Question

In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T.
(i) Prove that $\Delta$TPS$\sim$ $\Delta$TRQ
(ii) Find SP if TP$=18$cm, RQ$=4$cm and TR$=6$cm.
(iii) Find the area of PQRS if area of $\Delta$PTS$=27c{m}^2$.

Answer 1

We know that

$TR-TS=TQ.TP$

$\Rightarrow \frac{TR}{TP}=\frac{TQ}{TS}$

Also $\angle RTQ=\angle STP$

$\Rightarrow △TPS\sim △TRQ$

ii) So, $\frac{TP}{TR}=\frac{SP}{RQ}$

$\Rightarrow \frac{18}{6}=\frac{SP}{4}$

$\Rightarrow SP=12cm$

iii) $\frac{ar\left(△PTS\right)}{ar\left(△QTR\right)}={\left(\frac{TP}{TR}\right)}^2={\left(\frac{18}{6}\right)}^2=9$

$\Rightarrow \frac{27}{ar\left(△QTR\right)}=9\Rightarrow ar\left(△QTR\right)=3c{m}^2$

$ar\left(PQRS\right)=ar\left(△PTS\right)-ar\left(△QTR\right)=27-3=24c{m}^2$