Question

Attached.

Answer 1

I would write down all the facts I can think of about right triangles and isosceles triangles and perimeters...

Answer 2

particularly those facts that relate to the sides

Answer 3

Isoceles have two equal sides.

Right triangles use pytagorean theorem for lengths.

Answer 4

yes, now assign some variables to the legs (I like traditional ones, like x and y and z, but you can use whatever you want...)
and use the variables in the equations you know.
you are looking for 2 equations and 2 unknowns

Answer 5

So pythagorean theorem...

x^2 + y^2 = z^2

and x+y+z=24

What do i do from here?

Answer 6

you have an isosceles triangle. how would that affect these equations?

Answer 7

or wait.... x^2 + x^2 = y^2

and 2x+y=24!

I got itttt

Answer 8

you have the equations, now the algebra part.... to solve them

Answer 9

So I got x=6 and y=12...but they don't work in the pythag. theorem....

Answer 10

2x^2 = y^2 so sqrt(2)*x= y (square root both sides)
now use this for y in the 2nd equation
2x+ sqrt(2)x= 24
(2+sqrt(2))x= 24
x= 24/(2+sqrt(2))

Answer 11

did you follow that?

Answer 12

\[ 2x^2= y^2\]
\[ y= \sqrt{2}x \]
\[ 2x+y= 24 \]
\[2x+\sqrt{2}x=24\]

Answer 13

yeah I got that....now i'm not sure how to use it for y

Answer 14

\[ x= \frac{24}{2+\sqrt2} \]
people generally "rationalize" numbers like this, by multiplying top and bottom by 2 - \(\sqrt{2}\)
you get
\[x = \frac{24(2-\sqrt2)}{4-2} \]
simplify to
\[ x= 12(2-\sqrt2) \]
use the equation
\[ y= \sqrt2x\]
to find y (y will be the hypotenuse of your isosceles right triangle)