PQR is a triangle right-angled at P and M is a point on QR such that $P M ⊥ Q R$ . Show that $P M^{2} = Q M \times M R$ [3 MARKS]

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Solution

Figure: 1 Mark
Solution: 1 Mark
Proof: 1 Mark

The given figure shows $Δ P Q R$ right-angled at P. We draw PM perpendicular to QR.
Since in a right triangle, the perpendicular drawn from the right angle to the hypotenuse, divides the triangle into two similar triangles,

$∴ Δ P M Q \sim P M R$
$\Rightarrow \frac{A r e a o f Δ P M Q}{A r e a o f Δ P M R} = \frac{P M^{2}}{R M^{2}}$
$\Rightarrow \frac{\frac{1}{2} Q M \times P M}{\frac{1}{2} R M \times P M} = \frac{P M^{2}}{R M^{2}} \Rightarrow \frac{Q M}{R M} = \frac{P M^{2}}{R M^{2}} \Rightarrow P M^{2} = \frac{Q M \times R M^{2}}{R M} = Q M \times R M$
Hence, proved.

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