)
Two squares of sides 3 cm and 5 cm

PQRS ∼ UVWX
(ii) Two examples of pair of non-similar figures.
A quadrilateral and a rectangle.

A triangle and a parallelogram.

For any two right angled triangles.
The angles other than right angle need not be equal.
The corresponding sides need not be proportional.
Any two right angled triangles need not be similar.

On observing the given figures.
Their corresponding sides are proportional but their corresponding angles are not equal.
The shapes parallelogram and rectangle are not similar.

For a rhombus
1. All sides are equal.
2. The diagonals of a rhombus are perpendicular bisectors of one another.
3. Opposite angles are equal.
For a square
1. All sides are equal
2. All angles are 90^o.
3. Diagonals are equal and perpendicular bisectors of each other.
For both the shapes the angles are not equal, but corresponding sides are proportional. so they are neither similar nor congruent.

From (i) and (ii) we get
\(\begin{array}{r} A C^{2}+A B^{2}=A D^{2}+B C \cdot D E+\frac{1}{4} B C^{2}+A D^{2}- B C \cdot D E+\frac{B C^{2}}{4} \end{array}\)
\(=2 A D^{2}+2\left(\frac{B C^{2}}{4}\right)\)
\(A C^{2}+A B^{2}=2 A D^{2}+\frac{B C^{2}}{2}\)
\(b^{2}+c^{2}=2 p^{2}+\frac{a^{2}}{2}\)

Again in \(\triangle A B C, \angle A E D=\angle A E B=90^{\circ}\)
By Pythagoras theorem.
AB² = AE² + EB²
= AD² - DE² + (BD - DE)²
= AD² - DE² + BD² + DE² -2BD.DE
= AD² + BD² - 2BD.DE
\( \mathrm{AB}^{2} =\mathrm{AD}^{2}+\left(\frac{1}{2} B C\right)^{2}-2 \cdot \frac{1}{2} \mathrm{BC} \cdot \mathrm{DE} \)
\(\mathrm{AB}^{2} =\mathrm{AD}^{2}+\frac{1}{4} B C^{2}-\mathrm{BC} \cdot \mathrm{DE} \)
\(\mathrm{AB}^{2} =\mathrm{AD}^{2}-\mathrm{BC} \cdot \mathrm{DE}+\frac{1}{4} B C^{2} \ldots(2) \)
\(\mathrm{c}^{2} =\mathrm{p}^{2}-\mathrm{ax}+\frac{a^{2}}{4}\)  

AB = 4 cm,
AC = 6 cm,
BD  = 1.6 cm,
CD =  2.4 cm.

\( \frac{A B}{A C}=\frac{4}{6}=\frac{2}{3} \)
\(\frac{B D}{D C}=\frac{1.6}{2.4}=\frac{2}{3} \)
\(\frac{A B}{A C}=\frac{B D}{D C} \)
By the converse of the Angle Bisector theorem AD is the bisector of \(\angle\)A

From (1) and (2) we have
\(\frac{A Q}{A B} =\frac{A R}{A D} \)
\(\frac{A B}{A Q} =\frac{A D}{A R} \)
\(\frac{A Q+Q B}{A Q} =\frac{A R+R D}{A R} \)
\(1+\frac{Q B}{A Q} =1+\frac{R D}{A R} \)
\(\Rightarrow \frac{Q B}{A Q} =\frac{D R}{A R} \)

We have
\(\Delta \)QMO ~ \(\Delta \)RPN 
\(\frac { MQ }{ RP } =\frac { QO }{ RN } \) 
\(\frac{M Q}{Q R}=\frac{Q R}{R N}\)
[ \(\because\)OQRP is a square PR = QR and QO = QR]
QR x QR = MQ x RN
QR² = MQ x RN 

Also In \(\Delta \)QMO & \(\Delta \)RPN
\(\angle \)QMO~\(\angle \)RPN = 90^o
we have \(\Delta \) LOP ~ \(\Delta \)QMO and \(\Delta \)LOP ~ \(\Delta \)RPN
\(\Delta \)QMO ~ \(\Delta \)RPN

