(i) Man is at distance of 10 m from the initial position.
(ii) Right angled triangle pythagoras theorem.
(iii) Knowledge of direction and speed save the time.

As \(\triangle ABC\sim \triangle DFE\)
\(\angle D=\angle A={ 30 }^{ ° }\)
\(\angle C=\angle E={ 50 }^{ ° }\)
\(\angle B=\angle F={ 180 }^{ ° }-\left( { 50 }^{ ° }+{ 30 }^{ ° } \right) ={ 100 }^{ ° }\)
Now, \(\frac { AB }{ DF } =\frac { AC }{ DE } \)
\(\frac { 5 }{ 7.5 } =\frac { 8 }{ DE } \)
\(\Rightarrow DE=12cm,\angle F={ 100 }^{ ° }\)

In \(\triangle\) PQR, CA || PR
\(\therefore \quad \frac { PC }{ CQ } =\frac { RA }{ AQ } (By\quad BPT)\)
\(\Rightarrow \quad \frac { PC }{ 15 } =\frac { 20 }{ 12 } \)
\(\Rightarrow \quad PC=\frac { 20\times 15 }{ 12 } =25\)
\(\therefore \quad PC=25\quad cm\)
In \(\triangle\) PQR , CB || QR
\(\therefore \quad \frac { PC }{ CQ } =\frac { PB }{ BR } (By\quad BPT)\)
\(\Rightarrow \quad \frac { 25 }{ 15 } =\frac { 15 }{ BR } \)
\(\Rightarrow \quad BR=\frac { 15\times 15 }{ 25 } =9\quad cm\) 

Let \(CD\bot AB,\)
then CD=p
Area of \(\triangle\)ABC= \(\frac {1}{2}\) x base x height
\(\Rightarrow\)Area of \(\triangle\)ABC=\(\frac {1}{2}\)  x AB x CD= cp
Also, Area of \(\triangle\)ABC=\(\frac {1}{2}\) x BC x AC= ab
\(\frac {1}{2}\)cp=\(\frac {1}{2}\)ab
\(\Rightarrow\)cp=ab.

Given 2AB = DE and BC = 8 cm
\(\triangle\)ABC ~ \(\triangle\)DEF

So \(\frac { AB }{ BC } =\frac { DE }{ EF } \)
\(\Rightarrow d \frac { AB }{ 8 } =\frac { 2AB }{ EF } \)
\(\therefore EF=2\times 8=16 cm.\)

Given DE || BC, then
\(\frac { AD }{ AB } =\frac { DE }{ BC } \) (By Thales Theorem)
\(\Rightarrow \quad \frac { 3 }{ AD+DB } =\frac { x }{ 14 } \\ \Rightarrow \frac { 3 }{ 3+4 } =\frac { x }{ 14 } \\ \Rightarrow \quad \frac { 3 }{ 7 } =\frac { x }{ 14 } \)
\(\Rightarrow \) 7x=14 x 3
\(\therefore \quad x=\frac { 14\ \times 3 }{ 7 } =6\)

Let the triangles be \(\triangle\)ABC and \(\triangle\)DEF
\(\frac { ar(\triangle ABC) }{ ar(\triangle DEF) } =\left( \frac { 5 }{ 6 } \right) ^{ 2 }=\frac { 25 }{ 36 } \)
25 : 36

No, Angle included should be same.

In \(\triangle\)PAO and \(\triangle\)QBO
\(\angle\)A=\(\angle\)B=90⁰
\(\angle\)POA=\(\angle\)QOB   (Vertically Opposite Angle)
\(\triangle\)PAO~\(\triangle\)QBO,    (by AA)
\(\therefore \quad \frac { OA }{ OB } =\frac { PA }{ QB } \\ \Rightarrow \frac { 6 }{ 4.5 } =\frac { 4 }{ QB } \\ \Rightarrow \quad QB=\frac { 4\times 4.5 }{ 6 } \\ \therefore \quad QB=3\quad cm\)

