Question 12 Marks
If the sides of the triangle are in the ratio $1: \sqrt{2}: 1,$ show that is a right $-$ angled triangle.
Answer
View full question & answer→Let, the sides of the triangle be, $x: \sqrt{ 2} x$ and $x$.
Now,
$x^2+x^2=2 x^2=(\sqrt{2} x)^2 ....(i)$
Here $,$ in $(i)$ it is shown that a square of one side of the given triangle is equal to the addition of square of the other two sides.
This is nothing but Pythagoras theorem which states that in a right $-$ angled triangle$,$ the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Therefore $,$ the given triangle is a right $-$ angled triangle.
Now,
$x^2+x^2=2 x^2=(\sqrt{2} x)^2 ....(i)$
Here $,$ in $(i)$ it is shown that a square of one side of the given triangle is equal to the addition of square of the other two sides.
This is nothing but Pythagoras theorem which states that in a right $-$ angled triangle$,$ the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Therefore $,$ the given triangle is a right $-$ angled triangle.