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Tớ biết làm nè
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Biết làm cl í, tin người vcl:))
c. -Xét △ADC có: OM//DC (gt).
\(\Rightarrow\dfrac{MO}{DC}=\dfrac{AO}{AC}\) (định lí Ta-let).
\(\Rightarrow\dfrac{DC}{MO}=\dfrac{AC}{AO}\)
\(\Rightarrow\dfrac{DC}{OM}-1=\dfrac{OC}{AO}\) (1).
-Xét △BDC có: ON//DC (gt).
\(\Rightarrow\dfrac{ON}{DC}=\dfrac{BO}{BD}\) (định lí Ta-let).
\(\Rightarrow\dfrac{DC}{ON}=\dfrac{BD}{BO}\)
\(\Rightarrow\dfrac{DC}{ON}-1=\dfrac{OD}{BO}\)
-Xét △ABO có: AB//DC (gt).
\(\Rightarrow\dfrac{OD}{BO}=\dfrac{OC}{OA}=\dfrac{DC}{AB}\) (3)
-Từ (1), (2),(3) suy ra:
\(\dfrac{DC}{OM}-1=\dfrac{DC}{ON}-1=\dfrac{DC}{AB}\)
\(\Rightarrow\dfrac{DC}{OM}=\dfrac{DC}{ON}=\dfrac{DC}{AB}+1=\dfrac{AB+DC}{AB}\)
\(\Rightarrow\dfrac{1}{OM}=\dfrac{1}{ON}=\dfrac{AB+DC}{AB.DC}=\dfrac{1}{AB}+\dfrac{1}{CD}\)
a: Xét ΔAOB và ΔCOD có
\(\widehat{OAB}=\widehat{OCD}\)
\(\widehat{AOB}=\widehat{COD}\)
Do đó: ΔAOB∼ΔCOD
Suy ra: \(\dfrac{OA}{OC}=\dfrac{OB}{OD}=\dfrac{AB}{CD}\)
hay \(OA\cdot OD=OB\cdot OC\)
b: \(\dfrac{OA}{OC}=\dfrac{AB}{CD}\)
\(\Leftrightarrow OA=\dfrac{1}{2}\cdot6=3\left(cm\right)\)
Xét ΔOAB và ΔOCD có
\(\widehat{OAB}=\widehat{OCD}\)(hai góc so le trong, AB//CD)
\(\widehat{AOB}=\widehat{COD}\)
Do đó: ΔOAB đồng dạng với ΔOCD
=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}=\dfrac{AB}{CD}=\dfrac{6}{9}=\dfrac{2}{3}\)
\(\dfrac{OA}{OC}=\dfrac{2}{3}\)
=>\(OC=1,5OA\)
\(\dfrac{OB}{OD}=\dfrac{2}{3}\)
=>\(OD=3\cdot\dfrac{OB}{2}=1,5OB\)
AO+OC=AC
=>1,5OA+OA=OC
=>OC=2,5OA
=>\(\dfrac{OC}{OA}=2,5=\dfrac{5}{2}\)
=>\(\dfrac{OA}{OC}=\dfrac{2}{5}\)
OB+OD=BD
=>BD=1,5OB+OB=2,5OB
=>\(\dfrac{OB}{BD}=\dfrac{2}{5}\)
Xét ΔADC có MO//DC
nên \(\dfrac{MO}{DC}=\dfrac{AO}{AC}\)
=>\(\dfrac{MO}{9}=\dfrac{2}{5}=0,4\)
=>MO=0,4*9=3,6(cm)
Xét ΔBDC có ON//DC
nên \(\dfrac{ON}{DC}=\dfrac{BO}{BD}\)
=>\(\dfrac{ON}{9}=\dfrac{2}{5}\)
=>ON=0,4*9=3,6(cm)
MN=MO+ON
=3,6+3,6
=7,2(cm)
Tham khảo:
https://hoidapvietjack.com/q/685998/cho-hinh-thang-can-abcd-abcd-co-ab12cd-cho-ab6cm-bc5cm-a-tinh-chu-vi-hinh-thang-
cảm ơn e nha