Question

$AO:OB$

• A. $1:2$
• B. $3:2$
• C. $3:5$
• D. $3:8$

Answer 1

Answer: (B)

$3:2$

Given, $AD$ is a median of triangle $ABC$ and $O$ is a point on $AB$ such that area of $△AOD$ : area of $△ADC$ is $3:5$.
In $△ABC$, $AD$ is median.

So, median divides triangle into two triangle of equal area.
$\therefore$ Area of $△ADC=$ Area of $△ABD$
Given, area of $△AOD:$ Area of $△ADC=$ $3:5$,
Area of $△AOD:$ Area of $△ABD=$ $3:5$ (As Area of $△ADC=$ Area of $△ABD$ ) ....(1)
Drawing a perpendicular $DF$ from $D$ to $AB$.
Area of $△ABD=\frac{1}{2}\times$ base $\times$ altitude $=\frac{1}{2}\times AB\times DF$
Area of $△AOD=\frac{1}{2}\times$ base $\times$ altitude $=\frac{1}{2}\times AO\times DF$
From (1)
$\Rightarrow \frac{Areaof△AOD}{Areaof△ABD}=\frac{3}{5}\Rightarrow \frac{\frac{1}{2}\times AO\times DF}{\frac{1}{2}\times AB\times DF}=\frac{3}{5}\Rightarrow \frac{AO}{AB}=\frac{3}{5}\Rightarrow \frac{AO}{AO+OB}=\frac{3}{5}(As,AB=AO+OB)\Rightarrow 3\times (AO+OB)=5\times AO\Rightarrow 5AO-3AO=3OB\Rightarrow 2AO=3OB\Rightarrow \frac{AO}{OB}=\frac{3}{2}=3:2$