Respuesta :
Question is Incomplete, Complete question is given below.
Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.
Answer:
∆ABC is right angled triangle with right angle at B.
Step-by-step explanation:
Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.
We need to prove that triangle is the right angled triangle.
Let the triangle be denoted by Δ ABC with side as;
AB = (a - 1) cm
BC = (2√ a) cm
CA = (a + 1) cm
Hence,
[tex]{AB}^2 = (a -1)^2[/tex]
Now We know that
[tex](a- b)^2 = a^2+b^2 - 2ab[/tex]
So;
[tex]{AB}^2= a^2 + 1^2 -2\times a \times1[/tex]
[tex]{AB}^2 = a^2 + 1 -2a[/tex]
Now;
[tex]{BC}^2 = (2\sqrt{a})^2= 4a[/tex]
Also;
[tex]{CA}^2 = (a + 1)^2[/tex]
Now We know that
[tex](a+ b)^2 = a^2+b^2 + 2ab[/tex]
[tex]{CA}^2= a^2 + 1^2 +2\times a \times1[/tex]
[tex]{CA}^2 = a^2 + 1 +2a[/tex]
[tex]{CA}^2 = AB^2 + BC^2[/tex]
[By Pythagoras theorem]
[tex]a^2 + 1 +2a = a^2 + 1 - 2a + 4a\\\\a^2 + 1 +2a= a^2 + 1 +2a[/tex]
Hence, [tex]{CA}^2 = AB^2 + BC^2[/tex]
Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.
This proves that ∆ABC is right angled triangle with right angle at B.
