Madhya Pradesh Police Exam Model Paper 6

Given:
$\mathit{A}\mathit{B}+\mathit{B}\mathit{C}=14\mathit{c}\mathit{m}$
$\mathit{A}\mathit{C}+\mathit{B}\mathit{D}=20\mathit{c}\mathit{m}$
Formula:
Perimeter of the rectangle $=2\left(\mathit{l}+\mathit{b}\right)$
Rectangle has two equal diagonals.
(Diagonal)2 = (Length)2 + (Breadth)2
Calculation:
AC and BD are two diagonals of rectangle, then
$\mathit{A}\mathit{C}+\mathit{B}\mathit{D}=20$
$\Rightarrow 2\mathit{A}\mathit{C}=20\left[\because \mathit{A}\mathit{C}=\mathit{B}\mathit{D}\right]$
$\Rightarrow \mathit{A}\mathit{C}=20/2$
$\Rightarrow \mathit{A}\mathit{C}=10$
Let AB = x, then
$\mathit{B}\mathit{C}=\left(14\mathit{–}\mathit{x}\right)$
$\mathit{A}{\mathit{C}}^2=\mathit{A}{\mathit{B}}^2+\mathit{B}{\mathit{C}}^2$
$\Rightarrow 102={\mathit{x}}^2+{\left(14\mathit{–}\mathit{x}\right)}^2$
$\Rightarrow 100={\mathit{x}}^2+196+{\mathit{x}}^2\mathit{–}28\mathit{x}$
$\Rightarrow 100=2{\mathit{x}}^2+196\mathit{–}28\mathit{x}$
$\Rightarrow 2{\mathit{x}}^2\mathit{–}28\mathit{x}+96=0$
$\Rightarrow {\mathit{x}}^2\mathit{–}14\mathit{x}+48=0$
$\Rightarrow {\mathit{x}}^2\mathit{–}8\mathit{x}\mathit{–}6\mathit{x}+48=0$
$\Rightarrow \mathit{x}\left(\mathit{x}\mathit{–}8\right)\mathit{–}6\left(\mathit{x}\mathit{–}8\right)=0$
$\Rightarrow \left(\mathit{x}\mathit{–}6\right)\left(\mathit{x}\mathit{–}8\right)=0$
Taking,
$\left(\mathit{x}\mathit{–}6\right)=0$
$\Rightarrow \mathit{x}=6$ (not in option)
Taking,
$\left(\mathit{x}\mathit{–}8\right)=0$
$\Rightarrow \mathit{x}=8$
Hence, BC = 8 cm.