Question

In the given figure, PQRS is a parallelogram. If $PT:TQ=1:2,SA:AT=3:5,TB:BC:CR=1:2:3andar\left(PQRS\right)=1440c{m}^2,then\left[ar\right(PAS)+ar(QBC\left)\right]$ equals

• A.
  $240c{m}^2$
• B.
  $250c{m}^2$
• C.
  $300c{m}^2$
• D.
  $360c{m}^2$

Answer 1

The correct option is B

$250c{m}^2$
Let h be the height of the parallelogram PQRS

Draw$PU\perp STandQV\perp RT.Given,PT:TQ=1:2\Rightarrow PT=\frac{PQ}{3}andTQ=\frac{2PQ}{3}..\left(i\right)and,SA:AT=3:5\Rightarrow SA=\frac{3}{8}ST....\left(ii\right)and,TB:BC:CR=1:2:3\Rightarrow BC=\frac{2}{6}TR=\frac{1}{3}TR\dots \left(iii\right)Now,Ar\left(PQRS\right)=1440c{m}^2\Rightarrow PQ\times h=1440\dots \left(iv\right)$

We know that, if in a triangle ABC, D is a point on BC which divides it in the ratio of a : b, i.e.,

$BD:DC=a:b,then\frac{Ar\left(\Delta ABD\right)}{Ar\left(\Delta ACD\right)}=\frac{BD}{DC}=\frac{a}{b}.Now,SA=\frac{3}{8}ST\text{[From (ii)]}\Rightarrow \frac{Ar\left(\Delta PAS\right)}{Ar\left(\Delta PST\right)}=\frac{3}{8}\Rightarrow Ar(\Delta PAS=\frac{3}{8}Ar(\Delta PST)\Rightarrow Ar(\Delta PAS)=\frac{3}{8}\times (\frac{1}{2}\times PT\times h)=\frac{3}{8}\times (\frac{1}{2}\times PT\times h)=\frac{3}{6}\times \frac{1}{2}\times \frac{PQ}{3}\times h[From\left(i\right)]=\frac{1}{16}\times (PQ\times h)=\frac{1440}{16}[From\left(iv\right)]=90c{m}^{2}Similarly,BC=dfrac13TR[From\left(iii\right)]\Rightarrow Ar(\Delta QBC)=\frac{1}{3}Ar(\Delta TQR)=\frac{1}{3}\times (\frac{1}{2}\times TQ\times h)=\frac{1}{3}\times \frac{1}{2}\times \frac{2}{3}PQ\times h[From\left(i\right)]=\frac{1}{9}\times (PQ\times h)=\frac{1440}{9}[From\left(iv\right)]=160c{m}^2\therefore Ar(\Delta PAS)+Ar(\Delta QBC)=90+160=250c{m}^2$

Hence, the correct answer is option (b).