Question

ABC is an isosceles right angle triangle, right angled at B. The ratio of the sides AB: BC : AC is .

• A. $\sqrt{2}:1:1$
• B. $1:2:\sqrt{3}$
• C. $\sqrt{3}:2:1$
• D. $1:1:\sqrt{2}$

Answer 1

The correct option is D $1:1:\sqrt{2}$


Let us assume the length of side AB of the triangle to be x units.
Applying trignometric ratios to the sides, we get :
$sin{45}^{\circ }=\left(\frac{x}{AC}\right)\Rightarrow \frac{1}{\sqrt{2}}=\left(\frac{x}{AC}\right)\Rightarrow AC=x\sqrt{2}...\left(i\right)Similarly,tan{45}^{\circ }=\left(\frac{x}{BC}\right)\Rightarrow 1=\left(\frac{x}{BC}\right)\Rightarrow BC=x...\left(ii\right)$
So, the ratios of the sides of the triangle with angles ${45}^{\circ },{45}^{\circ }\&{90}^{\circ }=x:BC:AC$
$=x:x:x\sqrt{2}\left(from\right(i\left)\&\right(ii\left)\right)$
$=1:1:\sqrt{2}$

An alternate and shortcut method of solving this question is:

For the given triangle, as two angles are equal; the two sides opposite to these angles will also be equal.
And as the third angle is ${90}^{\circ }$, the triangle is right - angled triangle.
Let us assume the length of the equal sides is equal to x units.
So, length of the hypotenuse = $\sqrt{(x^2+x^2)}$ = $x\sqrt{2}$
So, Ratio of the sides of the triangle = x : x : $x\sqrt{2}$
$\Rightarrow$ Ratio of the sides of the triangle = 1 : 1 : $\sqrt{2}$