Ex.6.5 Q2 Triangles Solution - NCERT Maths Class 10

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\({PQR}\) is a triangle right angled at \(P\) and \(M\) is a point on \(QR\) such that \(PM \bot QR\). Show that \({{(\text{PM})}^{2}}\) = \({QM. MR}\)


Text Solution


As we know if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.


In \(\,\Delta PQR\,;\,\angle QPR={{90}^{0}}\) and \(PM\bot QR\)

In \(\,\Delta PQR\,\) and \(\,\Delta MQP\,\)

\(\angle Q P R=\angle Q M P=90^{\circ}\)

 \(\angle P Q R=\angle M Q P \) (Common Angles)

\(\begin{align} {\Rightarrow} \quad \Delta P Q R \sim \Delta M Q P\end{align}\) (AA Criterion)......-(1)

In \(\Delta PQR\) and \(\Delta MPR\)

\(\angle QMP=\angle PMR={{90}^{0}}\)

\(PRQ= RPM\) (Common Angle)

\(\begin{align}\Rightarrow\quad \,\Delta \,PQR&\sim{\ }\Delta \,MPR\end{align}\)  (AA Criterion).........(2)

From (1) and (2)

\(\Delta MQP\sim{\ }\Delta MPR\)

\(\begin{align}\frac{P M}{M R}&=\frac{Q M}{P M} \text{[Comparing sides opposite to equal angles]}\\&{\Rightarrow\quad P M^{2}=Q M \cdot M R}\end{align}\)


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