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, BlankCase2) Blank, Blank, blankWhat goes in these blanks?

Answer:Part 1) The base is [tex]25\ in[/tex]  and the height is [tex]10\ in[/tex] Part 2) The base is [tex]7.5\ in[/tex]  and the height is [tex]18.75\ in[/tex] Step-by-step explanation:case 1) Right isosceles triangle of the leftLetx------> the base of the rectangley----> the height of the rectangleRemember thatIn a right isosceles triangle the lengths of the legs of the triangle is the same [tex]y+x+y=45[/tex][tex]2y+x=45[/tex] ----> equation A[tex]\frac{x}{y} =\frac{5}{2}[/tex][tex]x=2.5y[/tex] -----> equation Bsubstitute equation B in the equation A[tex]2y+2.5y=45[/tex][tex]4.5y=45[/tex][tex]y=10\ in[/tex]Find the value of x[tex]x=2.5(10)=25\ in[/tex] case 2) Right isosceles triangle of the right Letx------> the base of the rectangley----> the height of the rectangleRemember thatIn a right isosceles triangle the lengths of the legs of the triangle is the same [tex]y+x+y=45[/tex][tex]2y+x=45[/tex] ----> equation A[tex]\frac{y}{x} =\frac{5}{2}[/tex][tex]y=2.5x[/tex] -----> equation Bsubstitute equation B in the equation A[tex]2(2.5x)+x=45[/tex][tex]6x=45[/tex][tex]x=7.5\ in[/tex]Find the value of y[tex]y=2.5(7.5)=18.75\ in[/tex]