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**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.

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