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

QUESTION

E is the mid - point of median AD. Show that ar$\dpi{120} \fn_jvn \large (\Delta BED)=\frac{1}{4}\: ar(\Delta ABC)$.

EXPLANATION
Explain TypeExplanation Content
Text

$\therefore$  ar($\Delta$ABD) = ar($\Delta$ADC)                [By theorem]

Or     $ar(\Delta ABD)=\frac{1}{2}\: ar(\Delta ABC)$    ......(1)

E is mid point of AD.

$\therefore$   BE is median.

$\therefore$  ar($\Delta$BDE) = $\frac{1}{2}$  ar($\Delta$ABD)       ....(2)

From (1) and (2),

$ar(\Delta BDE)=\frac{1}{2}\left ( \frac{1}{2}ar(\Delta ABC)\right )\\\\ ar(\Delta BDE)=\frac{1}{4}ar(\Delta ABC)$

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