Question

In the given figure, ∠QPR = 90°, seg PM ⊥ seg QR and Q–M–R, PM = 10, QM = 8, find QR.

Answer 1

We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

Here, seg PM ⊥ seg QR

$$\begin{gathered} \therefore {\mathrm{PM}}^2=\mathrm{QM}\times \mathrm{MR} \\ \Rightarrow {10}^2=8\times \mathrm{MR} \\ \Rightarrow 100=8\times \mathrm{MR} \\ \Rightarrow \mathrm{MR}=\frac{100}{8} \\ \Rightarrow \mathrm{MR}=\frac{25}{2} \\ \mathrm{Now}, \\ \mathrm{QR}=\mathrm{QM}+\mathrm{MR} \\ =8+\frac{25}{2} \\ =\frac{16+25}{2} \\ =\frac{41}{2} \\ =20.5 \end{gathered}$$

Hence, QR = 20.5