Only a right triangle has a hypotenuse. The third side of your triangle is
(7x+2y) - (2x+y) - (3x-5y) = 2x + 6y
The triangle will only be a right triangle for special values of x and y.
If i wanted to find the triangle and the perimeter is 7x+2y units.
OPPOSITE: 2x+y
ADJACENT: 3x-5y
WHAT IS THE HYPOTHENUSE?
Please help me.
3 answers
Oh...,ok...thank u
let the hypotenuse be h
we know
h + 3x - 5y + 2x + y = 7x + 2y
h = 2x-6y
then (2x+6y)^2 = (2x+y)^2 + (3x-5y)^2
4x^2 + 24xy + 36y^2 = 4x^2 + 4xy + y^2 + 9x^2 - 30xy + 25y^2
9x^2 - 50xy - 10y^2 = 0
Using the quadratic formula I got x = (50±√2860)/18 y
= appr. 5.75y or a negative which I will reject
So really there is an infinite number of solutions.
e.g. if y =1 then x = 5.75
side1 = 2x+y = 12.5
side2 = 3x-5y = 12.25
hypotenuse = 2x + 6y = 17.5
check: 17.5^2 = 306.25
12.5^2 + 12.25^2 = 306.31 close enough with the decimals carried.
If you let y = 2, then x = 11.5 etc.
we know
h + 3x - 5y + 2x + y = 7x + 2y
h = 2x-6y
then (2x+6y)^2 = (2x+y)^2 + (3x-5y)^2
4x^2 + 24xy + 36y^2 = 4x^2 + 4xy + y^2 + 9x^2 - 30xy + 25y^2
9x^2 - 50xy - 10y^2 = 0
Using the quadratic formula I got x = (50±√2860)/18 y
= appr. 5.75y or a negative which I will reject
So really there is an infinite number of solutions.
e.g. if y =1 then x = 5.75
side1 = 2x+y = 12.5
side2 = 3x-5y = 12.25
hypotenuse = 2x + 6y = 17.5
check: 17.5^2 = 306.25
12.5^2 + 12.25^2 = 306.31 close enough with the decimals carried.
If you let y = 2, then x = 11.5 etc.