Question

In figure 2.18, ∠QPR = 90°,seg PM⊥seg QR and Q-M-R,PM=10,QM=8, find QR.

Answer 1

Let the angle QPM = x
So,

$\tan(x) = \frac{8}{10} \\ \\ or. \: \: x = \tan ^{ - 1} ( \frac{4}{5} ) = 38.66$
So.. Angle MPR = (90 - x)

= (90 - 38.66) =51.34°

Now..
=> Tan (51.34°) = MR/PM

or, PM × tan (51.34°) = MR

$or. \: \: mr = 8 \times \tan(51 .34) \\ \\ or. \: \: mr = 8 \times 1.25 = 10$
Now.. QR = QM + MR =8 +10 =18 units
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Answer 2

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

∴ PM²=QM×MR

⇒10²=8×MR

⇒100=8×MR

⇒MR=100/8

⇒MR=25/2

Now,QR=QM+MR

=8+25/2

=16+25/2

=41/2

=20.5

Hence, QR = 20.5