math word problem

Question

A wood frame for pouring concrete has an interior perimeter of 14 meters. Its length is one meter greater than its width. The frame is to be braced with twelve-gauge steel cross-wires. Assuming an extra half-meter of wire is used at either end of a cross-wire for anchoring, what length of wire should be cut for each brace?

Answer 1

I'm assuming the form is a rectangle where the 4 corners, clockwise A, B, C, and D, are right angles.

Let the inside length be called L and the inside width be called W. The inside perimeter is L+L+W+W = 2L + 2W. But W=1 less than L. Thus Perimeter = 2L + 2(L-1) = 4L-2 = 14 meters. Simplifying, get 4L=16 (add 2 to both sides of previous equation). Therefore L=4 meters and W=L-1=3 meters. As a check, 4+4+3+3 = 14.

The braces are such that one connect corner A to corner C and the other connects corner B with corner D.

Each brace forms a right triangle with sides L, W and H, where H is the hypoteneus. The Pythagotean theorem says that for a right triangle H² = L² + W². Plugging in the values of L and W, H² = 16+9 = 25. Take the positive square root to get H, which is 5 meters. Since an additional 1/2 meter of bracing wire is added to each end, the total length of one wire is 5+1/2+1/2 = 6 meters.