In $$ △ABC $$ , medians $$ BE $$ and $$ CD $$ are perpendicular to each other and intersect at $$ Z $$ . If $$ BE = 8 $$ cm, $$ CD = 12 $$ cm, what is the area of $$ △ABC $$ ?

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Uniteasy Admin · Accepted Answer

$$ S_{△DEZ} = \dfrac{1}{2}DZ\cdotp EZ $$
$$ S_{△DBZ} = \dfrac{1}{2}DZ\cdotp BZ $$
$$ 	herefore    S_{△DBE} = S_{△DEZ}  + S_{△DBZ}  = \dfrac{1}{2}DZ(EZ+BZ) =  \dfrac{1}{2}DZ\cdotp BE $$

similarly, 

$$ S_{△CEZ} = \dfrac{1}{2}CZ\cdotp EZ $$

$$ S_{△CBZ} = \dfrac{1}{2}CZ\cdotp BZ $$

$$  S_{△CBE} = S_{△CEZ}  + S_{△CBZ}  = \dfrac{1}{2}CZ(EZ+BZ) =  \dfrac{1}{2}CZ\cdotp BE $$

$$ ∴ S_{▱DECB} = S_{△DBE}+  S_{△CBE}  = \dfrac{1}{2}(DZ+CZ)BE =  \dfrac{1}{2}CD\cdotp BE = 48 $$

$$ \because E $$ is the midpoint of $$ AC $$

$$ 	herefore   S_{△ABE} = S_{△CBE} $$

$$ \because D $$ is the midpoint of $$ AB $$

$$ 	herefore   S_{△AED} = S_{△BED} = \dfrac{1}{2}  S_{△ABE} $$

$$ 	herefore   S_{▱DECB} = 3S_{△BED} = 48 $$
$$ 	herefore  S_{△ABC} = 4S_{△BED} = 64 $$

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