In Delta ABC,mangle B = 90^circ and overline{ | vedclass

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(N/A) In $\Delta ABC$,$m\angle B = 90^\circ$.By Pythagoras theorem,$AC^2 = AB^2 + BC^2$.$AC^2 = 8^2 + 6^2 = 64 + 36 = 100$.$AC = 10$.The area of $\Del

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(N/A) In $\Delta ABC$,$m\angle B = 90^\circ$.
By Pythagoras theorem,$AC^2 = AB^2 + BC^2$.
$AC^2 = 8^2 + 6^2 = 64 + 36 = 100$.
$AC = 10$.
The area of $\Delta ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times AC \times BM$.
$\frac{1}{2} \times 8 \times 6 = \frac{1}{2} \times 10 \times BM$.
$48 = 10 \times BM \implies BM = 4.8$.
In $\Delta AMB$,$m\angle M = 90^\circ$. By Pythagoras theorem,$AB^2 = AM^2 + BM^2$.
$8^2 = AM^2 + (4.8)^2$.
$64 = AM^2 + 23.04$.
$AM^2 = 64 - 23.04 = 40.96$.
$AM = \sqrt{40.96} = 6.4$.
Since $AC = AM + CM$,$10 = 6.4 + CM$.
$CM = 10 - 6.4 = 3.6$.
Thus,$AM = 6.4$,$BM = 4.8$,and $CM = 3.6$.

In $\Delta ABC$,$m\angle B = 90^\circ$ and $\overline{BM}$ is an altitude. If $AB = 8$ and $BC = 6$,find $AM$,$BM$,and $CM$.

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