My Math Forum Area of a Quadrilateral vs. Area of a Kite

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 December 12th, 2016, 04:08 PM #1 Newbie   Joined: Dec 2016 From: United States Posts: 1 Thanks: 0 Area of a Quadrilateral vs. Area of a Kite In my recent quiz I was told to find the area of the given quadrilateral. It had three given sides: 12, 9, 16. The angle between the sides lengths 12 and 9 was 90 degrees. The angle between the sides lengths 9 and 16 was 126 degrees. To find the area, I used the formula for a kite. To find the diagonals, I used the pythagorean theorem for the diagonal that forms the hypotenuse of the triangle with the right angle and I used the law of cosines to find the diagonal opposite the angle of 126 degrees. Then, i used the formula for area of a kite ((1/2)d1•d2). I got the correct answer (exactly) but the problem is that the quadrilateral is NOT a kite. How did I get the right answer?? The quadrilateral is definitely not a kite but i still got the right answer using the formula for area of a kite.
 December 12th, 2016, 04:43 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,442 Thanks: 1202 the two methods of finding the area are close, but not exact ... using the diagonals, $d_1=15$, $d_2=\sqrt{9^2+16^2-2 \cdot 9 \cdot 16 \cdot \cos(126)}$, A = 168.7553587 using two triangles (the 9-12-15 and the 15-16 with the angle $\theta = 126-\arctan(4/3)$ between them) ... A = 168.6766078 (the true area) both round to 169 I imagine they are close because the diagonals are close to being perpendicular Thanks from topsquark and olivia1

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