Đáp án:
b) $\,{P_{ABC}} = 2\sqrt {10} + 2\sqrt 5$
c) ${S_{ABC}} = 5$
d) $\widehat {ABC} = {45^0}$
Giải thích các bước giải:
\(\begin{array}{l}
\overrightarrow {AB} = \left( {3;1} \right);\,\overrightarrow {AC} = \left( {1; - 3} \right);\,\overrightarrow {BC} = \left( { - 2; - 4} \right)\\
a) \Rightarrow AB = AC = \sqrt {10} ;BC = 2\sqrt 5 ;\,\overrightarrow {AB} .\overrightarrow {AC} = 3.1 + 1.\left( { - 3} \right) = 0\\
\Rightarrow AB \bot AC\\
\Rightarrow \Delta ABC\,vuong\,can\,tai\,A\\
c){S_{ABC}} = \dfrac{1}{2}AB.BC = 5\\
AH \bot BC \Rightarrow H\,la\,trung\,diem\,BC\\
AH = \dfrac{{BC}}{2} = \dfrac{{2\sqrt 5 }}{2} = \sqrt 5 \\
b)\,{P_{ABC}} = AB + AC + BC = 2\sqrt {10} + 2\sqrt 5 \\
AM = \dfrac{{BC}}{2} = \dfrac{{2\sqrt 5 }}{2} = \sqrt 5 \,\\
N\left( {\dfrac{{ - 1}}{2};\dfrac{1}{2}} \right) \Rightarrow BN = \sqrt {{{\left( {\dfrac{5}{2}} \right)}^2} + {{\left( {\dfrac{5}{2}} \right)}^2}} = \dfrac{{5\sqrt 2 }}{2} = CK\\
d)\,\cos \widehat {ABC} = \dfrac{{\overrightarrow {BC} .\overrightarrow {BA} }}{{BC.AB}} = \dfrac{1}{{\sqrt 2 }} \Rightarrow \widehat {ABC} = {45^0}
\end{array}\)
Em làm tương tự hai góc kia nhé.
