Question

In the figure below, $\angle PSR={{90}^{\circ }},PQ=10cm,QS=6cm\text{ and }RQ=9cm.$ Calculate the length of PR.

Answer 1

Hint: To find the value of PR, we will first consider the $\Delta PQS$ . Since the triangle is a right-angled triangle, we will use the Pythagoras theorem. From this, we will get the side PS. Hence, $RS=RQ+QS$ . Now, consider triangle PRS and apply Pythagoras theorem. Through this, we will get side PR.

Complete step-by-step solution

We have to find the value of PR. It is given that $\angle PSR={{90}^{\circ }}$ . Hence the triangles PRS and PQS are right-angled triangles.
Let us first consider the $\Delta PQS$. The sides PQ and QS are given. So we will use the Pythagoras theorem to find the side PS.
Pythagoras theorem states that the square of the largest side of a triangle will be equal to the sum of squares of the other two sides.
$\Rightarrow P{{Q}^{2}}=P{{S}^{2}}+Q{{S}^{2}}...(i)$
It is given that $PQ=10cm,QS=6cm\text{ }$
Let us substitute these values in equation (i). We will get
${{10}^{2}}=P{{S}^{2}}+{{6}^{2}}$
We can write this as
$P{{S}^{2}}=100-36=64$
Now, let us take the square root.
$PS=8cm$
Now, let us consider the triangle PRS. We can apply Pythagoras theorem here.
$\Rightarrow P{{R}^{2}}=R{{S}^{2}}+P{{S}^{2}}...(ii)$
It is given that $QS=6cm\text{ and }RQ=9cm.$
Hence, $RS=RQ+QS$
Let us now substitute the values. We will get
$RS=6+9=15cm$
Now, we can substitute these values in equation (ii).
We will get
$P{{R}^{2}}={{15}^{2}}+{{8}^{2}}$
Now let us solve this. We will get
$P{{R}^{2}}=225+64=289$
Let us take the square root. We get
$PR=17cm$
Hence, the value of $PR=17cm$.

Note: Pythagoras theorem must be thorough to solve these types of problems. Pythagoras theorem can be applied only when a triangle is a right-angled triangle. You may make an error in the Pythagoras theorem as $P{{Q}^{2}}=P{{S}^{2}}-Q{{S}^{2}}$. In the right-angle triangle, the largest side to be the hypotenuse.