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:
Answer & explanation
Correct answer: option 2
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.