Question

In given fig ST parallel PS=3 and SR =4 cm find the ratio of the area of triangle PST to the area of triangle PQR

Answer 1

As ST║RQ, ΔPRQ~ΔPST (By basic proportionality theorem)
⇒$\frac{area of ΔPQR}{area of ΔPST}$=$\frac{PS^{2} }{PR ^{2} }$
$\frac{area of ΔPQR}{area of ΔPST}$=$\frac{3^{2} }{7 ^{2} }$
tex] \frac{area of ΔPQR}{area of ΔPST} [/tex]=$\frac{9 }{49} }$

Answer 2

Answer:

9:49

Step-by-step explanation:

Given

ST || RQ

PS= 3 cm

SR = 4cm

Proof :--

ar(∆PST) /ar(∆PRQ) = (PS)²/(PR)²

ar(∆PST) /ar(∆PRQ) = 3²/(PS+SR)²

ar(∆PST) /ar(∆PRQ) = 9/(3+4)²= 9/7² = 9/49

Hence, the required ratio ar(∆PST) :ar(∆PRQ) = 9:49