In triangle PQR, points E and F are on sides PQ and PR respectively such that EF is parallel to QR. If PE = 2 cm and EQ = 3 cm, then area(ΔPQR) : area(ΔPEF) is equal to:

25 : 4

According to the question

PQ = (PE + EQ)

= 2 + 3

= 5 cm

From the basic proportionality theorem,

= \(\frac{PE}{PQ}\) = \(\frac{PF}{PR}\)

and, \(\angle\)PEF = \(\angle\)PQR   (Corresponding angle)

Therefore, \(\Delta \)PEF is similar \(\Delta \)PQR

Now,

\( { PQ}^{ 2} \)/\( { PE}^{ 2} \) = \( { 5}^{ 2} \)/\( { 2}^{ 2} \)

= \(\frac{25}{4}\)

Now,

\(\frac{area\;of\; PQR}{area \;of\;PEF}\) = \( { PQ}^{ 2} \)/\( { PE}^{ 2} \)

= \(\frac{25}{4}\)

Therefore, the required ratio is 25 : 4.