Let AB be the lamp-post and ED be the position of girl after 4s.
Given, height of the girl, ED = 100 cm
and height of the lamp-post, AB = 5 m = 500 cm
Distance of the girl from lamp-post after 4 s
= 1.9 x 4 = 7.6 m = 760 cm
[\(\because\)  distance = speed x time]
i.e. BD = 760 cm
Let DC = x cm
In \(\triangle CDE\) and \(\triangle CBA\),
\(\angle DCE=\angle BCA\) [common angle]
\(\angle CDE=\angle CBA\)  [each 90°]
\(\therefore \triangle CDE\sim \triangle CBA\) [by AA similarity criterion]
So, \(\frac { CD }{ CB } =\frac { DE }{ BA } \Rightarrow \frac { x }{ x+760 } =\frac { 100 }{ 500 } \)
\(\Rightarrow\)  5x = x + 760
\(\Rightarrow\)  4x = 760
\(\Rightarrow\)  x = 190 cm
Hence, the length of her shadow 4s is 190 cm.

\(\frac{16}{9}\)

 In \(\triangle ADE\) and \(\triangle ABC\),
\(\angle ADE=\angle ABC\)  [given]
\(\angle DAE=\angle BAC\)  [common angle]
So, \(\triangle ADE\sim \triangle ABC\)
[by AA similarity criterion]
Then, \(\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}\)
[since, corresponding sides of similar triangles are proportional]
\(\Rightarrow \frac{7.6}{7.2 + 4.2}=\frac{DE}{8.4}\)  [\(\because\)  AB = AE + BE]
\(\Rightarrow \frac{7.6}{11.4}\times 8.4=DE \Rightarrow\) DE = 5.6 m

Given \(\triangle NSQ\cong \triangle MTR\) and \(\angle1=\angle2\)
To prove \(\triangle PTS\sim \triangle PRQ\)
Proof Since, \(\triangle NSQ\cong \triangle MTR\)
\(\therefore\) SQ = TR  ... (i)
Also, \(\angle1=\angle2\)
\(\Rightarrow\) PT = PS .... (ii)
[since, sides opposite to equal angles are also equal]
From Eqs.(i) and (ii), \(\frac{PS}{SQ}=\frac{PT}{TR}\)
\(\Rightarrow ST\parallel QR\)
[by converse of basic proportionality theorem]
\(\therefore \angle 1=\angle PQR\)   and \( \angle 2=\angle PRQ\)
[\(\therefore\)  atternate exterior angle]
In \(\triangle PTS\) and  \(\triangle PRQ\),
\(\angle P=\angle P\)    [common angle]
\(\angle 1=\angle PQR\)  [proved above]
and \(\angle 2=\angle PRQ\)
\(\therefore \triangle PTS\sim \triangle PRQ\)
[by AAA similarity criterion]

Let \(\Delta A B C\) is an equilateral triangle of each side '2a' inscribed in a circle of radius 6 ern and centre O.
Draw AD⊥BC. Then, BD = DC [∵ In an equilateral triangle, perpendicular bisects the base]
Also, O lies on AD [∵ In an equilateral triangle circumcentre and centroid coincides]

Thus, OA : OD = 2 : 1
\(\Rightarrow \frac{6}{O D}=\frac{2}{1} \Rightarrow O D=3 \mathrm{~cm}, \text { then } A D=9 \mathrm{~cm}\)
Now, use pythagoras theorem in ΔABD.
\(6 \sqrt{3} \mathrm{~cm}\)

12 cm, 16 cm

9 : 7

In \(\triangle\)ABC and \(\triangle\)AOQ.

BC II PQ (Given)
\(\therefore\) \(\angle\)CBA = \(\angle\)AQP (Alternate angles)
\(\angle\)BAC = \(\angle\)PAQ (Vertically opposite angles)
\(\therefore\) \(\triangle\)ABC - \(\triangle\)AQP (AA Similarity)
\(\Rightarrow \frac { AB }{ AQ } =\frac { BC }{ QP } =\frac { AC }{ AP } \)
\(\Rightarrow \frac { 6.5 }{ AQ } =\frac { 8 }{ 4 } =\frac { AC }{ 2.8 } \)
\(\Rightarrow AQ=\frac { 6.5 }{ 2 } =3.25 cm,\)
and AC = 2 x 2.8 = 5.6 cm.

