In the right-angled triangle $P Q R,$ if $∠ Q = 90^{o}$
then $P Q^{2} + Q R^{2} =$ _________.

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Solution

In a right-angled triangle $P Q R,$
$P R$ is hypotenuse
$P Q$ is perpendicular
$Q R$ is base
Pythagoras' theorem states that, $I n a r i g h t - a n g l e d t r i a n g l e, t h e s q u a r e o f t h e h y p o t e n u s e s i d e i s e q u a l t o t h e s u m o f s q u a r e o f t h e o t h e r t w o s i d e s .$
Hence, $P Q^{2} + Q R^{2} = P R^{2}$

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In the right-angled triangle PQR,if ∠Q=90othen PQ2+QR2=_________.

In a right-angled triangle PQR,PR is hypotenusePQ is perpendicularQR is basePythagoras' theorem states that,Inaright-angledtriangle,thesquareofthehypotenusesideisequaltothesumofsquareoftheothertwosides.Hence, PQ2+QR2=PR2

In the right-angled triangle $P Q R,$ if $∠ Q = 90^{o}$
then $P Q^{2} + Q R^{2} =$ _________.