Let's go
Bài 1
đặt \(S_{ABO}=S_1\left(S_1>0\right);S_{CDO}=S_2\left(S_2>0\right);S_{BOC}=S_3\left(S_3>0\right);S_{ADO}=S_4\left(S_4>0\right)\)
Ta có
\(\left\{{}\begin{matrix}S_1+S_3=S_1+S_4=2,5\left(cm^2\right)\\S_2+S_3=S_2+S_4=5\left(cm^2\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2S_1+S_3+S_4=5\left(cm^2\right)\\2S_2+S_3+S_4=10\left(cm^2\right)\end{matrix}\right.\) \(\Rightarrow2\left(S_2-S_1\right)=5\left(cm^2\right)\) (*)
mặt khác ta chứng minh được \(\Delta AOB\text{ đồng dạng}\Delta COD\)
suy ra \(\dfrac{S_1}{S_2}=\left(\dfrac{AB}{CD}\right)^2=\dfrac{1}{4}\) (**)
(*);(**) \(\Rightarrow S_1=\dfrac{5}{6}\left(cm^2\right);S_2=\dfrac{10}{3}\left(cm^2\right);S_3=S_4=\dfrac{5}{3}\left(cm^2\right)\)
kẻ CF, DK vuông góc với AB
đặt AK=x(x>0) thì AF=4-x
ta có
\(\left(4-x\right)^2+2,5^2=6^2\)
\(\left[{}\begin{matrix}x=\dfrac{8+\sqrt{119}}{2}\left(n\right)\\x=\dfrac{8-\sqrt{119}}{2}\left(l\right)\end{matrix}\right.\)
vậy \(BD=\sqrt{\left(x+2\right)^2+2,5^2}\approx11,72cm\)
next
Bài 2: mình lười làm bài hình lắm nên mình tua qua luôn
finally
Bài 3: