In right triangle ABC, with right angle C, AB = 9 and BC = 5, how do you find the length of the missing side?

Hint: Draw a rough diagram of a right-angle triangle ABC right angled at C. Now, assume the side BC as base, AB as hypotenuse and AC as the perpendicular of the triangle. Apply the Pythagoras theorem given as: - \[{{h}^{2}}={{p}^{2}}+{{b}^{2}}\], where h = hypotenuse, p = perpendicular and b = base and substitute the values of h and b and calculate the value of p to get the answer. Use h = 9 and b = 5.

Complete answer:
Here, we have been provided with a right-angle triangle ABC which is right-angled at C. The length of AB is 9 units and that of BC is 5 units. We are asked to find the length of the missing side, that is AC.
Now, let us draw a diagram of the given situation. So, we have,
 
In the above figure we have drawn a right-angle triangle ABC right angled at C. We have considered BC as the base whose length is 5 units and AB as the hypotenuse whose length is 9 units. AC is the perpendicular whose length is unknown.
Now, we know that we can apply Pythagoras theorem in a right-angle triangle to determine the unknown side if the length of the other two sides are given. So, we have,
In \[\Delta ABC\],
AC = perpendicular = p
BC = base = b
AB = hypotenuse = h
Applying Pythagoras theorem given as: - \[{{h}^{2}}={{p}^{2}}+{{b}^{2}}\], we have,
\[\begin{align}
  & \Rightarrow {{9}^{2}}={{5}^{2}}+{{p}^{2}} \\
 & \Rightarrow {{p}^{2}}={{9}^{2}}-{{5}^{2}} \\
 & \Rightarrow {{p}^{2}}=81-25 \\
\end{align}\]
\[\Rightarrow {{p}^{2}}=56\]
Taking square root both the sides, we get,
\[\Rightarrow p=\sqrt{56}\]
Now, we can write \[56=2\times 2\times 2\times 7\] as the product of its primes. So, we have,
\[\Rightarrow p=\sqrt{2\times 2\times 2\times 7}\]
\[\Rightarrow p=2\sqrt{14}\] units
Hence, the length of the missing side is \[2\sqrt{14}\] units.

Note: One must remember the Pythagoras theorem to solve the above question. Note that here the unit of length is not given to us, so we considered it as ‘units’. This question can also be solved using trigonometry. What we can do is we will use the relation: - \[\cos \theta =\dfrac{b}{h}\] to calculate \[\cos \theta \] in the figure. Using this we will calculate \[\sin \theta \] by applying the identity: - \[\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }\] and then apply the relation: - \[\sin \theta =\dfrac{p}{h}\] to get the value of p.