In ∆ PQR, PQ2=PR2+QR2 then state which angle Will be the right angle
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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
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