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Since, \(\triangle ABN\sim \triangle ACM\)  [given]
\(\frac { ar\left( \triangle ABN \right) }{ ar\left( \triangle ACM \right) } =\frac { { AB }^{ 2 } }{ { AC }^{ 2 } } \) ... (i)
Similarity, \(\frac { ar\left( \triangle BCL \right) }{ ar\left( \triangle ACM \right) } =\frac { { BC }^{ 2 } }{ { AC }^{ 2 } } \) ... (ii) 

On adding Eqs.(i) and (ii), we get
\(\frac { ar\left( \triangle ABN \right) +ar\left( \triangle BCL \right) }{ ar\left( \triangle ACM \right) } =1\)

Given, \(\triangle ABC\sim \triangle QRP\), AB = 18 cm, BC = 15 cm

We know that, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
\(\therefore \frac { ar\left( \triangle ABC \right) }{ ar\left( \triangle QRP \right) } =\frac { { \left( BC \right) }^{ 2 } }{ { \left( RP \right) }^{ 2 } } \)
But \(\frac { ar\left( \triangle ABC \right) }{ ar\left( \triangle QRP \right) } =\frac { 9 }{ 4 } \)  given]
\(\Rightarrow \frac { { \left( 15 \right) }^{ 2 } }{ { \left( RP \right) }^{ 2 } } =\frac { 9 }{ 4 } \)  [\(\because \) BC = 15 cm]
\(\Rightarrow { \left( RP \right) }^{ 2 }=\frac { 225\times 4 }{ 9 } =100\)
\(\therefore \)  RP = 10 cm
[taking positive square root]

Given, in \(\triangle ABC\), \(DE\parallel BC\)
By Thales theorem, we get
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)

\(\Rightarrow \frac { 4x-3 }{ 3x-1 } =\frac { 8x-7 }{ 5x-3 } \)
[\(\because \)  AD = 4x - 3, DB = 3x - 1, AE = 8x - 7, EC = 5x - 3]
\(\Rightarrow \) (4x - 3)(5x - 3) = (8x - 7)(3x - 1)
\(\Rightarrow \)  20x² - 12x + 9 - 15x = 24x² - 21x - 8x + 7
\(\Rightarrow \)  4x² - 2x - 2 = 0
\(\Rightarrow \)  2x² - x - 1 = 0     [dividing both sides by 2]
\(\Rightarrow \)  2x² - 2x + x - 1 = 0   [by splitting the middle term]
\(\Rightarrow \)  2x(x - 1) + 1 (x - 1) = 0
\(\Rightarrow \)  (2x + 1) (x - 1) = 0   \(\therefore x=-\frac { 1 }{ 2 } \) or x = 1
If \(x=-\frac { 1 }{ 2 } \), then AD = \(4\times -\frac { 1 }{ 2 } -3=-5<0\)  [not possible]
Hence, x = 1 is the required value.

Given, \(\triangle ABC\sim \triangle EDF\)
Also, AB = 5 cm, AC = 7 cm, DF = 15 cm
and DE = 12 cm                         .... (i)
Since, \(\triangle ABC\sim \triangle EDF\)
\(\therefore \frac { AB }{ ED } =\frac { AC }{ EF } =\frac { BC }{ DF } \)
[ ∵ Corresponding sides of similar triangles are proportions]
\(\Rightarrow \frac { 5 }{ 12 } =\frac { 7 }{ EF } =\frac { BC }{ 15 } \)  [from Eq. (i)]

On taking first and second terms, we get
\(\frac { 5 }{ 12 } =\frac { 7 }{ EF } \Rightarrow EF=\frac { 7\times 12 }{ 5 } \) = 16.8 cm
On taking first and third terms, we get
\(\frac { 5 }{ 12 } =\frac { BC }{ 15 } \Rightarrow BC=\frac { 5\times 15 }{ 12 } \) = 6.25 cm
Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm.

Draw the figure according to the question and get two triangles. Then, show both triangles are similar by AAA similarity criterion and then calculate the required distance.
She is at 9 m from the base of the pole.

(i) Examples of similar figures:
(a) All squared
(b) All regular hexagons
(ii) Examples of non-similar figures:
(a) Two isosceles triangles of different angle measures
(b) Two rhombus of different angle measures.

In \(\triangle OPQ\),  \(AB\parallel PQ\) [given]
\(\therefore \quad \frac { OA }{ AP } =\frac { OB }{ BQ } \)  ... (i)
[by basic proportionality theorem]
Also, in \(\triangle OPR\), \(AC\parallel PR\)  [given]
\(\therefore \quad \frac { OA }{ AP } =\frac { OC }{ CR } \)  ... (ii)
From Eqs. (i) and (ii), 
\(\frac { OB }{ BQ } =\frac { OC }{ CR } \Rightarrow \quad BC\parallel QR\)
[by converse of basic proportionality theorem]
Hence proved.

In \(\triangle APR\) and \(\triangle ABD\),
\(\angle APR\) = \(\angle ABD\)
\(\angle ARP\) = \(\angle ADB\)
\(\therefore \triangle ABC\sim \triangle ABD\)
[by AA criteria of similarity]

\(\frac{PR}{BD}=\frac{AR}{BD}\)   ... (i)
Similarity, we can prove
\(\frac{AR}{AD}=\frac{RQ}{DC}\)  ... (ii)
From Eqs.(i) and (ii), 
\(\frac{PR}{BD}=\frac{RQ}{DC}\)   [\(\because\) AD is median]
PR = RQ
So, median AD bisects PQ.

(ii) Similarity of triangles
(iii) Trees are helpful to maintain the balance in the environment. They should be saved at any cost.

Since G is the mid-point of PQ,
\(\therefore\)   PG=PQ
\(\Rightarrow \frac { PG }{ GQ } =1\)
According to the question,
GH || QR
\(\therefore \frac { PG }{ GQ } =\frac { PH }{ HR } \)
\(\Rightarrow 1=\frac { PH }{ HR } \)
\(\therefore\) PH=HR.
Hence, H is the mid-point of PR.

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

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

As DE || BC
\(\therefore \frac { AD }{ DB } =\frac { AE }{ EC } \)
\(\Rightarrow \frac { x }{ x+1 } =\frac { x+3 }{ x+5 } \)
\(\Rightarrow \)x²+5x=x²+4x+3
\(\Rightarrow \)x=3

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

12 cm, 16 cm