In the given figure, O is a point inside a $\triangle\text{PQR}$ such that $\angle\text{PQR}=90^\circ,\text{OP}=6\text{cm}$ and $\text{OR}=8\text{cm}.$ If $\text{PQ}=24\text{cm}$ and $\text{QR}=26\text{cm},$ prove that $\triangle\text{PQR}$ is right-angled.
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In $\triangle\text{PQR},\angle\text{QPR}=90^\circ,\text{PQ}=24\text{cm},$ and $\text{QR}=26\text{cm}^2$ In $\triangle\text{POR},\text{PO}=6\text{cm},\text{QR}=8\text{cm},$ and $\angle\text{POR}=90^\circ$
In $\triangle\text{POR},$ $\text{PR}^2=\text{PO}^2+\text{OR}^2$ $\text{PR}^2=(6^2+8^2)\text{cm}^2$ $=(36+64)\text{cm}^2=100\text{cm}^2$ $\text{PR}=\sqrt{100}\text{cm}=10\text{cm}$ In $\triangle\text{PQR},$ By Pythagoras theoram, we have $\text{QR}^2=\text{QP}^2+\text{PR}^2$ $(26)^2\text{cm}^2=\Big(24^2+10^2\Big)\text{cm}^2$ $676\text{cm}^2=(576+100)\text{cm}^2$ $676\text{cm}^2=676\text{cm}^2$ Hence, $\text{QR}^2=\text{QP}^2+\text{PR}^2$(sum of square of two sides equal to square of greatest side) Hence, $\triangle\text{PQR}$ is a right triangle which ois right angled at P.
In $\triangle\text{POR},$ $\text{PR}^2=\text{PO}^2+\text{OR}^2$ $\text{PR}^2=(6^2+8^2)\text{cm}^2$ $=(36+64)\text{cm}^2=100\text{cm}^2$ $\text{PR}=\sqrt{100}\text{cm}=10\text{cm}$ In $\triangle\text{PQR},$ By Pythagoras theoram, we have $\text{QR}^2=\text{QP}^2+\text{PR}^2$ $(26)^2\text{cm}^2=\Big(24^2+10^2\Big)\text{cm}^2$ $676\text{cm}^2=(576+100)\text{cm}^2$ $676\text{cm}^2=676\text{cm}^2$ Hence, $\text{QR}^2=\text{QP}^2+\text{PR}^2$(sum of square of two sides equal to square of greatest side) Hence, $\triangle\text{PQR}$ is a right triangle which ois right angled at P.
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