Question

In the adjoining figure, $ABC$ and $DBC$ are two triangles on the same base $BC,AL\perp BC$ and $DM\perp BC$. Then, $\frac{area\left(△ABC\right)}{area\left(△DBC\right)}$ is equal to:

• A. $\frac{AO}{OD}$
• B. $\frac{{AO}^2}{{OD}^2}$
• C. $\frac{AO}{AD}$
• D. $\frac{{OD}^2}{{AO}^2}$

Answer 1

The correct option is A $\frac{AO}{OD}$

Given: $AL\perp BC$ and $DM\perp BC$
In $△ALO$ and $△MOD$
$\angle AOL=\angle MOD$ (Vertically opposite angles)
$\angle ALO=\angle DMO$ (Each ${90}^{\circ }$)
$\angle LAO=\angle MDO$ (Third angle)
Thus, $△ALO\sim △DMO$ (AAA rule)
Hence, $\frac{AO}{OD}=\frac{AL}{DM}$ (Corresponding sides)...(I)
Area of triangle = $\frac{1}{2}base\times height$
$\frac{A\left(△ABC\right)}{A\left(△DBC\right)}=\frac{\frac{1}{2}BC\times AL}{\frac{1}{2}BC\times DM}$
$\frac{A\left(△ABC\right)}{A\left(△DBC\right)}=\frac{AL}{DM}$
$\frac{A\left(△ABC\right)}{A\left(△DBC\right)}=\frac{AO}{OD}$ (from I)