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In fig. 1, S and T are points on the sides of PQ and PR, respectively of $\vartriangle$PQR, such that PT$=2$ cm, TR$=4$ cm and ST is parallel to QR. Find the ratio of the ratio of the area of $\vartriangle$PST and $\vartriangle$PQR.
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Given: In fig. 1 $\vartriangle$ PQR PT$=2$ cm, TR$=4$ cm and ST$\parallel$QR

To do: to find the ratio of area(ΔPST) and area(ΔPQR)

$\ ( \vartriangle PST) ∶area( \vartriangle PQR) =?$

Solution: 
As given $\displaystyle PT=2\ cm,\ TR=4\ cm\ $ and $ST\| QR$

$In\ \vartriangle PST\ and\ \vartriangle PQR$

$\angle PST=\angle PQR$

$\angle PTS=\angle PRQ$

$\angle P=\angle P$

$\Rightarrow \ ΔPST\sim ΔPQR\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( By\ AAA\ similarity)$

$\ \therefore \ \frac{PS}{PQ} =\frac{PT}{TR} =\frac{ST}{QR}$

$\ Also\ \ \frac{Area( ΔPST)}{Area( ΔPQR) \ } =\left(\frac{PS}{PQ}\right)^{2}$$=\left(\frac{PT}{TR}\right)^{2} =\left(\frac{ST}{QR}\right)^{2}$

$\therefore \frac{Area( ΔPST)}{Area( ΔPQR) \ }=\left(\frac{2}{4}\right)^{2}=\frac{1}{2}$

Hence the ratio of the areas of $\vartriangle PST$ and $\vartriangle PQR$  is 1:2.
 

Updated on: 10-Oct-2022

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