What is the ratio of the length of the diagonal of a square to the length of the hypotenuse of the isosceles right triangle having the same area?
Pick a side length for the square. (It doesn't matter what it is, since you'll be finding the ratio. You might find it helpful to use a value of "1".)
Noting that an isosceles right triangle with vertical height "h" has horizontal base "b", with b = h. Use this, along with A = (1/2)bh and the area of the square, to find the value of h (and b).
Use the Pythagorean Theorem to find the length of the diagonal of the square and the hypotenuse of the triangle.
If you get stuck, please reply showing your work and reasoning so far. Thank you!
Let A be the area for both the square and the isosceles right triangle.
For the square, Let x be its diagonal,
since, one side of the square = $\displaystyle \sqrt{A}$, by Pythagarus Theorem,
$\displaystyle \sqrt{A}^2 + \sqrt{A}^2 = x^2$
$\displaystyle 2A = x^2$
$\displaystyle x = \sqrt{2A}$
For the isosceles right triangle, let y be its hypotenuse and z be its two other sides,
$\displaystyle A = z^2 / 2$
$\displaystyle z = \sqrt{2A}$
and then $\displaystyle z^2 + z^2 = y^2$, hence
$\displaystyle 2A + 2A = y^2$
$\displaystyle y = \sqrt{4A}$
$\displaystyle y = 2\sqrt{A}$
In the end, x:y = $\displaystyle \frac{\sqrt{2A}}{2\sqrt{A}}$
x:y = 1:$\displaystyle \sqrt{2}$