In \(\Delta \)LOP & \(\Delta \)PRN,
we have
\(\angle \)PLO = \(\angle\)NRP = 90°
and \(\angle\)LPO = \(\angle\)PNR (corresponding angles)
By AA criterion of similarity
\(\Delta \)LOP~\(\Delta \)RPN

In \(\triangle\)PST and ​​\(\triangle\)PQR
\(\frac { PS }{ PQ } =\frac { 2 }{ 2+3 } =\frac { 2 }{ 5 } ,\frac { PT }{ PR } =\frac { 2 }{ 2+3 } =\frac { 2 }{ 5 } \)
Thus, \(\frac { PS }{ PQ } =\frac { PT }{ PR } \) and \(\angle\)P is is common
Therefore, by SAS similarity,
\(\triangle\)PST~\(\triangle\)PQR

From the figure, D is the mid point of BC.

We have \(\angle A E D=90^{\circ}\)
Given BC = a, AC = b, AB = c, ED = x, AD = P and AE = h
In \(\triangle\)AEC, by Pythagoras theorem
AC² =  AE² + EC²
AC² = AE² + (ED + DC)²
AC² = AE² + ED² + DC² + 2.E.D. DC
AC² = (AE² + ED²) + DC² + 2 ED . DC
AC² = AD² + DC² + 2 ED. DC
\(\mathrm{AC}^{2}=\mathrm{AD}^{2}+\left(\frac{1}{2} B C\right)^{2}+2\left(\frac{1}{2} B C\right) D E\)
[D is the mid point of BC, BD,= DC]
\( A C^{2} =A D^{2}+\mathrm{BC} \cdot \mathrm{DE}+\frac{1}{4} B C^{2} \ldots .(1) \)
\(\text { i.e., } b^{2} =\mathrm{P}^{2}+\mathrm{ax}+\frac{1}{4} a^{2} \)
\(\mathrm{~b}^{2} =\mathrm{p}^{2}+\mathrm{ax}+\frac{a^{2}}{4}\)

\(\Delta AEC\quad \Delta ADB\)
\(\angle AEC=\angle ADB={ 90 }^{ 0 }\) 
\(\angle C A E=\angle B A D\)
[common By AA similarity criteria] 
 \(\Delta AEC\sim \Delta ADB\)
(ii) Their corresponding sides are Proportional
\(\frac{C A}{A B}=\frac{C E}{D B}\)
Hence proved.

Since the tangents at a point to a circle is
perpendicular to the radius through the point of contact
\(\therefore \angle O P O^{\prime}=90^{\circ}\)
OP² + OP² = (OO')²
[By Pythagoras theorem]
3² + 4² = (OO')²
9+16 = (OO')²
25 = (OO')²
OO' = 5cm
Since the line joining the centres of two intersecting circles is perpendicular bisector of their common chord.
\(\mathrm{OR} \perp \mathrm{PQ} \text { and } O^{\prime} \mathrm{R} \perp \mathrm{PQ}\)
AIso PR = QR
Let OR = x, then O'R = 5 - x
AIso at PR = QR = y cm
\(\text { In } \triangle O R P \text { and } \Delta O^{\prime} R P\)
Applying Pythagoras theorem
OP² = OR³ + RP² and O'P'² = O'R² + RP²
\(3^{2}=x^{2}+y^{2} and 4^{2}=(5-x)^{2}+y_{i}^{2} \)
\(Subtracting \Rightarrow 4^{2}-3^{2}=\left\{(5-x)^{2}+y^{2}\right\}-\left(x^{2}+y^{2}\right) \)
\(16-9=25-10 x+x^{2}+y^{2}-x^{2}-y^{2} \)
7 - 25 = 10x
10x = 25 - 7
10x = 18
x = 1.8 cm
3² = x² + y²
\(y=\sqrt{9-(1.8)^{2}}=\sqrt{5.76}\)
y = 2.4cm
Hence PR = QR = 2.4 cm
PQ = 2y = 4.8 cm

Let o be the center of concentric circles and APB be the chord of length 16 cm of the larger circle touching the smaller circle at p.
Then OP \(\perp\) AB and p is the midpoint of AB.
AP = PB = 8 cm
In LOPA, we have
OA² = OP² + AP² [By Pythagoras Theorem]
OA² = 6² + 8²
OA² = 36 + 64
OA² = 100
OA = 10cm
Radius of the larger circle in 10 cm

