Angles $Q$ and $R$ of a $∆PQR$ are $25^{\circ}$ and $65^{\circ}$. Write which of the following is true:
$(i).\ PQ^2+QR^2=RP^2$
$(ii).\ PQ^2+RP^2=QR^2$
$(iii).\ RP^2+QR^2=PQ^2$

Given: Angles $Q$ and $R$ of a $∆PQR$ are $25^{\circ}$ and $65^{\circ}$.

To do: To write the truth of the following:

$(i).\ PQ^2+QR^2=RP^2$

$(ii).\ PQ^2+RP^2=QR^2$

$(iii).\ RP^2+QR^2=PQ^2$

Solution:


$∠PQR+∠QRP +∠RPQ = 180°$            [By angle sum property of a triangle]

$25^{\circ}+65^{\circ}+\angle RPQ=180^{\circ}$

$90^{\circ} +\angle RPQ =180^{\circ}$

$\angle RPQ = 180^{\circ}-90^{\circ}$

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

Thus $\Delta PQR$ is a right-angle triangle.

Hence according to Pythagoras Theorem $(ii)$ option $PQ^2+RP^2=QR^2$ is correct.

Updated on: 10-Oct-2022

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