Question

In the given figure, $\angle$B = ${90}^{\circ }$, $XY||BC$, $AB=12cm$, $AY=8cm$ and $AX:XB=1:2$. Find the length of $AC$ and $BC$.

• A. $AC=12cm$ and $BC=10\sqrt{3}cm$.
• B. $AC=24cm$ and $BC=12\sqrt{3}cm$.
  
• C. $AC=19cm$ and $BC=10\sqrt{3}cm$.
• D. $AC=34cm$ and $BC=14\sqrt{3}cm$.

Answer 1

The correct option is B

$AC=24cm$ and $BC=12\sqrt{3}cm$.

Given: $XY\parallel BC$, $\angle B={90}^{\circ }$, $AX:XB=1:2$

And, $AB=12cm$, $AY=8cm$

Now,

$\frac{AX}{XB}=\frac{1}{2}$
$\Rightarrow \frac{AX}{AB-AX}=\frac{1}{2}$
$\Rightarrow 2AX=AB-AX$
$\Rightarrow 3AX=AB$

$\Rightarrow 3AX=12cm$
$\Rightarrow AX=4cm$
Now, In $△AXY$ and $△ABC$
$\angle XAY=\angle BAC$ (Common)
$\angle AXY=\angle ABC$ (Corresponding angles)
$\angle AYX=\angle ACB$ (Corresponding angles)
Thus, $△AXY\sim △ABC$ ($AAA$ rule)

Hence, $\frac{AX}{AB}=\frac{AY}{AC}$
$\Rightarrow \frac{4}{12}=\frac{8}{AC}$

$\Rightarrow AC=\frac{8\times 12}{4}$

$\Rightarrow AC=24$

Hence, $AC=24$ cm
Now, applying Pythagoras theorem in $△ABC$,
$A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow {24}^2={12}^2+B{C}^2$
$\Rightarrow B{C}^2={24}^2-{12}^2$
$\Rightarrow B{C}^2=432$

$\Rightarrow BC=\sqrt{432}$
$\Rightarrow BC=12\sqrt{3}$

Hence, $BC=12\sqrt{3}cm$