# Ex.6.4 Q3 Triangles Solution - NCERT Maths Class 10

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## Question

In Fig. below, $$ABC$$ and $$DBC$$ are two triangles on the same base $$BC.$$ If $$AD$$ intersects $$BC$$ at $$O,$$ show that \begin{align}\frac{{\rm {Area\, of\,}}{\Delta ABC}}{{\rm {Area\,of\,}}{\Delta DBC}} = \frac{{AO}}{{DO}}\end{align}

Diagram

## Text Solution

Reasoning:

AA criterion

Steps:

In $$\,\Delta ABC$$

Draw $$\,AM \bot BC$$

In$$\,\,\Delta DBC$$

Draw $$\,DN \bot BC$$

Now in $$\,\,\Delta AOM,\,\,\Delta DON$$

\begin{align}{\angle AMO}&={\angle D N O}={90^{\circ}} \\ {\angle A O M}&={\angle D O N(\text { Vertically opposite angles })} \\ {\Rightarrow} \qquad {\Delta A O M} &\sim {\Delta D O N \text { (AA criterion) }} \\ {\Rightarrow \qquad\frac{A M}{D N}}&={\frac{O M}{O N}=\frac{A O}{D O} \ldots (1)}\end{align}

Now,

\begin{align} Area\,\,of\Delta ABC&=\frac{1}{2}\times base\times height \\ & =\frac{1}{2}\times BC\times AM \\ Area\,\,of\,\Delta DBC& =\frac{1}{2}\times BC\times DN \\\frac{Area\,of\,\Delta ABC}{Area\,of\,\Delta DBC}&=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times BC\times DN}\\\frac{Area\,of\,\Delta ABC}{Area\,of\,\Delta DBC}&=\frac{AM}{DN}\\\frac{Area\,of\,\Delta ABC}{Area\,of\,\Delta DBC}&=\frac{AO}{DO} \quad (\text{from}\,\text{(1)}) \end{align}

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