KVPY SA Exam 05-Nov-2017 Question Paper

$\mathit{C}\mathit{M}||\mathit{N}\mathit{D}$ and $\mathit{C}\mathit{N}||\mathit{M}\mathit{D}$
Therefore CMDN is a rectangle
$\Rightarrow \mathit{C}\mathit{M}=\mathit{N}\mathit{D}=4$ and $\mathit{C}\mathit{N}=\mathit{M}\mathit{D}=5$
$\ln ∆\mathit{C}\mathit{M}\mathit{D}$
By Pythagoras theorem
${\mathit{C}\mathit{D}}^2={\mathit{C}\mathit{M}}^2+{\mathit{M}\mathit{D}}^2$
$\mathit{C}\mathit{D}=\sqrt{41}$
$∆\mathit{A}\mathit{M}\mathit{D}~∆\mathit{D}\mathit{N}\mathit{B}$
$\frac{\mathit{y}}{5}=\frac{4}{\mathit{x}}=\frac{\mathit{z}}{\mathit{t}}$
$\mathit{x}\mathit{y}=20$.....(i)
In ∆BCD
${\left(\mathit{y}+5\right)}^2=41+{\mathit{z}}^2$.....(ii)
In ∆ADC
${\left(\mathit{x}+4\right)}^2=41+{\mathit{t}}^2$.....(iii)
⇒ $\frac{{\left(\mathit{y}+5\right)}^2-41}{{\left(\mathit{x}+4\right)}^2-41}={\left(\frac{\mathit{z}}{\mathit{t}}\right)}^2$
⇒ $\frac{{\left(\frac{20}{\mathit{x}}+5\right)}^2-41}{{\left(\mathit{x}+4\right)}^2-41}={\left(\frac{4}{\mathit{x}}\right)}^2$ (use (i))
⇒ $\frac{{\left(20+5\mathit{x}\right)}^2-41{\mathit{x}}^2}{{\left(\mathit{x}+4\right)}^2-41}=16$
⇒ $\frac{400+25{\mathit{x}}^2+200\mathit{x}-41{\mathit{x}}^2}{{\mathit{x}}^2+16+8\mathit{x}-41}=16$
⇒ $\mathit{–}16{\mathit{x}}^2+200\mathit{x}+400$ $=16{\mathit{x}}^2+128\mathit{x}-400$
⇒ $32{\mathit{x}}^2-72\mathit{x}-800=0$
⇒ $4{\mathit{x}}^2-9\mathit{x}-100=0$
$\mathit{x}=\frac{25}{4}$
so $\mathit{y}=\frac{20\times 4}{25}=\frac{16}{5}$
Hence AC = x + 4 = 41/4 and BC = 5 + y = 41/5