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

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

    
The correct correspondence will be \(\triangle QRP\sim \triangle SMT\) No

\(\triangle ABC\) and \(\triangle PQR\)  are similar
\(\frac { AB }{ PQ } =\frac { BC }{ QR } \Rightarrow \frac { x }{ 4.8 } =\frac { 2 }{ 6.4 } \Rightarrow x=1.5\)
 and \(\frac { AC }{ PR } =\frac { BC }{ QR } \Rightarrow \frac { 4 }{ y } =\frac { 2 }{ 6.4 } \Rightarrow x=12.8\)
x = 1.5 cm and y = 12.8 cm

Draw the figure according to the given conditions and use Pythagoras theorem to find the value of x, then required saved distance will be equal to the difference of (AC + BC) and 26.
8 km.

Diagonals of a rhombus bisect each other at right angles.

So, OA = OC = 8 cm
and OB = OD = 6 cm
Now use pythagoras theorem in  ΔAOB
10 cm.

Apply Pythagoras theorem, and use the result p - q = 1

Use Pythagoras theorem, to find the distance of the foot of the ladder from the building.

= 8 m

We know that perpendicular draw from a vertex to its opposite base bisects the base in an equilateral triangle.

\(BD=\frac { BC }{ 2 } =\frac { a }{ 2 } \)
In \(\triangle ADB,\angle ADB={ 90 }^{ ° }\)
AB² = AD² + BD²
[using Pythagoras theorem]
\({ a }^{ 2 }={ AD }^{ 2 }+\left( \frac { a }{ 2 } \right) ^{ 2 }\)
\(\Rightarrow AD=\frac { \sqrt { 3 } }{ 2 } \)
\(\triangle ABC\) and \(\triangle ADE\) are equilateral triangles and hence equiangular.
\(\therefore \triangle ABC\sim \triangle ADE\)
\(\frac { ar\left( \triangle ADE \right) }{ ar\left( \triangle ABC \right) } =\frac { { AD }^{ 2 } }{ { AB }^{ 2 } } =\frac { \left( \frac { \sqrt { 3 } }{ a } \right) ^{ 2 } }{ { a }^{ 2 } } =\frac { 3 }{ 4 } \)

Prove \(\triangle ADB\sim \triangle ADC\), then \(\frac { BD }{ AD } =\frac { AD }{ CD } \)
\(\Rightarrow \)  AD² = BD.CD