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

In ∆PQR, ∠QPR = 90° and [Given]

seg PM ⊥ seg QR

∴ PM^{2} = OM × MR [by Theorem of geometric mean]

∴ 10^{2} = 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.``” `

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