Question:
In the given figure, O is a point inside a ∆PQR such that ∠POR = 90°, OP = 6 cm and OR = 8 cm. If PQ = 24 cm and QR = 26 cm, prove that ∆PQR is right-angled.
Solution:
Applying Pythagoras theorem in right-angled triangle POR, we have:
$\mathrm{PR}^{2}=\mathrm{PO}^{2}+\mathrm{OR}^{2}$
$\Rightarrow \mathrm{PR}^{2}=6^{2}+8^{2}=36+64=100$
$\Rightarrow \mathrm{PR}=\sqrt{100}=10 \mathrm{~cm}$
In ∆ PQR,
$\mathrm{PQ}^{2}+\mathrm{PR}^{2}=24^{2}+10^{2}=576+100=676$
and $\mathrm{QR}^{2}=26^{2}=676$
$\therefore \mathrm{PQ}^{2}+\mathrm{PR}^{2}=\mathrm{QR}^{2}$
Therefore, by applying Pythagoras theorem, we can say that ∆PQR is right-angled at P.