$P Q R$ is a triangle right angled at $P$ and $M$ is a point on $Q R$ such that $P M ⊥ Q R$ show that $P M^{2} = Q M . M R$ .

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Solution

If a perpendicular is drawn from the vertex of the right angle to the
hypotenuse then triangle on high side of the perpendicular are same.
So, $Δ P M R \sim Δ Q M P$
If two triangles are similar then the ratio is
$\begin{matrix} \frac{P M}{Q M} = \frac{M R}{M P} P M \times M P = M R \times Q M P M = M R \times Q M \end{matrix}$

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Similar questions

$P Q R$ is a triangle right angled at $P$ and $M$ is a point on $Q R$ such that $P M ⊥ Q R$ .

Show that $P M^{2} = Q M . M R$

P M ⊥ Q R

P M^{2} = Q M \times M R

P M ⊥ Q R

P M^{2} = Q M \times M R

S

T

P R

Q R

△ P Q R

∠ P = ∠ R T S

△ R P Q \sim △ R T S

Δ

P M^{2} + R N^{2} = 5 M N^{2}

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