# Rectangles

## Rectangles

So far on this site I've seen probably 20 or 30 problems saying something like,

The perimeter of a rectangle is x units. If the length is z more than the width, find the area of the rectangle.

The truth is, they're not that hard! Once this is changed into algebraic form it becomes a simple algebra problem. I'll show a few examples of such problems:

Example 1:

The perimeter of a rectangle is 68. If the length is 4 more than twice the width, find the dimensions and the area of a rectangle.

Solution:

Let the dimensions of the rectangle be L and W. Since the perimeter is 68, then 2(L+W) = 68 --> L + W = 34. Also, the length is 4 more than twice the width, i.e. L = 2W + 4. Therefore we have the system of equations

L + W = 34

L = 2W + 4

Plugging L = 2W + 4 = 34 into the first equation we have:

3W + 4 = 34 --> W = 10.

From this, we obtain L = 24, so the area is 240 sq. units.

Example 2:

Find all rectangles with integer dimensions such that the area is numerically three times the perimeter.

Solution: Again, just another straightforward algebra problem. If the dimensions are L and W, then

LW = 6(L + W) = 6L + 6W

Moving all L terms to one side,

LW - 6L = 6W

L+=+6W%2F%28W-6%29

Now, it suffices to find all positive integers W such that L is also a positive integer. To do this, suppose W = X + 6. Then,

L+=+%286X+%2B+36%29%2FX+=+6+%2B+36%2FX (X must also be positive)

Hence, X can equal any positive factor of 36, namely {1, 2, 3, 4, 6, 9, 12, 18, 36}. Adding 6 to each of these terms to obtain W, we have

W = {7, 8, 9, 10, 12, 15, 18, 24, 42}

Solving for L, we obtain:

L = {42, 24, 18, 15, 12, 10, 9, 8, 7} (which is obviously symmetrical to W)

Therefore, all ordered pairs (L, W) assuming L>=W, that satisfy are (42, 7), (24, , (18, 9), (15, 10), and (12, 12).

The moral of this is, when you see a problem like this (or most any other problem), turn it into an algebra problem that you can easily solve!

The perimeter of a rectangle is x units. If the length is z more than the width, find the area of the rectangle.

The truth is, they're not that hard! Once this is changed into algebraic form it becomes a simple algebra problem. I'll show a few examples of such problems:

Example 1:

The perimeter of a rectangle is 68. If the length is 4 more than twice the width, find the dimensions and the area of a rectangle.

Solution:

Let the dimensions of the rectangle be L and W. Since the perimeter is 68, then 2(L+W) = 68 --> L + W = 34. Also, the length is 4 more than twice the width, i.e. L = 2W + 4. Therefore we have the system of equations

L + W = 34

L = 2W + 4

Plugging L = 2W + 4 = 34 into the first equation we have:

3W + 4 = 34 --> W = 10.

From this, we obtain L = 24, so the area is 240 sq. units.

Example 2:

Find all rectangles with integer dimensions such that the area is numerically three times the perimeter.

Solution: Again, just another straightforward algebra problem. If the dimensions are L and W, then

LW = 6(L + W) = 6L + 6W

Moving all L terms to one side,

LW - 6L = 6W

L+=+6W%2F%28W-6%29

Now, it suffices to find all positive integers W such that L is also a positive integer. To do this, suppose W = X + 6. Then,

L+=+%286X+%2B+36%29%2FX+=+6+%2B+36%2FX (X must also be positive)

Hence, X can equal any positive factor of 36, namely {1, 2, 3, 4, 6, 9, 12, 18, 36}. Adding 6 to each of these terms to obtain W, we have

W = {7, 8, 9, 10, 12, 15, 18, 24, 42}

Solving for L, we obtain:

L = {42, 24, 18, 15, 12, 10, 9, 8, 7} (which is obviously symmetrical to W)

Therefore, all ordered pairs (L, W) assuming L>=W, that satisfy are (42, 7), (24, , (18, 9), (15, 10), and (12, 12).

The moral of this is, when you see a problem like this (or most any other problem), turn it into an algebra problem that you can easily solve!

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