Question
In a right triangle $ABC, ÐB = 90^\circ $. If $AC = 13 \ cm, BC = 5 \ cm$, find $AB$
✓
Answer
$\triangle\text{ABC}$is right−angled at $B.$
Therefore, by applying pythagoras theoram, we obtain
$ A C^2=A B^2+B C^2 $
$ (13 \mathrm{~cm})^2=(A B)^2+(5 \mathrm{~cm})^2 $
$ A B^2=\left(13 \mathrm{~cm}^2\right)-(5 \mathrm{~cm})^2=(169-25) \mathrm{cm}^2 $
$ =144 \mathrm{~cm}^2 $
$\text{AB}=(\sqrt{144})\text{cm} = (\sqrt{12\times12})\text{cm}$
$\text{AB} = 12 \text{cm} $
Therefore, by applying pythagoras theoram, we obtain
$ A C^2=A B^2+B C^2 $
$ (13 \mathrm{~cm})^2=(A B)^2+(5 \mathrm{~cm})^2 $
$ A B^2=\left(13 \mathrm{~cm}^2\right)-(5 \mathrm{~cm})^2=(169-25) \mathrm{cm}^2 $
$ =144 \mathrm{~cm}^2 $
$\text{AB}=(\sqrt{144})\text{cm} = (\sqrt{12\times12})\text{cm}$
$\text{AB} = 12 \text{cm} $
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