Points P and Q are on the sides AB and BC respectively of a triangle ABC, right angled at B. If AQ = 11 cm, PC = 8 cm, and AC = 13 cm, then find the length (in cm) of PQ.

From right angled triangle ABC,

$$\left(x+y\right)^2+\left(w+z\right)^2=13^2$$

$$\left(x+y\right)^2+\left(w+z\right)^2=169$$.........(1)

From right angled triangle ABQ,

$$\left(x+y\right)^2+w^2=11^2$$

$$\left(x+y\right)^2+w^2=121$$..........(2)

From right angled triangle PBC,

$$x^2+\left(w+z\right)^2=8^2$$

$$x^2+\left(w+z\right)^2=64$$...........(3)

Solving (2) + (3) - (1), we get

$$x^2+w^2=121+64-169$$

$$x^2+w^2=16$$.........(4)

From right angled triangle PBQ,

PB$$^2$$ + BQ$$^2$$ = PQ$$^2$$

$$x^2+w^2$$ = PQ$$^2$$

PQ$$^2$$ = 16

PQ = 4 cm

Hence, the correct answer is Option C