posted by Candice .
Let ABCD be a square with side legnth 7cm. Another square KLMN is inscribed into ABCD such that its vertices lie on the sides of ABCD.
Find the lenghts of parts of ABCD sides which are divided by the vertices of KLMN. If [the area of KLMN]: [area of ABCD] = 25:49
you will have 4 congruent triangles.
each will have legs of x and 7-x and a hypotenuse of 5
thus x^2 + (7-x)^2 = 25
solve as a quadratic, it factors nicely, and you should have sides of 3 and 4
How would you know that the hypotenuse is length 5? Does it have something to do with the ratio?