math challenge question conseutive odd integers right triangle probably pythagorean thorem?

n = other leg
n + 2 = hypotenuse (consecutive odd integer)

20^2 + n^2 = (n+2)^2

and solve for n.

When we say consecutive odd integers, we mean only from the odd integers, like 41, 43, 45, 47 are consecutive.  The even integers in-between aren't counted.

Let the length of the unknown leg be
2x-1, then the hypotenuse is 2x+1. &
400+(2x-1)^2=(2x+1)^2=>
8x=400=>
x=50
So, the length of the hypotenus is
2(50)+1=101

First, the odd integers are... 1,3,5,7,9 etc...

Examples of consecutive odd integers would be 5 and 7,  21 and 23, 99 and 101 etc...

You can define them as...

x and x + 2


So using the Pythagorean Theorem...
20^2 + x^2 = (x + 2)^2

You know that the hypotenuse is x + 2 since the hypotenuse must be the longest side

So...

Multiply out the square...
400 + x^2 = x^2 + 4x + 4

Subtract x^2 from each side...
400 = 4x + 4

Divide each side by 4...
100 = x + 1

Subtract 1 from each side...
99 = x

The first leg was defined as 20
The second leg is x or 99
The hypotenuse is x + 1 or 101

Jenny,

A small favor: next time you post, use a spell checker.

Now, consecutive odd integers are 1,3,5,7,...They differ by 2, not 1.

An odd number can be written as 2n+1 where n is any number. How come? Well, 2n is always even and if you add 1 to an even number you get an odd one.

Next, the consecutive odd integer to 2n+1 is 2n+3 (2 away)

Finally Pythagoras' theorem applied to this problem

(2n+3)^2=(2n+1)^2+20  Solve for n

let x be the 2nd leg and x+2 be the hypotenuse
given that 20 is the first leg; x is the 2nd and third is the hypotenuse, we can come up with the phytagorean theorem
c²=a²+b² (3 sides of a triangle)
c=hypotenuse, a and b are the sides

(x+2)²= 20² + x²

x²  + 4x + 4 = 400 + x²

by transposition;
x²-x² + 4x=400-4
4x= 396
x=396/4
x=99  ( this is the second leg)

since hypotenuse is x+2 and the value of x=99;

hypotenuse = 99 + 2
                  = 101