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The diagonals of a kite measure 16 units and 21 units. What is the perimeter of this kite if the sides are integers?
(A) 48 units
(B) 42 units
(C) 62 units
(D) 54 units
(E) 46 units

One diagonal of the kite biscects the other and the diagonals are perpendicular to each other. We can see that four right triangles will be formed the hypotueneses of whom will form the sides of the kite.

I can sense that the diagonal that will be bisected will be the one with length 16, otherwise we would have 10.5 as one of the sides of a right triangle. So, one of the sides of the right triangle has to be 8. I feel that pythagorean triplets will feature somewhere in the solution, but dont know how to go about solving this.

Am in a bit of a hurry now, so will try to revisit this and post an answer if I can crack it. Seems like a good problem though.

Kite: One of the diagonals bisect. Both diagonals intersect at right angles.

Since the sides (AB,BC,CD,DA) are all integer values, the diagonal that measures 16 gets bisected: 8 and 8.

AO = OC = 8; OB = x; OD = 21-x.



We have four right triangles. When x = 6, we have FOUR right triangles whose sides are as follows:

AOB and COB: 6, 8, 10 (AB and BC are 10 each)

AOD and COD: 8, 15, 17 (when x = 6, 21-x = 15; AD and DC are 17 each)

It helps to remember common Pythagorean triplets: (3,4,5), (5,12,13), (7,24,25), (8,15, 17), (9,40,41) etc..

Therefore the perimeter is 17 + 17 + 10 + 10 = 54. Answer: (D)

NOTE: If the diagonal that measures 21 gets bisected, we get 10.5 and 10.5. These non-integer values will not give integer values for the sides of the kite.