Selective Question

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

Steven Zheng Steven Zheng posted 3 weeks ago


Answer

  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 Steven Zheng posted 3 weeks ago

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