Question

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

Answer 1

In ΔPQR,

`{:(∠"QPR" = 90°),("seg PM ⊥ seg QR"):}        }"Given"`

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.

∴ PM² = QM × MR       ...(Theorem of geometric mean)

∴ 10² = 8 × MR

∴ 100 = 8 × MR      

∴  MR = `100/8`

 ∴  MR = 12.5

Now,

QR = QM + MR        ...(Q-M-R)

∴ QR = 8 + 12.5

∴ QR = 20.5