Posted by **Anonymous** on Tuesday, June 13, 2006 at 5:43pm.

A kite has an area of 32. One diagonal is 4 times the other one. The endpoints of one of the diagonals (one of the vertices of the kite) is (3,6). Find other three possibilites for the other three vertices of the kite.

Can someone help me with this problem?

I can tell you what the two diagonals are, but there are more than three places to plot the other three vertices. They will depend upon which corner of the kite is at (3,6) and also depend upon where the two diagonals intersect.

Let a be the long diagonal's length and b be the short diagonal's length. The kite is symmetric about the long diagonal but not necessarily the short one. Then, no matter where the diagonals intersect (perpendicular to each other, for a kite), the area of the kite will be

32 = A = 2*(1/2) a (b/2)= ab/2 = 2 b^2.

Therefore b = 4 and a = 16.

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