Question

In the given figure, $ABC$ is a right angled triangle with $\angle BAC={90}^o$.
Find the ratio of the area of $△ADB$ is to area of $△CDA$.

Answer 1

b'Area of right angled triangle is given by,

$\frac{1}{2}\times base\times height$

Area of triangle ADB= $\frac{1}{2}\times DB\times AD=\frac{1}{2}\times 18\times AD$

Area of triangle CDA= $\frac{1}{2}\times CD\times AD=\frac{1}{2}\times 8\times AD$

So, the ratio will be,

$\begin{array}{c}\frac{\frac{1}{2}\times 18\times AD}{\frac{1}{2}\times 8\times AD}=\frac{18}{8}=\frac{9}{4}\end{array}$

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