Question

(D) Given :ln Delta PQR,PQ^ 2 +QR^ 2 =PR^ 2 AABC is constructed such that PQ = AB , QR = BC , and angle B=90^ Prove that: triangle PQR is a right-angled triangle.​

Answer 1

Answer:

This is converse of Pythagoras theorem

We can prove this contradiction sum

$\sf \: q {}^{2} =p {}^{2} +r {}^{2}$

in ΔPQR while triangle is not a rightangle

Now consider another triangle ΔABC we construct ΔABC AB=qCB=b and C is a Right angle

$\boxed{ \sf \: By \: the \: Pythagorean \: theorem \: (AC){}^{2} =p{}^{2}+r{}^{2}}$

$\boxed{ \sf \: But \: we \: know \:  p {}^{2} +r {}^{2} =q {}^{2}  and q=PR}$

$\boxed{ \sf \: So (AB) {}^{2} =p {}^{2} +r {}^{2} =(SR) {}^{2}}$

Since PQ and AB are length of sides we can take positive square roots

AC=PQ

All the these sides ΔABC are congruent to ΔPQR

So they are congruent by sss theorem