In a $ \Delta \mathrm{PQR}, \mathrm{PR}^{2}-\mathrm{PQ}^{2}=\mathrm{QR}^{2} $ and $ \mathrm{M} $ is a point on side $ \mathrm{PR} $ such that $ \mathrm{QM} \perp \mathrm{PR} $.
Prove that $ \mathrm{QM}^{2}=\mathrm{PM} \times \mathrm{MR} $.

AcademicMathematicsNCERTClass 10

Given:

In a \( \Delta \mathrm{PQR}, \mathrm{PR}^{2}-\mathrm{PQ}^{2}=\mathrm{QR}^{2} \) and \( \mathrm{M} \) is a point on side \( \mathrm{PR} \) such that \( \mathrm{QM} \perp \mathrm{PR} \).

To do:

We have to prove that \( \mathrm{QM}^{2}=\mathrm{PM} \times \mathrm{MR} \).

Solution:


$\mathrm{PR}^{2}-\mathrm{PQ}^{2}=\mathrm{QR}^{2}$

$P R^{2}=P Q^{2}+QR^{2}$

This implies,

$\triangle P Q R$ is a right angled triangle with right angle at $Q$.

In $\triangle Q M R$ and $\triangle P M Q$,

$\angle M =\angle M=90^{\circ}$

$\angle M Q R =\angle Q P M=90^{\circ}-\angle R$

Therefore, by AA similarity,

$\triangle Q M R  \sim \triangle P M Q$

Using the property of area of similar triangles, we get, $\frac{\operatorname{ar}(\Delta Q M R)}{\operatorname{ar}(\Delta P M Q)} =\frac{(Q M)^{2}}{(P M)^{2}}$

$\frac{\frac{1}{2} \times R M \times Q M}{\frac{1}{2} \times P M \times Q M}=\frac{(Q M)^{2}}{(P M)^{2}}$

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

Hence proved. 

raja
Updated on 10-Oct-2022 13:28:00

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