Given TB is the tangent from the external point T and OB-radius
\(\angle OBT={ 90 }^{ 0 }\)
\( \angle A B T =65^{\circ} \)
\(\therefore \angle O B A =\angle O B T-\angle A B T \)
= 90^o - 65^o = 25^o
Since OA = OB [radius]
\(\triangle\)ABO is an isosceles triangle.
Angles opposite to equal sides are equal.
\( \therefore \angle O B A=\angle B A O=25^{\circ} \)
\(Now\ in\ \triangle A O B, \)
\(\angle A O B+\angle O B A+\angle O A B =180^{\circ} \)
\(\angle A O B+25^{\circ}+25^{\circ} =180^{\circ}\)   [sum of angles of a triangle]
\(\angle\)AOB +50^o = 180^o
\(\angle\)AOB = 180^o - 50^o = 130^o
\(\angle\)AOB = 130^o

Given PQ is the tangent from the point P outside the circle and OQ is the radius.
We know that tangent meet the radius perpendicularly
\( \therefore \angle P Q O =90^{\circ} \)
\(\angle P O Q =180-120=60^{\circ} \)
\( [\because\ \angle\ P O Q \ and\ \angle P O R\ are \ linear\ pair \ of\ angles]\)
\( In \ \triangle P O O \angle P O O+\angle P O O+\angle O P Q=180^{\circ}\)
[sum of angles of a triangle]
60^o + 90^o + \(\angle\)OPQ = 180o
\(\angle\)OPQ = 180^o - 150^o
\(\angle\)OPQ = 30^o

We know that the tangents drawn from are external point to a circle are equal.
Therefore AD AF = x 
BD = BE = y
and CE = CF = z 
Now, AB = 12 cm, BC = 8 cm, and CA = 10 cm.
x + y = 12,y + z = 8 and z + x = 10
(x + y) + (y + z) + (z + x) = 12 + 8 + 10
2(x + y + z) = 30
x+  y + z = 15
Now, x + y = 12 and x + y + z = 15
12 + z  = 15
Z = 3
y + z = 8 and x + y + z = 15
\(x+8=15\Rightarrow x=7\)
and z + x  = 10 and x + y + z = 15
\(10+y=15\Rightarrow y=5\)
Hence, AD = x = 7cm,
BE = y = 5 cm and
CF = z = 3 cm

Given \(\triangle\)LMN is a right angled triangle with \(\angle\)L = 90^o
By Pythagoras theore
Since PL ∥ OQ,PL= r = OP
we have  NR=NP=NL−PL
[ NR and NP are tangents]
= (6 - r)cm
MR = MQ = ML- LQ = (8 - r) cm
[MR and MQ are tangents]
NM = NR+ RM
= (6 - r + 8 - r)cm
= (14 - 2r) cm
Now NM² = NL² + LM² [By pythagoras Theorem]
(14-2r)2 =  82+62
(14-2r)2 = 64+36

Let 'O' be the center of the circle AP be the tangent.
Given OA = 25 cm; AP = 24 cm. Op is the radius.
Tangent and radius through the point are perpendicular to each other.
In the right triangle \(\triangle\)OPA,
OA² = OP² + PA²
25² = OP² + 24²
625 = OP² + 576
OP² = 625 - 576 = 49
OP = 7cm
Radius  = 7cm

OA = 4 cm, OB = 5 cm; also OA\(\bot \)BC.
OB² = OA² + AB²
5² =  4² + AB² gives AB² = 9
Therefore AB = 3 cm
BC = 2AB hence BC = 2 x 3 = 6 cm

AN = AM = 3 cm (Tangents drawn from same external point are equal)
BN = BL = 4 cm
CL = CM = AC - AM = 9 - 3 = 6 cm
Gives BC = BL + CL = 4 + 6 = 10 cm

\(\angle\)OPQ = 90^o - 50^o = 40^o (angle between the radius and tangent is 90^o)
OP = OQ (Radii of a circle are equal)
\(\angle\)OPQ = \(\angle\)OQP = 40^o (\(\triangle\)OPQ is isosceles)
\(\angle POQ={ 180 }^{ 0 }-\angle OPQ-\angle OQP\)
\(\angle\)POQ  = 180^o - 40^o- 40^o = 100^o.

