# prove that a triangle with sides 2n, n^2-1 and n^2+1 is right angled. continued?

a) prove prove that a triangle with sides 2n, n^2-1 and n^2+1 is right angled.

b) we can generate Pythagorean triplets by substituting various values of n in the above triplets. draw a table for values of n from 2 to 12.

c) Prove that a triangle with sides 4n, n^2-4 and n^2=4 is right angled.

d) Draw a table like in (b)

Relevance
• 10 years ago

a)

2n, n^2 -1, n^2 + 1

to prove it is a right triangle, we must satisfy the Pythagorean Theroem

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

4n^2 + n^4 - 2n^2 + 1 = n^4 + 2n^2 + 1

2n^2 + n^4 + 1 = n^4 + 2n^2 + 1

since both LHS and RHS are equivalent, this is a right triangle

now create a table of values for 2n, n^2 -1 and n^2 + 1 starting from n=2 to n=12

n=2, 2n = 4, n^2 -1 = 3, n^2 + 1 = 5

n=3, 2n = 6, n^2 -1 = 8, n^2 + 1 = 10

n=4, 2n = 8, n^2 -1 = 15, n^2 + 1 = 17

n=5, 2n = 10, n^2 -1 = 24, n^2 + 1 = 26

n=6, 2n = 12, n^2 -1 = 35, n^2 + 1 = 37

n=7, 2n = 14, n^2 -1 = 48, n^2 + 1 = 50

n=8, 2n = 16, n^2 -1 = 63, n^2 + 1 = 65

n=9, 2n = 18, n^2 -1 = 80, n^2 + 1 = 82

n=10, 2n = 20, n^2 -1 = 99, n^2 + 1 = 101

n=11, 2n = 22, n^2 -1 = 120, n^2 + 1 = 122

n=12, 2n = 24, n^2 -1 = 143, n^2 + 1 = 145

repeat for part c and part d

voila