Question

In $△$ ABC, AC$=5$, PC$=3$, and BP$=4\sqrt{3}$. What is the length of $\overline{AB}$?

Answer 1

Given

In $△$$APC$,

$AC$ $=$ $5$

$PC$ $=$ $3$

By Pythagoras theorem,

${AP}^2$ $+$ ${PC}^2$ $=$ ${AC}^2$

${AP}^2$ $+$ $3^2$ $=$ $5^2$

${AP}^2$ $+$ $9$ $=$ $25$

${AP}^2$ $=$ $25$ $-$ $9$

${AP}^2$ $=$ $16$

$AP$ $=$ $\sqrt{16}$

$AP$ $=$ $4$

In $△$$APB$,

$AP$ $=$ $4$

$BP$ $=$ $4\sqrt{3}$

To find $AB$,

By Pythagoras theorem,

${AP}^2$ $+$ ${BP}^2$ $=$ ${AB}^2$

${AB}^2$ $=$ $4^2$ $+$ ${\left(4\sqrt{3}\right)}^2$

${AB}^2$ $=$ $16$ $+$ $48$

${AB}^2$ $=$ $64$

$AB$ $=$ $\sqrt{64}$

$AB$ $=$ $8$

Therefore, the length of $\overline{AB}$ is ${}^{\prime }{8}^{\prime }$.