Given OP = 5 cm, radius r = 3 cm
To find the length of tangent PT.
In right angled \(\triangle\)OTP
OP² = OT² + PT² gives PT² = 25 - 9 = 16
Length of the tangent PT = 4cm

Let \(\triangle\)ABC be the right triangle with \(\angle\)B = 90 ^o
Let the shortest side of the right triangle be x m.
Then Hypotenuse AC = (2x + 6) m
Third side BC = [(2x + 6) - 2] m
= (2x +  6 - 2)m
= (2x +  4) m
By Pythagoras theorem
AC² = AB² + BC²
(2x+ 6)² - x² + [2x + 4]²
4x² +36 + 24x = x² + 4x² + 16+ 16x
4x² + 24x + 36 = 5x² + 16x + 16
5x² + 16x+ 16- 4x² -24x- 36 = 0
x² - 8x - 20 = 0
(x + 2)(x - 10) = 0
x = -2 or x = 10
Length cannot be negative.
x = 10m
AB = 10m
BC = 2x + 4 = 2(10) + 4
= 24m
AC = 2x + 6 = 2(10) + 6 = 26m
The sides of the triangle are 10 m, 24m, 26m.

XY + CZ = 17cm
XZ + YW = 26cm
We know that diagonals if a rectangle bisect each other and the diagonals have equal length.
\(\therefore \text { Each diagonal }=\frac{26}{2}=13 \mathrm{~cm}\)
i.e., XZ = 13 cm and YW = 13 cm
Also given XY + YZ = 17 cm
Squaring on both sides (XY + YZ)² = 17²
\((\mathrm{XY})^{2}+(\mathrm{YZ})^{2}+2 \times(\mathrm{XY}) \times(\mathrm{YZ})=289\)
By Pythagoras theorem (XY)² + (YZ)² = XZ²
\(\therefore[\mathrm{XZ}]^{2}+2(\mathrm{XY}) \times(\mathrm{YZ})=289\)
132 + 2 x length x breadth = 289
2 x Area = 289 - 169
\(\text { Area }=\frac{289-169}{2}=\frac{120}{2}\)
The possible length and breadth are
(1,60) (2,30) (3,20) (4, 15), (5, 12) (6, 10).
In this pair the length and breadth should satisfy Pythagoras theorem for diagonal.
5,12 is the possible length and breadth.

Let the initial position of the man be A and his final position.be C
Since the man goes 18 m east and then 24 m north, LABC is a right angled triangle with \(\angle\)B =90^o; AB = 18 m and BC = 24m.
By Pythagoras theorem, we have
AC² - AB² + BC²
AC² = 18² + 24²
AC² = 324 + 576
AC² = 900 = 30 x 30
AC = 30m
His current distance from starting point = 30 m

Let x be the length of the ladder. BC = 4 ft, AC = 7 fit.
By Pythagoras theorem we have, AB² = AC² + BC²
x² = 7² + 4² gives x² = 49 + 16
x² = 65, Hence \(x=\sqrt { 65 } \)
The number \(\sqrt { 65 } \) is between 8 and 8.1.
82 = 64 < 65.61 = 8.1²
Therefore, the length of the ladder is approximately 8.1ft

Distance between the insect and the foot of the lamp post BD = 8 m
The height of the lamp post, AB = 6 m
After moving a distance of x m, let the insect be at C
Let, AC = CD = x . Then BC = BD − CD = 8 − x
In \(\triangle\)ABC, \(\angle\)B = 90^o
AC² = AB² + BC² gives x² = 62 + (8 - x)2
x² = 36 + 64 − 16x + x²
16x = 100 then x = 6.25
Then, BC = 8 − x = 8 − 6.25 = 1.75m
Therefore the insect is 1.75 m away from the foot of the lamp post.

AB = 5 cm,
AC = 10 cm,
BD  = 1.5 cm,
CD =  3.5 cm.

