Question

In ΔABC, AB = BC = AC/√2. m∠B..,select a proper option (a), (b), (c) or (d) from given options so that the statement becomes correct.
(a) is acute
(b) is obtuse
(c) is right angle
(d) cannot be obtained

Answer 1

given, $AB=BC=\frac{AC}{\sqrt{2}}$
Let $AB=BC=\frac{AC}{\sqrt{2}}=x$
AB = x , BC = x and AC = √2x

here we can see that
AC² = (√2x )² = 2x²
AB² + BC² = x² + x² = 2x²
so, AB² + BC² = AC²
from Pythagoras theorem,
we can say that ∆ABC is right angle triangle where B is right angle.

option (c) correct.

Answer 2

Hi ,

In ∆ABC

AB = BC = AC/√2

AB² + BC²

= ( AC/√2 )² + ( AC/√2 )²

= AC²/2 + AC²/2

= AC²

In ∆ABC , square of one side is equal to

the sum of the squares of the other two

sides , then the angle opposite to the

first side is a right angle .

Therefore ,

∆ABC is right angle triangle .

I hope this helps you.

: )