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

As we know the sum of the angles of any triangle is always 180. So,

$\angle P + \angle Q + \angle R = 180^0$

$\angle P + 25^0 + 65^0 = 180^0$

$\angle P = 180^0- 25^0 - 65^0$

$\angle P = 90^0$

Now. Since PQR is a right-angled triangle with right angle at P. So

$(Hypotenus)^2=(Base)^2+(Perpendicular)^2$

$(QR)^2=(PQ)^2+(RP)^2$

$\small PQ^2+RP^2=QR^2$

Hence option (ii) is correct.

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