Question

In a $\Delta ABC$, line CD is perpendicular to AB. $\angle CAD={45}^{\circ }$ and AB = 20 cm. The area of triangle ABC is $40c{m}^2$. The length of AC will be equal to

• A. 4 cm
• B. $4\sqrt{2}cm$
• C. 8 cm
• D. $8\sqrt{2}cm$

Answer 1

The correct option is B $4\sqrt{2}cm$

Given $CD\perp AB,\angle CDA={90}^{\circ }$

In $\Delta ADC,$

$sin{45}^{\circ }=\frac{CD}{AC}$

$\frac{1}{\sqrt{2}}=\frac{CD}{AC}$

$AC=\sqrt{2}CD$ ...(i)

Now, area of $△ABC=\frac{1}{2}\times AB\times CD$

$\Rightarrow 40c{m}^2=\frac{1}{2}\times 20\times CD$

$\Rightarrow CD=4cm$

On substituting the value of CD in equation (i), we get

$AC=\sqrt{2}\times 4=4\sqrt{2}cm$