ABCD is a kite. AB=393 meters; BC=524 meters, AD=314 meters, find CD, rounded to nearest meter.
ALSO:::
Diagonal BD=236
Diagonal AC=655
Thank!
2007-01-11 19:22:45 · 6 answers · asked by Spearfish 5 in Science & Mathematics ➔ Mathematics
OK. It's not the usual kite shape, and it probably wouldn't fly. But it's a quadrilateral, and the problem can be solved.
I developed a solution by placing the points on a set of x-y axes and solving for the x-y coordinates of each vertex.
In brief, here is the process:
Let A be the origin (0, 0).
Place B along the x axis at (393, 0).
Find C so that it is 524 m from B and 655 m from A.
Answer: C is at (393, 524). Note that ABC is a right triangle.
Find D so that it is 314 m from A and 236 m from B.
Answer: D is at (125.64, 287.77)
Finally, find the distance from C to D.
Answer: 356.77 m (which, to the nearest meter, is 357 m).
2007-01-11 20:02:44 · answer #1 · answered by actuator 5 · 0⤊ 0⤋
Quadrilateral ABCD can't be a kite. The definition of a kite is:
A planar convex quadrilateral consisting of two adjacent sides of length "a" and the other two sides of length "b".
Here is a link.
http://mathworld.wolfram.com/Kite.html
For a true kite with diagonals of length p and q, the area A, is:
A = ½pq
2007-01-12 03:42:48 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
Scale down to some reasonable size and draw a picture.
2007-01-12 03:53:27 · answer #3 · answered by becky 2 · 0⤊ 0⤋
i think the answer is around 700
i tried a lot so either correct or wrong pl give it as best
2007-01-12 03:40:14 · answer #4 · answered by SAI H 2 · 0⤊ 0⤋
tapessium
2007-01-12 03:36:11 · answer #5 · answered by Lynda L 3 · 0⤊ 1⤋
it will never fly
2007-01-12 03:30:05 · answer #6 · answered by da rinse mode 4 · 0⤊ 1⤋