Answer:
option: C
Step-by-step explanation:
To find the side PQ ; we need to first find the length of the given line segment which is perpendicular to side PR; let name it as QS.
Now as ΔQSR is an right angled triangle.
and the length of the side QR and SR is given , so using Pythagorean theorem we have
![QR^{2}=SR^{2}+QS^{2}](https://tex.z-dn.net/?f=QR%5E%7B2%7D%3DSR%5E%7B2%7D%2BQS%5E%7B2%7D)
![5^{2}=3^{2}+QS^{2}\\ \\QS^{2}=5^{2}-3^{2}\\\\QS^{2}=25-3=16=4^2](https://tex.z-dn.net/?f=5%5E%7B2%7D%3D3%5E%7B2%7D%2BQS%5E%7B2%7D%5C%5C%20%5C%5CQS%5E%7B2%7D%3D5%5E%7B2%7D-3%5E%7B2%7D%5C%5C%5C%5CQS%5E%7B2%7D%3D25-3%3D16%3D4%5E2)
⇒ QS=4
Now again ΔQSP is an right angled triangle; so using Pythagorean theorem in ΔQSP we have
![PQ^{2}=QS^{2}+PS^{2}\\ \\PQ^{2}=4^{2}+6^{2}=16+36=52](https://tex.z-dn.net/?f=PQ%5E%7B2%7D%3DQS%5E%7B2%7D%2BPS%5E%7B2%7D%5C%5C%20%5C%5CPQ%5E%7B2%7D%3D4%5E%7B2%7D%2B6%5E%7B2%7D%3D16%2B36%3D52)
This means ![QS=\sqrt{52}=2\sqrt{13}](https://tex.z-dn.net/?f=QS%3D%5Csqrt%7B52%7D%3D2%5Csqrt%7B13%7D)
Hence, the length of side PQ is
.
Hence, option C is correct.