Question

Let ABC be a triangle such that AB= (x - 1) cm, AC = 2Vx cm, BC = (x +1) cm. Then :
​

Answer 1

$In \:\triangle ABC , \\AB = (x-1) \:cm , \\AC = \frac{2}{x} \:cm , \\BC = (x+1) \:cm$

$AB^{2} = (x-1)^{2} \\= x^{2} - 2x + 1 \: --(1)$

$BC^{2} = (x+1)^{2} \\= x^{2} + 2x + 1 \: --(2)$

$CA^{2} = \big( \frac{2}{x} \big)^{2} \\= \frac{4}{x^{2}}\: --(3)$

$AB^{2} + BC^{2} \\= x^{2}-2x+1 + x^{2}+2x+1 \\= 2x^{2} + 2 \\\neq CA^{2}$

$BC^{2} + CA^{2} \\= x^{2} + 2x + 1 + \frac{4}{x^{2}}\\\neq AB^{2}$

$CA^{2} + AB^{2} \\= \frac{4}{x^{2}} + x^{2} - 2x + 1 \\\neq BC^{2}$

$\underline { \pink{Pythagoras \: Theorem :}}$

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides .

$Here, AB^{2} + BC^{2} \neq AC^{2}$

$BC^{2} + CA^{2} \neq AB^{2}$

$CA^{2} + AB^{2} \neq BC^{2}$

Therefore.,

$\triangle ABC \:is \:not \: a \:right \:angled \: triangle .$

$\red { None \:of \:these }$

•••♪