Question

ABC is an isosceles triangle with AC = BC. If $A{B}^2=2A{C}^2$, then:

• A. ABC is an acute angled triangle.
• B. $\angle ACB={60}^{\circ }$
• C. ABC is a right angled triangle
• D. $\angle ACB={90}^{\circ }$

Answer 1

Answer: (C), (D)

The correct options are
C ABC is a right angled triangle
D $\angle ACB={90}^{\circ }$

Given that $\Delta ABC$ is an isosceles triangle having AC = BC and $A{B}^2=2A{C}^2$
In $\Delta ACB$,
AC = BC (Given)
$A{B}^2=2A{C}^2$ (Given)
$A{B}^2=A{C}^2+A{C}^2$
$=A{C}^2+B{C}^2$ [Since, AC = BC]
Hence, by Pythagoras theorem, $\Delta ABC$ is a right angled triangle.