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ΔMNP vuông tại M
=>\(\widehat{MNP}+\widehat{P}=90^0\)
=>\(\widehat{N}=90^0-45^0=45^0\)
Xét ΔMNP vuông tại M có \(tanP=\dfrac{MN}{MP}\)
=>\(\dfrac{10}{MP}=tan45=1\)
=>MP=10(cm)
ΔMNP vuông tại M
=>\(MN^2+MP^2=NP^2\)
=>\(NP=\sqrt{10^2+10^2}=10\sqrt{2}\left(cm\right)\)
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a.
Ta có: \(\widehat{BAE}+\widehat{BAC}+\widehat{CAF}=180^0\)
\(\Rightarrow\widehat{BAE}+90^0+\widehat{CAF}=180^0\)
\(\Rightarrow\widehat{BAE}+\widehat{CAF}=90^0\) (1)
Lại có \(BE\perp d\Rightarrow\Delta BAE\) vuông tại E
\(\Rightarrow\widehat{BAE}+\widehat{ABE}=90^0\) (2)
(1);(2) \(\Rightarrow\widehat{CAF}=\widehat{ABE}\)
Xét hai tam giác ABE và CAF có:
\(\left\{{}\begin{matrix}\widehat{ABE}=\widehat{CAF}\\\widehat{AEB}=\widehat{CFA}=90^0\end{matrix}\right.\)
\(\Rightarrow\Delta ABE\sim\Delta CAF\left(g.g\right)\)
\(\Rightarrow\dfrac{AE}{CF}=\dfrac{BE}{AF}\Rightarrow AE.AF=BE.CF\)
b.
\(S_{ABC}=\dfrac{1}{2}AB.AC\Rightarrow AC=\dfrac{2S_{ABC}}{AB}=\dfrac{2.24}{6}=8\left(cm\right)\)
Áp dụng hệ thức lượng:
\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\Rightarrow AH=\dfrac{AB.AC}{\sqrt{AB^2+AC^2}}=\dfrac{6.8}{\sqrt{6^2+8^2}}=4,8\left(cm\right)\)


\(=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\left(\sqrt{3}-1\right)}=\sqrt{2}\left(\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\right)\)
\(=\sqrt{2}\left(\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\right)\)
\(=\sqrt{2}\left(\dfrac{6}{9-3}\right)=\sqrt{2}\)