Question
In a right triangle $ABC$, $\angle$$B = 90^\circ$. If $AB = 6\ cm, BC = 8\ cm$, find $AC$.
✓
Answer
In the right triangle $ABC$,
$\because $ $\angle$$B = 90^\circ . . . .$ [given]
$\therefore$ By Pythagoras theorem
$AC^2= AB^2+ BC^2$
$\therefore$ $AC^2= 6^2+ 8^2$
$\therefore$ $AC^2= 36 + 64$
$\therefore$ $AC^2= 100$
$\therefore$ AC = $\sqrt {100} $
Therefore, $\sqrt {100} = 10$.
Hence, $AC$ is equal to $10\ cm$.
$\because $ $\angle$$B = 90^\circ . . . .$ [given]
$\therefore$ By Pythagoras theorem
$AC^2= AB^2+ BC^2$
$\therefore$ $AC^2= 6^2+ 8^2$
$\therefore$ $AC^2= 36 + 64$
$\therefore$ $AC^2= 100$
$\therefore$ AC = $\sqrt {100} $
Therefore, $\sqrt {100} = 10$.
Hence, $AC$ is equal to $10\ cm$.
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