Xavier Aptitude Test 2014 Paper

(d) Let the breadth and length of the rectangle ABCD be3x and y respectively.
∴ 3xy = 90 ⇒ xy = 30V is midpoint of WR. PW | | EV ⇒ EV PW / 2 = =$\frac{\mathit{x}}{4}$
Similarly, FV = WQ/2 =$\frac{\mathit{x}}{4}$
∴ EF = EV + VF =$\frac{\mathit{x}}{2}$
$\because \mathrm{\Delta }\mathit{M}\mathit{P}\mathit{A}\sim \mathit{\Delta }\mathit{M}\mathit{E}\mathit{V}$
$\therefore$ Height of $\mathrm{\Delta }\mathit{M}\mathit{P}\mathit{A}$ with respect to base AP: Height (h_ ${}_1^{})$ of $\mathit{\Delta }\mathit{M}\mathit{E}\mathit{V}$ with respect to base
$\mathrm{}\mathit{E}\mathit{V}=\mathrm{}\mathit{A}\mathit{P}:\mathit{E}\mathit{V}=\mathrm{}\mathit{x}:\frac{\mathit{x}}{4}=4:1$
Hence, if height of $\mathrm{\Delta }\mathit{M}\mathit{P}\mathit{A}=4\mathrm{}\mathit{k},$ then height $\mathrm{}\left(\mathrm{}{\mathit{h}}_1\mathrm{}\right)$ of$\mathrm{\Delta }\mathit{M}\mathit{E}\mathit{V}=\mathrm{}\mathit{k}$
$\mathrm{}\therefore 4\mathrm{}\mathit{k}+\mathrm{}\mathit{k}=4\mathit{k}=\mathrm{}\frac{\mathit{y}}{2}$
$\therefore \mathit{k}=\frac{\mathit{y}}{10}$
$\therefore$ Height $\left(\mathrm{}{\mathit{h}}_1\mathrm{}\right)$ of $\mathrm{\Delta }\mathit{M}\mathit{E}\mathit{V}=\mathrm{}\frac{\mathit{y}}{10}$
Area $\left(\mathrm{\Delta }\mathit{M}\mathit{E}\mathit{V}\right)=\mathrm{}\frac{1}{2}\times \mathit{E}\mathit{V}\times {\mathit{h}}_1$
$\mathrm{}=\mathrm{}\frac{1}{2}\times \frac{\mathrm{}\mathit{x}}{4}\times \frac{\mathrm{}\mathit{y}}{10}=\frac{30}{80}=0.375$
Also, $\mathrm{\Delta }\mathit{V}\mathit{F}\mathit{N}\sim \mathit{\Delta }\mathit{C}\mathit{R}\mathit{N}$
∴ Height $({\mathit{h}}_2$) ofΔVFN with respect to base VF : Height of ΔCRN with respect to base CR = VF : CR = x/4 : 3x/ 2 = 1:6
Hence, if height (${\mathit{h}}_2$) of ΔVFN = m, then height of ΔCRN= 6 m
∴ m + 6m = 7m = y/2
∴ m = y/14
∴ Height (${\mathit{h}}_2$) of ΔVFN = y/14
$\mathrm{}\mathit{A}\mathit{r}\mathit{e}\mathit{a}\left(\mathrm{\Delta }\mathit{V}\mathit{F}\mathit{N}\right)=\mathrm{}\frac{1}{2}\times \mathit{F}\mathit{V}\times {\mathit{h}}_2$
$\mathrm{}=\frac{1}{2}\times \frac{\mathrm{}\mathit{x}}{4}\times \frac{\mathrm{}\mathit{y}}{14}=\mathrm{}\frac{30}{112}\approx 0.27$
Area $\left(\mathrm{}\square \mathit{P}\mathit{Q}\mathit{F}\mathit{E}\right)=\mathrm{}\frac{1}{2}\times \left(\mathit{P}\mathit{Q}+\mathrm{}\mathit{E}\mathit{F}\right)\times \frac{\mathrm{}\mathit{y}}{2}$
$=\mathrm{}\frac{1}{2}\times \left(\mathrm{}\mathit{x}+\mathrm{}\frac{\mathrm{}\mathit{x}}{2}\mathrm{}\right)\times \frac{\mathrm{}\mathit{y}}{2}=\mathrm{}\frac{3\mathit{x}\mathit{y}}{8}=\frac{90}{8}$
$=11.25$
$=11.25-0.375+0.27\approx 11.145$