Question

In given triangle \[\Delta ABC\] if \[m\angle B = {90^0}\] and \[AB = BC\] then \[AC:BC = \_\_\_\_\_\_\_\]
A. \[\sqrt 2 :1\]
B. \[1:2\]
C. \[1:\sqrt 2 \]
D. \[1:3\]

Answer 1

Hint: In \[\Delta ABC\] since \[m\angle B = {90^0}\] and \[AB = BC\], it is a right-angled isosceles triangle. Drawing the diagram of the triangle it will give us a clear picture of what we have to find out. So, use this concept to reach the solution of the problem.

Complete step-by-step answer:
Given in \[\Delta ABC\], \[m\angle B = {90^0}\]and \[AB = BC\]
So, clearly it is a right-angled isosceles triangle.
Let \[AB = BC = x\]

From the diagram, according to the Pythagoras’s Theorem,
 \[ \Rightarrow A{C^2} = A{B^2} + B{C^2}\]
Since \[AB = BC = x\], we have
\[ \Rightarrow A{C^2} = {x^2} + {x^2} = 2{x^2}\]
Rooting on both sides, we have
\[
   \Rightarrow \sqrt {A{C^2}} = \sqrt {2{x^2}} \\
  \therefore AC = \sqrt 2 x \\
\]
Therefore, the required ratio is given by
\[
   \Rightarrow AC:BC = \sqrt 2 x:x \\
  \therefore AC:BC = \sqrt 2 :1 \\
\]
Thus, the correct option is A. \[\sqrt 2 :1\].

Note: A right angled isosceles triangle consists of two equal sides and the corresponding angle is \[{90^0}\] and the other angles are acute angles. Here we have not considered the negative value of \[AC\]since the length of the side of a triangle is always positive.