Question

Giúp với ạ, mình cần gấp

Answer 1

a) Vì \(AB\parallel CD\) \(\Rightarrow\angle HDC=\angle DAB=\angle KBC\) \((AD\parallel BC)\)

Xét \(\Delta CBK\) và \(\Delta CDH:\) Ta có: \(\left\{{}\begin{matrix}\angle CBK=\angle CDH\\\angle CKB=\angle CHD=90\end{matrix}\right.\)

\(\Rightarrow\Delta CBK\sim\Delta CDH\left(g-g\right)\Rightarrow\dfrac{CK}{CH}=\dfrac{CB}{CD}=\dfrac{CB}{AB}\)

Ta có: \(\angle KCH=\angle BCH+\angle KCD-\angle BCD=90+90-\angle BCD\)

\(=180-\angle BCD=\angle ABC\)

Xét \(\Delta CKH\) và \(\Delta BCA:\) Ta có: \(\left\{{}\begin{matrix}\angle KCH=\angle ABC\\\dfrac{CK}{CH}=\dfrac{CB}{AB}\end{matrix}\right.\)

\(\Rightarrow\Delta CKH\sim\Delta BCA\left(c-g-c\right)\)

b) Vì \(\Delta CKH\sim\Delta BCA\left(c-g-c\right)\Rightarrow\dfrac{HK}{AC}=\dfrac{CK}{CB}=sinCBK\)

mà \(\angle CBK=\angle BAD\Rightarrow\dfrac{HK}{AC}=sinBAD\Rightarrow HK=AC.sinBAD\)

c) \(\angle ABC=120\Rightarrow\angle CBK=60\Rightarrow sinCBK=sin60\)

\(\Rightarrow\dfrac{CK}{CB}=\dfrac{\sqrt{3}}{2}\Rightarrow CK=CB.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{2}.10=5\sqrt{3}\)

Ta có: \(cosCBK=\dfrac{BK}{BC}\Rightarrow cos60=\dfrac{BK}{BC}\Rightarrow BK=\dfrac{1}{2}.BC=\dfrac{1}{2}.10\)

\(=5\left(cm\right)\)

Lại có: \(\left\{{}\begin{matrix}sinCDH=sin60=\dfrac{CH}{CD}\\cosCDH=cos60=\dfrac{HD}{CD}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}CH=\dfrac{\sqrt{3}}{2}.8=4\sqrt{3}\left(cm\right)\\HD=\dfrac{1}{2}.8=4\left(cm\right)\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}AH=AD+DH=10+4=14\left(cm\right)\\AK=AB+BK=8+5=13\left(cm\right)\end{matrix}\right.\)

Ta có: \(S_{AKCH}=S_{AKC}+S_{AHC}=\dfrac{1}{2}.AH.CH+\dfrac{1}{2}.AK.KC\)

\(=\dfrac{1}{2}.14.4\sqrt{3}+\dfrac{1}{2}.13.5\sqrt{3}=\dfrac{121}{2}\sqrt{3}\)