Maths / Areas Of Parallelograms And Triangles / Median and Altitude of Triangle

QUESTION

In the given figure, AD is the median of $\dpi{120} \fn_jvn \large \Delta ACD,\;BE$ is the median $\dpi{120} \fn_jvn \large \Delta ABD\;and\;AO$ is the median of $\dpi{120} \fn_jvn \large \Delta ABE.$

Show that $\dpi{120} \fn_jvn \large ar( AOB)=\frac {1}{8}ar(ABC).$

EXPLANATION
Explain TypeExplanation Content
Text

$\therefore \: \: \: \: \: \: \: \: \: \: \: ar(\Delta ABD)=ar(ADC)$

Also, BE is a median

$\therefore \: \: \: \: \: \: \: \: \: \: \: ar(\Delta ABO)=ar(AEO)$

$\therefore \: \: \: \: \: \: \: \: \: \: \: ar(\Delta ABO)=ar(ADC)$

Now,

$ar(ABC)=ar(ABD)+ar(ADC)$

$=2(ar(ABD))$

$=2(ar(ABE)+ar(BED))$

$=2(2ar(ABE))$

$=2(ar(ABO)+ar(AOE))$

$=4(2ar(ABO))$

$ar(ABC)=8ar(ABD)$

$\Rightarrow$         $ar(AOB)=\frac {1}{8}ar(ABC).$

Hence proved

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