In \(\triangle ADC\) and \(\triangle ACB\),
\(\angle ADC=\angle ACB\)  [each 90°]
\(\angle DAC=\angle CAB\) [common angle]
So, \(\triangle ADC\sim \triangle ACB\)
[by AA similarity criterion]
Then, \(\frac{AD}{AC}=\frac{AC}{BA}\)
[since, corresponding sides of similar triangles are proportional]
\(\Rightarrow \) AC² = AB x AD     ..... (i)
Similarity, \(\triangle BDC\sim \triangle BCA\)
\(\therefore \frac{BD}{BC}=\frac{BC}{AB}\)
[since, corresponding sides of similar triangles are proportional]
\(\Rightarrow \) BC² = AB x BD     ..... (ii)
On dividing Eq (ii) by Eq (i), we get
\(\frac { { BC }^{ 2 } }{ { AC }^{ 2 } } =\frac { BD }{ AD } \)

Prove \(\triangle PNK\) and \(\triangle LMK\), similar then, \(\frac { NK }{ MK } =\frac { PN }{ LM } \)
\(\frac { c }{ b+c } =\frac { x }{ a } \Rightarrow x=\frac { ac }{ b+c } \)

In \(\triangle PRQ\) and \(\triangle STQ\),
\(\angle PRQ=\angle STQ\) [each 90°]
\(\angle PQR=\angle SQT\)  [common angle]
So, \(\triangle PRQ\sim \triangle STQ\)  [by AA similarity criterion]
Then, \(\frac { QR }{ QT } =\frac { QP }{ QS } \)
[since, corresponding sides of similar triangles are proportional]
\(\Rightarrow \)  QR x QS = QP x QT

\(DP\parallel BC\) and \(EQ\parallel AC\)
\(\frac { AD }{ DB } =\frac { AP }{ PC } \) and \(\frac { BE }{ EA } =\frac { BQ }{ QC } \)
\(\frac { AD }{ DB } =\frac { AP }{ PC } \) and \(\frac { AD }{ DB } =\frac { BQ }{ QC } \)
[EA = ED + DA = ED + BE = BD \(\Rightarrow \) AD = BE]
\(\Rightarrow \frac { AP }{ PC } =\frac { BQ }{ QC } \)
In \(\triangle ABC\), P and Q divides sides CA and CB respectively, in the same ratio
\(\Rightarrow PQ\parallel AB\)

Prove \(\triangle ABC\) and \(\triangle AMP\)  are similar.
then take ratio \(\frac{AC}{AP}=\frac{BC}{MP}\)

In \(\triangle ABC, DE\parallel BC\)
Let EC = x cm
\(\Rightarrow \frac { AD }{ DB } =\frac { AE }{ EC } \)

[by basic proportionality theorem]
\(\Rightarrow \frac { 3 }{ 4 } =\frac { 6 }{ x } \Rightarrow \)  x = 8 cm
\(\therefore \)  EC = 8 cm

In \(\triangle ABC\), \(\frac { AD }{ DB } =\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)

and \(\frac { AE }{ EC } =\frac { 8 }{ 12 } =\frac { 2 }{ 3 } \)
So, \(\frac { AD }{ DB } =\frac { AE }{ EC } \)
Hence, \(DE\parallel BC\)
[ by converse of basic proportionallity theorem]

Given, AB = 6.5 cm and PQ = 10.4 cm

Since, \(\triangle ABC\sim \triangle PQR\)
\(\frac { AB }{ PQ } =\frac { BC }{ QR } =\frac { AC }{ PR } =\frac { 6.5 }{ 10.4 } =\frac { 65 }{ 104 } \)
[ \(\because \) Corresponding sides of similar triangles are proportional]
i.e. \(AB=\frac { 65 }{ 104 } PQ,BC=\frac { 65 }{ 104 } QR,AC=\frac { 65 }{ 104 } PR\)
Also given, perimeter of \(\triangle ABC\) = 60
\(\therefore AB+BC+AC=60\)
\(\Rightarrow \frac { 65 }{ 104 } \left( PQ+QR+PR \right) =60\)
\(PQ+QR+PR=\frac { 60\times 104 }{ 65 } \) = 96 cm
Hence, periemeter of \(\triangle PQR\) is 96 cm.

\(\frac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)