Question

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Let ABC be a triangle such that AB=(x –1) cm, AC=2√x cm, BC=(x +1) cm. Then
(a) A = 90°
(b) B=90°
(c) C = 90°
(d) none of these​

Answer 1

Given :

•Let ABC be a triangle such that :

•AB=(x-1)cm

•AC=2√x cm

•BC=(x+1)cm

To Find :

• The angle between them =?

Solution :

Let ABC be a triangle of sides

•AB=(x-1)cm

•AC=2√x cm

• BC=(x+1)cm

Here, BC > AB

By using Pythagoras theorem :

$\\ \implies\sf{(AB)^{2}+(AC)^{2} }$

$\\ \implies\sf{(x-1)^{2}+(2 \sqrt{x} )^{2} }$

$\\ \implies\sf{x^{2}+1-2x+4x}$

$\\ \implies\sf{x^{2}+1+2x}$

$\\ \implies\sf{(x+1)^{2} }$

$\\ \implies\boxed{\sf{BC^{2} }}$

$\\\bf{\therefore \:( BC)^{2} =(AB)^{2}+(AC)^{2} }$

Here, BC is a hypotenuse

In triangle ABC , $\sf{\angle A}$ is a right angled triangle.

$\bf{\therefore \: \angle A=90 \degree}$

The required options is (a)