Question

✓ 50 Points

Q. Prove the following result:

" A triangle with sides that can be written in the form n^2 + 1 , n^2 -1 and 2n (where n > 1) is right angled"

Show, by means of a counter Example, that the converse is false.

✓ Well explained answer required.

Answer 1

According to Pythagoras theorem ,
any traingle ABC is a right angled traingle when all three sides AB , BC, CA, of traingle
Follow ,
$AB^2 = BC^2 + CA^2 \\ AC^2 = BC^2 + AB^2 \\ BC^2 = AC^2 + AB^2$
If we let A is right angle in ABC traingle,
then, $BC^2 = AC^2 + AB^2$
now, given three sides are given .
Is this follow Pythagoras triplet ?
Let be check ,
here, $n^2-1, n^2+1, 2n$
If we put n = 1 then, two sides are equal to zero
So, n > 1 put n =2
$n^2-1=(2)^2-1=3 \\ n^2+ 1=(2)^2+1=5 \\ 2n=2 (2)=4$
Hence, for all n > 1
$n^2+ 1$ is biggest sides,
Now, we can assume
BC = n^2+ 1
CA = n^2-1
AB = 2n
use, Pythagoras triplet ,
$(n^2+1)^2 = (n^2-1)^2+ (2n)^2 \\ n^4+2n^2+1=n^4-2n^2+1+4n^2 \\ n^4+2n^2+1=n^4+2n^2+1$
Hence , these sides, follow the Pythagoras triplet so, traingle must be right angle traingle .\\

Answer 2

hi mate,

Answer :

n^2+1 - 2n = (n-1)^2 if n>1

then (n-1)^2 >0

therefore n^2+1 > 2n

similarily (n^2+1) - (n^2-1) = 2

therefore n^2+1 > n^2-1

so n^2+1 > n^2-1.

i hope it helps you.