CAT Exam Model Paper 7 with solutions for free online practice

FG || BC, DE||AB and IH || AC
AsFP || BE and BF||EP, FBEP is a parallelogram
Similarly, ADPI & PGCH are also parallelograms

∆DPG, ∆IFP and ∆PEH are similar to ∆ABC.
If the area (in sq. cm) of ∆DPG, ∆IFP and ∆PEH are 1, 9 and 25 respectively then we can say their corresponding sides are in the ratio 1 : 3 : 5. Let the lengths (in units) be x, 3x and 5x for the sides PG, FP and EH respectively.
Also BC = BE + EH + HC = FP + EH + PG
BC = 3x + 5x + x = 9x
∆DPG is similar to a ∆ABC and the ratio of the areas of similar triangles is equal to the ratio of the squares of their corresponding sides.
So $\frac{\mathrm{Area}\left(∆\mathrm{DPG}\right)}{\mathrm{Area}\left(∆\mathrm{DBC}\right)}=\frac{{\left(\mathit{x}\right)}^2}{{\left(9\mathit{x}\right)}^2}$
$\Rightarrow \frac{1}{\mathrm{Area}\left(∆\mathrm{ABC}\right)}={\left(\frac{1}{9}\right)}^2$
⇒ Area (∆ABC) = 81 sq. cm