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)

2 Answers

Relevance
  • 10 years ago
    Favorite Answer

    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

    • Log in to reply to the answers
  • 3 years ago

    impossible the main important area is often the hypotenuse so a million² + (?3)² = 3² a million+3 = 9 4?9 this is if fact be told impossible for a precise triangle to have those 3 facets Now, a precise triangle could have facets: a million, ?3, and a pair of this is a 30-60-ninety precise triangle yet a million, ?3, 3 is impossible as a precise triangle

    • Log in to reply to the answers
Still have questions? Get answers by asking now.