\( \frac{A B}{A C}=\frac{5}{10}=\frac{1}{2} \)
\(\frac{B D}{D C}=\frac{1.5}{3.5}=\frac{3}{7} \)
\(\frac{A B}{A C} \neq \frac{B D}{D C} \)
By the converse of the angle bisector theorem, AD is not a bisector of \(\angle\)A

Assume BD=y
BC-CD =y
6-CD =y
CD =6-y

In, \(\Delta ABD,\frac { BD }{ sin\theta } =\frac { AB }{ sin\propto } \Rightarrow \frac { 10 }{ sin\propto } \)
\(\Rightarrow sin\propto =\frac { 10 }{ y } sin\theta \)
In \(\Delta ACD,\frac { CD }{ sin\theta } =\frac { AC }{ sin\left( 180-\propto \right) } \)
\(\Rightarrow \frac { 6-y }{ sin\theta } =\frac { 14 }{ sin\propto } \)
Substituting (1) in (2),
\(\frac { 6-y }{ sin\theta } =\frac { 14 }{ \frac { 10 }{ y } sin\theta } \Rightarrow 6-y=\frac { 14y }{ 10 } \)
\(\Rightarrow y\left( 1+\frac { 14 }{ 10 } \right) =6\Rightarrow y\left( \frac { 24 }{ 10 } \right) =6\)
\(\Rightarrow y=\frac { 60 }{ 24 } \Rightarrow y=2.5\)
\(\therefore\) BD = 2.5 km and CD = 3.5 cm

In \(\triangle\)ACB, 
PQ || CB
Using Basic Proportionality theorem, we have
\(\frac{A Q}{A B}=\frac{A P}{A C}\)
Again in \(\triangle\)ACD PR || CD
Using Basic Proportionality theorem
\(\frac{A P}{A C}=\frac{A R}{A D}\)
From (1) and (2)
\(\frac{A Q}{A B}=\frac{A P}{A C}=\frac{A R}{A D} \)
Thus we have \(\frac{A R}{A D}=\frac{A Q}{A B} \)

Given AB = 12 cm, AD = 8 cm, AE = 12 cm, AC = 18 cm
By basic proportionality theorem \(\frac{A D}{D B}=\frac{A E}{E C}\)
\(\frac{A D}{D B}=\frac{8}{A B-A D}=\frac{8}{12-8} \)
\(\frac{A D}{D B}=\frac{8}{4}=2 \)
\(\frac{A E}{E C}=\frac{12}{A C-A E} \)
\(\frac{A E}{E C}=\frac{12}{18-12}=\frac{12}{6}=2 \)
From (1) and (2)
\(\frac{A D}{D B}=\frac{A E}{E C}\)
DE || BC

Join AC such that it meets PQ at G.
AB || DC and PQ || DC
PQ || AB
IN ADC
PG || DC
\(\frac{A P}{P D}=\frac{A G}{G C}\)
In CAB
\(\frac{A G}{G C}=\frac{B Q}{Q C}\)
From (1) and (2) we get
\(\frac{A P}{P D} =\frac{B Q}{Q C} \)
\(\frac{A P}{18} =\frac{35}{15} \)
\(A P =\frac{35 \times 18}{15} \)
AP = 42cm
Now AD = AP + PD
= 42 + 18 = 60 cm
AD = 60 cm

Let BD =  x cm, then DC = (6 – x)cm
AD is bisector of\(\angle\) A
Therefore by Angle Bisector Theorem
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac { 10 }{ 14 } =\frac { x }{ 6-x } \quad \frac { 5 }{ 7 } =\frac { x }{ 6-x } \)
So, 12x = 30 we get, \(x=\frac { 30 }{ 12 } =2.5\)
Therefore, BD = 2.5 cm, DC = 6−x = 6−2.5 = 3.5 cm

In \(\triangle\)ABC, AD is the bisector of \(\angle\)A
Therefore by Angle Bisector of \(\angle\)A
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac{4}{3}=\frac{6}{A C}\) gives 4AC = 18. Hence, AC \(=\frac{9}{2}=4.5 \mathrm{~cm}\)

We have AB = 56.cm, AD = 14. cm, AC = 72. cm and AE = 18.cm.
BD = AB - AD = 5.6 –1.4 = 4.2 cm
and EC = AC – AE = 7.2–1.8 = 5.4 cm
\(\frac { AD }{ DB } =\frac { 1.4 }{ 4.2 } =\frac { 1 }{ 3 } \) and \(\frac { AE }{ EC } =\frac { 1.8 }{ 5.4 } =\frac { 1 }{ 3 } \)
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)
Therefore, by converse of Basic Proportionality Theorem, we have DE is parallel to BC. Hence proved.

