Question

$P$ is a point inside the $\Delta ABC$ . Three straight lines parallel to $AB,BC$ and $AC$ are drawn which divides the triangle into six regions. If the area of three triangles formed in these regions are $9,16$ and $25$ square units, then the area (in sq. units) of $\Delta ABC$ is

• A. $81$
• B. $256$
• C. $144$
• D. $64$

Answer 1

Answer: (C)

$\because \Delta ABC,\Delta TUP,\Delta PDE,\Delta SPR$ all are similar
$\therefore$ $ratioofareas={\left(ratioofsides\right)}^2$
Let, $PT=\alpha =AS$
$\therefore$ $\frac{SR}{PT}=\frac{4}{5}\Rightarrow SR=\frac{4\alpha }{5}$
$\frac{PE}{PT}=\frac{3}{5}\Rightarrow PE=\frac{3\alpha }{5}$
$\Rightarrow AC=AS+SR+RC$
$=\alpha +\frac{4\alpha }{5}+\frac{3\alpha }{5}$
$=\frac{12\alpha }{5}$
$\therefore \frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PDE\right)}=\frac{{\left(AC\right)}^2}{{\left(PE\right)}^2}=\frac{12^2\cdot {\left(\alpha \right)}^2}{5^2\cdot \left(\frac{3^2{\left(\alpha \right)}^2}{5^2}\right)}$
Hence, $ar\left(\Delta ABC\right)=16ar\left(\Delta PDE\right)=144sq.units$