﻿ In △ABC , medians BE and CD are perpendicular to each other and intersect at

#### Multiple Choice Question (MCQ)

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 ?

1. ×

24

2. 64

3. ×

48

4. ×

32

Collected in the board: Triangle

Steven Zheng posted 2 years ago

1. S_{△DEZ} = \dfrac{1}{2}DZ\cdotp EZ

S_{△DBZ} = \dfrac{1}{2}DZ\cdotp BZ

\therefore 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

\therefore S_{△ABE} = S_{△CBE}

\because D is the midpoint of AB

\therefore S_{△AED} = S_{△BED} = \dfrac{1}{2} S_{△ABE}

\therefore S_{▱DECB} = 3S_{△BED} = 48

\therefore S_{△ABC} = 4S_{△BED} = 64

Steven Zheng posted 2 years ago

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