Three squares whose sides lengths are integers are placed overlapping as shown in the diagram (link). If BC = CD and the shaded area is 31, determine the area of the largest square.
Here is the picture. http://i58.photobucket.com/albums/g254/poker5495/POWIV.jpg
10 points will go to the best answer.
2007-04-08 13:13:42 · 2 answers · asked by poker5495 4 in Science & Mathematics ➔ Mathematics
AC^2-AB^2=31The smallest number satisfying this is
AC= 16 and AB = 15.
So BC=CD = 1
So AD = 18So area of largets square = 18^2 = 324 units ^2
2007-04-08 13:32:38 · answer #1 · answered by ironduke8159 7 · 1⤊ 0⤋
By knowing that the difference between two consecutive squares is an odd integer you can find that 16 squared minus 15 squared equals the shaded region 31.
This means that the distance BC = 1 = CD
Therefore the area of the largest square is (16+1) squared = 17 squared = 289
2007-04-08 20:33:26 · answer #2 · answered by scubasteve 1 · 0⤊ 0⤋