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10' +-------+ | | | | | | 15' | | 20' | | | | 30' | +-----------------------+ | | | | 5' +-------------------------------+ 40'Any area with square corners (where sides all meet at 90 degree angles, also known as right angles) can be divided into a number of rectangles or squares and then those rectangles or squares can be measured individually, then combined to determine the total area.

In the above example, two rectangles (10' x 15 and 5' x 40) can be used describe the entire area:

10' +-------+ | | | | |These two values can then compute the area by multiplying the sides of each square or rectangle and adding the results, as inA| 15' | | | | | | +-------+-----------------------+ |B| | | 5' +-------------------------------+ 40'

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10 x 15 = 150 square feet (in Area A)
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5 x 40 = 200 square feet (in Area B)

Combined, the described L-shaped area contains 350 square feet.

Areas with more complex shapes would use additional rectangles to measure the total size in the same way.

100' +---------+ | \ | \ 50' | \ | \ | \ | \ +----------------+ 175'In such cases, it is usually possible to divide the space up so that the angled area can be measured separately, like this:

100' +---------+ | |\ | | \ 50' | | \ |The section on the right side of this area (area "B") is now a triangle with one corner that has a 90 degree angle (a right angle), the lower left corner of the triangle in this example.A|B\ | | \ | | \ +---------+------+ 100' 75'

The size of Area A (which is now a rectangle), is computed as shown earlier.

Because one corner of the triangle has a 90 degree angle, the area of the triangle can always be envisioned as a square or rectangle by imagining a second triangle of the same size, positioned upside down and placed next to the real triangle, like this:

100' +---------+ - - -+ | |\ | | | \ 50' | | \ C | |A|B\ | | \ | | | \ +---------+------+ 100' 75'

In this example, Area C is imaginary, but viewed together with the real triangle in Area B, the combination gives you a rectangle, with each triangle always containing half of the area.

Once you have only squares and rectangles to work with, you can easily compute the size of the area where the pair of triangles are, by multiplying the sides of the imaginary rectangle together. After doing that, divide the overall size of the imaginary square or rectangle by two so that you are left with only the area within the real triangle (Area B).

100 x 50 = 5,000 square feet (In Area A) (75 x 50) / 2 = 1,875 square feet (In Area B) 5,000 + 1,875 = 6,875 square feet total for the entire area.Note that the angle of the sloping side of the triangle and the length of that side of the triangle were not needed to compute the area of the triangle. The only two size values of the triangle that are needed using this method are the lengths of the two sides that meet at a right angle, and these are usually the easiest values to obtain.

The same process can be used to compute the size of an area that is irregular in many places, by dividing it up into as many squares, rectangles and triangles that are necessary to represent the area. As long as all boundaries are not curved* and you can get one of the corners of each triangle to be a 90 degree angle, this simple method works.

* Areas with curved boundaries require more complex mathematics to solve precisely, and computing area for such objects is not described here.

Here is a very strangely shaped area that we need to measure:

100' +---------+ | | | | 100' | | + -- 200' | | / \ | | | / \ 60' | | 50' / \| 100' | +----+ +-----------+ | |- 100'-| / | / | / | / | / +----------------------------+ 300'First, divide the area up into however many squares and rectangles that it takes to represent the the space inside the boundary, then create imaginary triangles to make squares or rectangles out of the real triangles.

In the case of the pyramid-shaped area, it can be divided into two triangles and then each given its own imaginary matching triangle to form two rectangles. (The solution method shown for the pyramid-shaped area always works, and works even if the two halves of the original triangle are not exactly symmetrical. There are more examples for how to handle triangles in a moment.)

100' +---------+ | | | | 100' |In this drawing, sections F, G and H are imaginary. The computation of square feet is then:A| +- -+ - + 100' | | |F /|\ G| | | | / | \ 60' | | |/C|D\| | | 50' | +- - - - -+----+- -+- -+-----+-----+ | | 50'|50'|50'| 50' |E/| | / 100' |B| / | 100' | / | |/ H | +----------------------------+ - - + | 300' | 50' |

100 x 100 = 10,000 square feet in AreaA100 x 300 = 30,000 square feet in AreaB(60 x 50) / 2 = 1,500 square feet in AreaC(60 x 50) / 2 = 1,500 square feet in AreaD(50 x 100) / 2 = 2,500 square feet in AreaECombined, 10,000 + 30,000 + 1,500 + 1,500 + 2,500 = 45,500 square feet total

For example, here is an uneven triangle that has no 90 degree right angle:

|--- 66' ------+-- 33'-| + -- / \ | / \ | / \ / \ 50' / \ / \ | / \ | +-----------------------+ -- 99'As described above, one solution for this situation is to divide the triangle into two sections, then create identical imaginary triangles to go with each divided triangle, forming two rectangles. Compute the area of each rectangle, and divide the result by two to find the area of the original triangle.

|--- 66' ------+-- 33'-| +- - - - - - - -+- - - -+ -- | / |\ | | C / | \ D | | / | \ | / | \ 50' | /(In this example, areas C and D are imaginary.) The original triangle has an area of 1,650 + 825 = 2,475 square feet.A|B\ | / | \ | | / | \| | +---------------+-------+ -- 99'

The Square Root can be computed using a variety of time-consuming techniques, or you can push the "square root" button on your calculator and let it perform this iterative computation.

For example, if you have a 400 square foot area, taking the square root of that value gives you 20. So if the area happened to be a square, each side of the area is 20 feet long. Multiplying 20 by 20 gets you back to 400.

It is not possible to compute the original size of the sides of a non-square area unless you already know the original length of one side, or the size ratio between two sides, such as knowing that two sides are three times longer than the other two sides.

It is important that all lengths are in the same unit of measurement, so be sure to convert all units of measurement that are used to the same unit type. For example, if one side of the area is in yards, but another side is in feet and inches, you need to convert the values to all be the same unit of measure before trying to compute the area.

For example, let's say we want to compute the surface area of a section
of road three miles long and 20 feet wide. Here, converting the miles length
into feet gives us 15,840 feet (3 x 5,280). The area would then be
15,840 x 20 = 316,800 square feet.

(Conversions between units of length or distance measurement can be found
in a chart in the Related Topics section, below.)

Once a common unit of length measurement is used, the area can be
computed, and that result can be converted into some other unit of
*area* measurement. If you computed the area using meters for
the lengths, the result would be in square meters, but you might want to
convert that result to square feet. To do that, use the conversion
table for units of area measurement, which can be found in the
**Related Topics** section, below.

**Length and Distance Measurement Conversion Reference** (HTML)

**U.S. Capacity Measurement Conversions** (HTML)

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