$\triangle\text{ABC}$ is an isosceles triangle with $AB = AC = 13\ cm$. The length of altitude from $A$ on $BC$ is $5\ cm$. Find $BC$.
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Given: $\triangle\text{ABC}$ is an isosceles triangle with $AB = AC = 13\ cm^2$ Const:
Draw altitude from A to BC $(\text{AL}\perp\text{BC} ).$
Now, $AL = 5\ cm$
In $\triangle\text{ALB},$ $\angle\text{ALB}=90^\circ$
$\therefore\text{AB}^2=\text{AL}^2+\text{BL}^2$(by pythagoras theorem)
$\therefore13^2=(5)^2+\text{BL}^2$ $(169-25)\text{cm}^2=\text{BL}^2$
$\text{BL}^2=144\text{cm}^2$ $\text{BL}=\sqrt{144}\text{cm}=12\text{cm}$ In $\triangle\text{ALC},$
$\angle\text{ALC}=90^\circ$ $\text{AC}^2=\text{AL}^2+\text{LC}^2$
$\Rightarrow\text{LC}^2=\Big(\text{AC}^2-\text{AL}^2\Big)$
$=\Big[(13)^2-(5)^2\Big]\text{cm}^2$ $=(169-25)\text{cm}^2$
$=144\text{cm}^2$
$=\sqrt{144}=12\text{cm}$ $\therefore\text{BC}=\text{BL}+\text{LC}$
$=(12+12)\text{cm}=24\text{cm}$
Draw altitude from A to BC $(\text{AL}\perp\text{BC} ).$
Now, $AL = 5\ cm$
In $\triangle\text{ALB},$ $\angle\text{ALB}=90^\circ$
$\therefore\text{AB}^2=\text{AL}^2+\text{BL}^2$(by pythagoras theorem)
$\therefore13^2=(5)^2+\text{BL}^2$ $(169-25)\text{cm}^2=\text{BL}^2$
$\text{BL}^2=144\text{cm}^2$ $\text{BL}=\sqrt{144}\text{cm}=12\text{cm}$ In $\triangle\text{ALC},$
$\angle\text{ALC}=90^\circ$ $\text{AC}^2=\text{AL}^2+\text{LC}^2$
$\Rightarrow\text{LC}^2=\Big(\text{AC}^2-\text{AL}^2\Big)$
$=\Big[(13)^2-(5)^2\Big]\text{cm}^2$ $=(169-25)\text{cm}^2$
$=144\text{cm}^2$
$=\sqrt{144}=12\text{cm}$ $\therefore\text{BC}=\text{BL}+\text{LC}$
$=(12+12)\text{cm}=24\text{cm}$
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