Question

in ∆ PQR if PQ2=PR2+QR2 state with reason whether ∆ PQR is a right angled triangle or not​

Answer 1

In ∆ PQR,

Base is PQ

Perpendicular is PR

Hypotenuse is QR

So, by Pythagoras Theorem,

if (base)²+(perpendicular)²=(hypotenuse)²

The triangle will always be a right-angled triangle.

But here, we can see that it is given,

(perpendicular)² + (hypotenuse)² = (base)²

∴ ∆PQR is not right-angled.

Hope it helps! Please mark brainliest!

Answer 2

This is converse of Pythagoras theorem We can prove this contradiction sum q

2

=p

2

+r

2

in APQR while triangle is not a rightangle Now consider another triangle AABC we construct AABC AB=qCB=b and C is a Right angle

By the Pythagorean theorem (AC)

2

=p

2

tr

2 N +

But we know p

2

+r

2

=q

2

and q=PR

So (AB)

2

=p

2

2

=(SR)

2

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

AC=PQ

All the these sides AABC are congruent to APQR

So they are congruent by sss theorem