Re: o/t algebra help

Posted by MarkB_MI on November 13, 2008 at 14:08:13 from (74.126.2.95):

In Reply to: o/t algebra help posted by glennster on November 12, 2008 at 18:55:48:

OK, I realize it's a little late now, but here's how you solve it (as opposed to guessing).

First, consider how the canvas must be cut up. The only way that you can use all the canvas is to cut the two smaller pieces from the same side of the canvas, leaving the rest of the canvas for the same piece. Note that this means that the short side of the middle-sized piece is the same as the side of the small, square piece. So we only need to solve for the two smaller pieces.

So, let's say that:

The length of one side of the uncut canvas = a = 18

The length of the side of the small, square piece = b

The length of the long side of the middle-size piece = c

The length of the short side of the large piece = d

We know that the area of the mid-size piece is twice the area of the small piece:

2b^2 = bc

Dividing both sides of the equation by b, we have 2b - c = 0.

We also know that b + c = 18;

So we have two equations with two variables; we can solve that by adding the two equations:

2b - c = 0

b + c = 18

------------

b = 6

Solving for c, since b + c = 18, c = 12.

Checking our answers:

2b^2 = 2 * 6 * 6 = 72

bc = 6 * 12 = 72

Now to solve for d:

d = a - b = 18 - 6 = 12

So the dimensions are:

6 X 6

6 X 12

12 X 18

This post was edited by MarkB_MI at 14:11:28 11/13/08.

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