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 [by Theorem of geometric mean]
∴ 10² = 8 × MR
∴ MR = \(100\over8\)
= 12.5
Now, QR = QM + MR [Q – M – R]
= 8 + 12.5
∴ QR = 20.5 units

 

{no need to write just for explaination  Theorem of geometric mean }
Statement- “ the geometric mean of the two segments equals the altitude.”