In figure xp/py=xq/qz=3if the area of …

Given  {tex}\frac{{XP}}{{PY}} = \frac{{XQ}}{{QZ}}{/tex} 
​{tex}\Rightarrow {/tex}​ PQ {tex}\parallel{/tex} YZ  .....(i) [By converse of B.P.T]
In ​{tex}\triangle {/tex}​ XPQ and ​{tex}\triangle {/tex}​ XYZ we have
[ ​{tex}\angle{/tex}​ XPQ = ​{tex}\angle{/tex}​ Y [From (i) corresponding angles]
​{tex}\angle{/tex}​ X = ​{tex}\angle{/tex}​X [common]
​{tex}\therefore {/tex}​ ​{tex}\triangle {/tex}​ XPQ ​{tex}\cong{/tex}​ {tex}\triangle {/tex}XYZ [By AA similarity]
​{tex}\therefore {/tex}​ {tex}\frac{{ar\left( {\triangle XYZ} \right)}}{{ar\left( {\triangle XPQ} \right)}}=\frac{{X{Y^2}}}{{X{P^2}}}{/tex}
{tex}\frac { P Y } { X P } = \frac { 1 } { 3 }{/tex}
We have {tex}\Rightarrow \frac { P Y } { X P } + 1 = \frac { 1 } { 3 } + 1{/tex}
{tex}\Rightarrow \frac { P Y + X P } { X P } = \frac { 4 } { 3 }{/tex} 
​​​{tex}\Rightarrow {/tex}​ {tex}\frac{{XY}}{{XP}} = \frac{4}{3}{/tex}
Substituting in (i), we get
{tex}\frac{{ar\left( {\triangle XYZ} \right)}}{{ar\left( {\triangle XPQ} \right)}} = {\left( {\frac{4}{3}} \right)^2} = \frac{{16}}{9}{/tex}
​{tex}\Rightarrow {/tex}​{tex}\frac{{32}}{{ar\left( {XPQ} \right)}} = \frac{{16}}{9}{/tex}
{tex}ar\left( {XPQ} \right) = \frac{{32 \times 9}}{{16}} = 18c{m^2}{/tex}
Area of quadrilateral PYZQ = 32 - 18 = 14 cm²