Đáp án:
b) ${S_{ABC}} = 5$
c) $\,{P_{ABC}} = 2\sqrt {10} + 2\sqrt 5$
d) $\widehat {JAB} \approx {94^0}$
Giải thích các bước giải:
\(\begin{array}{l}
\overrightarrow {AB} = \left( { - 1; - 3} \right);\,\overrightarrow {AC} = \left( {2; - 4} \right);\,\overrightarrow {BC} = \left( {3; - 1} \right)\\
a) \Rightarrow AB = BC = \sqrt {10} ;AC = 2\sqrt 5 ;\,\overrightarrow {AB} .\overrightarrow {BC} = \left( { - 1} \right).3 + \left( { - 3} \right).\left( { - 1} \right) = 0\\
\Rightarrow AB \bot BC\\
\Rightarrow \Delta ABC\,vuong\,can\,tai\,B\\
b){S_{ABC}} = \dfrac{1}{2}AB.BC = 5\\
BH \bot AC \Rightarrow H\,la\,trung\,diem\,AC\\
BH = \dfrac{{AC}}{2} = \dfrac{{2\sqrt 5 }}{2} = \sqrt 5 \\
c)\,{P_{ABC}} = AB + AC + BC = 2\sqrt {10} + 2\sqrt 5 \\
BK = \dfrac{{AC}}{2} = \dfrac{{2\sqrt 5 }}{2} = \sqrt 5 \,\\
d)\,\cos \widehat {JAB} = \dfrac{{\overrightarrow {AJ} .\overrightarrow {AB} }}{{AJ.AB}} = \dfrac{{ - 2}}{{\sqrt {970} }} \Rightarrow \widehat {JAB} \approx {94^0}
\end{array}\)
