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

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Last updated date: 23rd Jul 2024
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Answer
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Hint: We have to apply Pythagoras theorem in triangle APC and find the side AP. Pythagoras theorem states the square of the hypotenuse is equal to the sum of the square of the other two sides. Similarly, we have to apply Pythagoras theorem in triangle APB, substitute the values of AP and BP in the equation and find the side AB.

Complete step by step solution:
We have to find the length of the side AB. We can see from the figure that angle APC is a right-angle, that is, $\angle APC=90{}^\circ $ . Therefore, we can apply Pythagoras theorem in $\Delta APC$ . We know that for a right-angled triangle, Pythagoras theorem states the square of the hypotenuse is equal to the sum of the square of the other two sides. Therefore, in triangle APC, we can apply Pythagoras theorem. The hypotenuse of $\Delta APC$ is AC.
$\Rightarrow A{{C}^{2}}=P{{C}^{2}}+A{{P}^{2}}$
We are given that $AC=5\text{ units and }PC=3\text{ units}$ . We have to substitute these values in the above equation.
$\begin{align}
  & \Rightarrow {{5}^{2}}=A{{P}^{2}}+{{3}^{2}} \\
 & \Rightarrow 25=A{{P}^{2}}+9 \\
\end{align}$
Let us take 9 to the LHS from the RHS.
$\begin{align}
  & \Rightarrow 25-9=A{{P}^{2}} \\
 & \Rightarrow A{{P}^{2}}=16 \\
\end{align}$
We have to take the square root on both sides.
$\begin{align}
  & \Rightarrow AP=\sqrt{16} \\
 & \Rightarrow AP=4\text{ units}...\left( i \right) \\
\end{align}$
From the given figure, we can see that $\angle APB=90{}^\circ $ . We have to apply Pythagoras theorem in the $\Delta APB$ . From the figure, we can infer that AB is the hypotenuse.
$\Rightarrow A{{B}^{2}}=A{{P}^{2}}+B{{P}^{2}}$
We are given that $BP=4\sqrt{3}$ . Let us substitute this value and (i) in the above equation.
$\Rightarrow A{{B}^{2}}={{4}^{2}}+{{\left( 4\sqrt{3} \right)}^{2}}$
We know that ${{\left( \sqrt{a} \right)}^{2}}=a$ . Therefore, the above equation becomes
$\begin{align}
  & \Rightarrow A{{B}^{2}}=16+\left( 16\times 3 \right) \\
 & \Rightarrow A{{B}^{2}}=16+48 \\
 & \Rightarrow A{{B}^{2}}=64 \\
\end{align}$
We have to take the square root on both sides.
$\begin{align}
  & \Rightarrow AB=\sqrt{64} \\
 & \Rightarrow AB=8\text{ units} \\
\end{align}$
Therefore, the length of $\overline{AB}$ is $8$ units.

Note: Students must note that they can only apply Pythagoras theorem on a triangle if the triangle is a right-angled one. They have a chance of making a mistake by writing Pythagoras theorem as the square of the hypotenuse is equal to the difference of the square of the other two sides. They must never forget to write the units at the end even if the unit is not specified in the question.