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

 Geometry Geometry Math Forum

 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: 2,805 Thanks: 1449 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

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Tangeton Geometry 8 February 24th, 2016 11:50 AM shankarathreya New Users 3 July 2nd, 2014 11:38 AM panky Algebra 2 December 18th, 2011 09:39 PM michary91 Algebra 2 September 29th, 2010 05:34 AM meph1st0pheles Algebra 4 February 25th, 2010 11:40 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top