Determine whether the triangle having sides (a − 1) cm,
Question:

Determine whether the triangle having sides $(a-1) \mathrm{cm}, 2 \sqrt{a} \mathrm{~cm}$ and $(a+1) \mathrm{cm}$ is a right angled triangle.

Solution:

Let

$A=(a-1)$

$B=\sqrt{2} a$

 

$C=(a+1)$

Larger side is $C=(a+1)$

We know that any number plus 1 is always greater than that number minus 1 and product of 2 and its square root.

For example : If a = 36

$a-1=35$

$a+1=37$

$\sqrt{2} a=12$

If $a=5$

$a-1=4$

 

$a+1=6$

$\sqrt{2} a=4.47$

In order to prove that the given sides forms a right angled triangle we have to prove that $A^{2}+B^{2}=C^{2}$.

Let us solve the left hand side first.

$A^{2}+B^{2}=(a-1)^{2}+(\sqrt{2} a)^{2}$

$=a^{2}-2 a+1+4 a$

$=a^{2}+2 a+1$

Now we will simplify the right hand side as shown below,

$C^{2}=(a+1)^{2}$

$=a^{2}+2 a+1$

We can see that left hand side is equal to right hand side.

Therefore, the given sides determined the right angled triangle.

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