hình học

D

duong8b3_hp

N

ngocbich74

mình tính theo S của ABCD hử ?

+Ta có $S_{ABCD}$=BC.DC (1)

a,$S_{BDE}$=$\dfrac{DE.BC}{2}$=$\dfrac{\dfrac{DC}{2}.BC}{2}$

=$\dfrac{DC.BC}{4}$ =$\dfrac{S_{ABCD}}{4}$

b,+Ta có $S_{HEC}$=$\dfrac{HC.EC}{2}=\dfrac{\dfrac{BC}{2}. \dfrac{DC}{2}}{2}$

$=\dfrac{DC.BC}{8}=\dfrac{S_{ABCD}}{8}$

+Ta có $S_{ICK}$=$\dfrac{IC.KC}{2}=\dfrac{\dfrac{BC}{4}. \dfrac{DC}{4}}
{2}$

$=\dfrac{S_{ABCD}}{32}$

\Rightarrow$S_{EHIK}$=$S_{HEC}-S_{IKC}=\dfrac{S_{ABCD}}{8}-\dfrac{S_{ABCD}}
{32}$

$=\dfrac{3S_{ABCD}}{32}$


 
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