A rectangle is inscribed in a right isosceles triangle, such that two of its vertices lie on the hypotenuse, and two other on the legs. What are the lengths of the sides of the rectangle, if their ratio is 5:2, and the length of the hypotenuse is 45 in? (Two cases)
Case 1) Blank, Blank, Blank
Case2) Blank, Blank, blank
What goes in these blanks?

Refer to the attachment for the diagram of the problem.

Let us call the two identical sides of the isosceles triangle with s. Since we know that the hypotenuse measures 45 inches, we can find s by using the Pythagorean Theorem:
[tex] 45^{2} = s^{2}+ s^{2} [/tex]
[tex] 45^{2} = 2s^{2}[/tex]
[tex] 45^{2} = 2s^{2}[/tex]
[tex] 1012.5 = s^{2}[/tex]
[tex] s=31.82[/tex]

We can notice that the figure will have similar triangles with the smaller triangle having one length equal to 5x. We then compute for the equivalent hypotenuse of the smaller triangle (denoted as y).
[tex] \frac{31.82}{5x} = \frac{45}{y} [/tex]
[tex]y=7.07x[/tex]

We can then solve for the value of x by applying the Pythagorean Theorem in the smaller triangle.
[tex] (7.07x)^{2} = (5x)^{2}+ (31.82-2x)^{2} [/tex]
[tex] 50x^{2} = 25x^{2}+ 1012.5-127.28x+4x^{2} [/tex]
[tex] 0= -21x^{2}-127.28x+1012.5[/tex]
[tex]x=4.55,-10.61[/tex]

We ignore the negative value since there is no negative side. x is therefore 4.55 thus the dimensions of the rectangle would be 22.75 and 9.1. For the other case, the dimensions would simply be inverted since we would be dealing with a rectangle with its width longer than its length.