In \(\Delta \)LOP & \(\Delta \)QMO,
\(\angle \)OLP = \(\angle \)MQO 90°
and \(\angle \)LOP =\(\angle \)OMQ (corresponding angles)
(by AA criterion of similarity)
\(\Delta \)LOP ~ \(\Delta \)QMO

Given \(\Delta ABC\sim \Delta APQ\)
Their corresponding sides are proportional
\(\frac{A C}{A P}=\frac{C B}{P Q}=\frac{A B}{A Q} \)
\(\frac{A C}{2.8}=\frac{8}{4}=\frac{6.5}{A Q} \)
Taking
\(\frac{A C}{2.8}=\frac{8}{4} \)
\(A C=\frac{8 \times 2.8}{4}=5.6 \mathrm{~cm} \)
AC = 5.6 cm
Also taking \(\frac{8}{4} =\frac{6.5}{A Q} \)
\(\mathrm{AQ} =\frac{6.5}{8} \times 4=3.25 \mathrm{~cm} \)
AQ = 3.25 cm

Given o = \(\triangle\)ABC and \(\triangle\)ADE
By AA criteria for similarity
\(\triangle A B C \sim \triangle A D E\)
Their corresponding sides are proportional
\(\therefore \frac{B C}{D E} =\frac{A C}{A E}=\frac{A B}{A D} \)
\(\frac{12}{D E} =\frac{3+2}{A E}=\frac{A B}{3} \)
Also in right ABC , o
using Pythagoras theorem, we have
AB² = BC² + AC² = 12² + 5²
= 144 + 25 = 169
AB = 13 cm
Now from (1), we get
\(\frac{12}{D E}=\frac{13}{3} \)
\(\mathrm{DE}=\frac{12 \times 3}{13}=\frac{36}{13} \)
DE = 2.77 cm
Also from (1) \(\frac{5}{A E}=\frac{13}{3} \)
\(\mathrm{AE}=\frac{5 \times 3}{13}=\frac{15}{13}=1.15 \mathrm{~cm} \)

Given = O
PR and SQ intersect at T.
In \(\triangle\)QPT and \(\triangle\)RST
< QSR = o
Given
[Vertically opposite angles]
By AA similarity criteria
\(\triangle P Q T \sim \triangle S R T\)
Their corresponding sides are proportional 
\(\frac{P T}{S T}=\frac{T Q}{T R}\)
PT x TR = ST x TQ
Hence proved

Let DE be the vertical stick and AB is the tower,
DE = 6 m, EF = 400 cm = 4 m, BC = 28 m
From DFE and ACB
Using similarity criteria
\(\frac{A B}{D E}=\frac{B C}{E F}\)
\(\frac{A B}{6}=\frac{28}{4} \)
\(A B=\frac{28 \times 6}{4}=42 \mathrm{~m} \)
Height of the tower = 42 m

(i) In ABC and ADE <A is common
\( \frac{A E}{E C}=\frac{2}{3 \frac{1}{2}}=\frac{\frac{2}{7}}{2}=\frac{2 \times 2}{7}=\frac{4}{7} \)
\(\frac{A D}{D B}=\frac{3}{5} \)
\(Here\ \frac{4}{7} \pm \frac{3}{5} \)
\(\frac{A E}{E C} \neq \frac{A D}{D B} \)
The corresponding sides are not proportional.
ABC and ADE are not similar
(ii) In CPQ and CAB <C is common
<PQC = 180^o - 110^o = 70^o
[ <PQC and <PQB are liner pair of angles]
<ABC = 70^o
<BAC = <QPC
[ sum of three angles of a triangle are 180]
<PCQ = 180^o - (< QPC + 70^o)
<ABC = <PQC = 70^o
<C common and <BAC = <QPC
By AAA similarity criteria, <ABC <PQC
Corresponding sides are Proportional
\(\frac{A B}{P Q} =\frac{B C}{Q C} \)
\(\frac{5}{\sqrt{x}} =\frac{6}{3} \)
[BC = BQ + QC = 3 + 3 = 6]
\(x=\frac{5}{6} \times 3=2.5\)
x = 2.5

