Question

Answer 1

VD1:

ÁP dụng hệ thức lượng cho `ΔABC` vuông tại `A`

`=> 1/(AH^2) = 1/(AB^2) + 1/(AC^2)`

`=> 1/(a^2) = 1/(AB^2) `\(+\dfrac{1}{\left(a\sqrt{2}\right)^2}\)

`=> 1/(AB^2) =1/(a^2)` \(-\dfrac{1}{2a}\)

`=>`\(\dfrac{1}{AB^2}=\dfrac{2}{2a^2}-\dfrac{1}{2a^2}=\dfrac{1}{2a^2}\)

`=> AB^2 = 2a^2`

`=>`\(AB=\sqrt{2a^2}=a\sqrt{2}\)

diện tích tam giác `ΔABC` là 

`\(a\sqrt{2}.a\sqrt{2}=2a^2\)

 

Answer 2

VD1 :

\(AC^2=AH^2+HC^2\Rightarrow HC^2=AC^2-AH^2=2a^2-a^2=a^2\Rightarrow HC=a\)

\(AC^2=HC.BC\Rightarrow BC=\dfrac{AC^2}{HC}=\dfrac{2a^2}{a}=2a\)

\(S_{ABC}=\dfrac{1}{2}.AH.BC=\dfrac{1}{2}.a.2a=a^2\left(đvdt\right)\)

Answer 3

VD2:

diện tích tam giác `ΔABC` là 

\(a.a\sqrt{2}=a^2\sqrt{2}\)

ÁP dụng hệ thức lượng cho `ΔABC` vuông tại `A`

\(\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{AH^2}\)

\(\dfrac{1}{a^2}+\dfrac{1}{\left(a\sqrt{2}\right)^2}=\dfrac{1}{AH^2}\)

\(\dfrac{1}{a^2}+\dfrac{1}{2a^2}=\dfrac{1}{AH^2}\)

\(\dfrac{1}{AH^2}=\dfrac{1}{a^2}+\dfrac{1}{2a^2}=\dfrac{2}{2a^2}+\dfrac{1}{2a^2}=\dfrac{3}{2a^2}\)

\(=>2a^2=3AH^2\)

\(=>AH^2=\dfrac{2a^2}{3}\)

`=> AH = \(\dfrac{a\sqrt{2}}{\sqrt{3}}\)

 

Answer 4

VD2 : 

a) \(S_{ABC}=\dfrac{1}{2}AB.AC=\dfrac{1}{2}.a.a\sqrt{2}=\dfrac{a^2\sqrt{2}}{2}\)

b) \(BC^2=AB^2+AC^2=a^2+2a^2=3a^2\Rightarrow BC=a\sqrt{3}\)

\(S_{ABC}=\dfrac{1}{2}AH.BC\Rightarrow AH=\dfrac{2S_{ABC}}{BC}=\dfrac{2.\dfrac{a^2\sqrt{2}}{2}}{a\sqrt{3}}=\dfrac{a\sqrt{6}}{3}\)