Question

In a right-angled triangle PQR, If PQ2
= PR2
+ QR2, then the right angle is
a) <PRQ
b) <PQR
c) <RQP
d) <RPQ​
explain the answer​

Answer 1

Step-by-step explanation:

This is converse of Pythagoras theorem

We can prove this contradiction sum 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

By the Pythagorean theorem (AC)

2

=p

2

+r

2

But we know p

2

+r

2

=q

2

and q=PR

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

Answer 2

Answer:

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