Since the ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have
\(\frac { Area(\Delta ABC) }{ Area(\Delta DEF) } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \) gives \(\frac { 54 }{ Area(\Delta DEF) } =\frac { { 3 }^{ 2 } }{ { 4 }^{ 2 } } \)
\(Area(\Delta DEF)=\frac { 16\times 54 }{ 9 } =96{ cm }^{ 2 }\)

The ratio of the corresponding sides of similar triangles is same as the ratio of their perimeters.
Since, \(\Delta ABC\sim \Delta PQR\)
\(\frac { AB }{ PQ } =\frac { BC }{ QR } =\frac { AC }{ PR } =\frac { 36 }{ 24 } \)
\(\frac { AB }{ PQ } =\frac { 36 }{ 24 } \ \frac { AB }{ 10 } =\frac { 36 }{ 24 } \)
\(AB=\frac { 36\times 10 }{ 24 } =15cm\)

\(\Delta AOQ\) and \(\Delta BOP,\angle OAQ=\angle OBP=90^{ 0 } \)
\(\angle AOQ=\angle BOP\) (Vertically opposite angles)
Therefore, by AA Criterion of similarity,
\(\Delta AOQ\sim \Delta BOP\)
\(\frac { AO }{ BO } =\frac { OQ }{ OP } =\frac { AQ }{ BP } \)
\(\frac { 10 }{ 6 } =\frac { AQ }{ 9 } \) gives \(AQ=\frac { 10\times 9 }{ 6 } =15cm\)

\(\Delta \ CAB\) and \(\Delta CED\),\(\angle C\) is common, \(\angle A=\angle CED\)
Therefore, \(\Delta CAB\sim \Delta CED\)
Hence, \(\frac { CA }{ CE } =\frac { AB }{ DE } =\frac { CB }{ CD } \)
\(\frac { AB }{ DE } =\frac { CB }{ CD } \quad \frac { 9 }{ x } =\frac { 10+2 }{ 8 } ,x=\frac { 8\times 9 }{ 12 } =6\) cm.

In \(\Delta BAC\) and \(\Delta PRQ,\quad \frac { AB }{ RQ } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
\(\frac { BC }{ QP } =\frac { 6 }{ 12 } =\frac { 1 }{ 2 } ;\frac { CA }{ PR } =\frac { 3\sqrt { 3 } }{ 6\sqrt { 3 } } =\frac { 1 }{ 2 } \)
Therefore, \(\frac { AB }{ QP } =\frac { BC }{ QP } =\frac { CA }{ PR } \)
By SSS similarity, we have \(\triangle\)BAC~\(\triangle\)QRB
\(\angle\)P =\(\angle\)C (since the corresponding parts of similar triangle)
\(\angle\)P =\(\angle\)C 180^o -\((\angle A+\angle B)={ 180 }^{ 0 }-({ 90 }^{ 0 }+{ 60 }^{ 0 })\)
\(\angle\)P = 180^o - 150^o = 30^o

In Is \(\triangle\)ABC ~ \(\triangle\)PQR
\(\frac { PQ }{ AB } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } ;\frac { QR }{ BC } =\frac { 4 }{ 10 } =\frac { 2 }{ 5 } \)
Since \(\frac { 1 }{ 2 } \neq \frac { 2 }{ 5 } ,\frac { PQ }{ AB } \neq \frac { QR }{ BC } \)
The corresponding sides are not proportional.
Therefore \(\triangle\)ABC  is not similar to \(\triangle\)PQR.

In \(\triangle\)PST and \(\triangle\)PQR,
\(\frac { PS }{ PQ } =\frac { 2 }{ 2+1 } =\frac { 2 }{ 3 } ,\frac { PT }{ PR } =\frac { 4 }{ 4+2 } =\frac { 2 }{ 3 } \)
Thus, \(\frac { PS }{ PQ } =\frac { PT }{ PR } \) and \(\angle\)P is common
Therefore, by SAS similarity,
\(\triangle\) PST~\(\triangle\)PQR