Question

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

Answer 1

In ∆PQR, ∠QPR = 90° and [Given] 

seg PM ⊥ seg QR

∴ PM² = OM × MR [Theorem of geometric mean] 

∴ 10² = 8 × MR

∴ MR = \(\frac{100}{8}\)

= 12.5

Now, QR = QM + MR [Q – M – R] 

= 8 + 12.5 

∴ QR = 20.5 units