In Delta PQR,overline{QM} is an altitude. If | vedclass

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(N/A) Given: In $\Delta PQR$,$\overline{QM} \perp \overline{PR}$. Let $PM = x$. Then $PQ = 2x$ and $RM = 3x$.In right-angled $\Delta QMP$,by Pythagora

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(N/A) Given: In $\Delta PQR$,$\overline{QM} \perp \overline{PR}$. Let $PM = x$. Then $PQ = 2x$ and $RM = 3x$.
In right-angled $\Delta QMP$,by Pythagoras theorem: $PQ^2 = PM^2 + QM^2 \implies (2x)^2 = x^2 + QM^2 \implies 4x^2 = x^2 + QM^2 \implies QM^2 = 3x^2$.
In right-angled $\Delta QMR$,by Pythagoras theorem: $QR^2 = QM^2 + RM^2 \implies QR^2 = 3x^2 + (3x)^2 \implies QR^2 = 3x^2 + 9x^2 = 12x^2$.
Now,consider the sides of $\Delta PQR$: $PR = PM + RM = x + 3x = 4x$. So,$PR^2 = (4x)^2 = 16x^2$.
Check if $PQ^2 + QR^2 = PR^2$: $PQ^2 + QR^2 = (2x)^2 + 12x^2 = 4x^2 + 12x^2 = 16x^2$.
Since $PQ^2 + QR^2 = PR^2$,by the converse of Pythagoras theorem,$\Delta PQR$ is a right-angled triangle with $\angle PQR = 90^\circ$.

In $\Delta PQR$,$\overline{QM}$ is an altitude. If $PQ = 2PM$ and $RM = 3PM$,prove that $\Delta PQR$ is a right-angled triangle.

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