In a trapezium PQRS, PQ is parallel to RS and diagonals PR and QS intersect at O. If PQ = 4 cm. SR = 10 cm, then what is area(ΔPOQ) : area(ΔSOR) ?

4 : 25

In \(\Delta \)POQ and \(\Delta \)SOR

\(\angle\)RSO = \(\angle\)PQO (Alternate angles)

\(\angle\)SRO = \(\angle\)QPO (Alternate angles)

\(\Delta \)POQ is similar to  \(\Delta \)SOR

Now,

\(\frac{area\;of\;POQ}{area \;of\; SOR}\) = (\(\frac{PQ}{SR}\)) ²  

= (\(\frac{4}{10}\)) ²  = \(\frac{16}{100}\) = \(\frac{4}{25}\)

Therefore, required ratio 4 : 25.