Our formula for the area of any triangle can be explained if we take a look at a “right” triangle.

Just look at the images below—a rectangle that comes with the same base and height as our original triangle.

**The next lesson:**Area and Circumference of a Circle

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The following transcript is provided for your convenience.

The area of a right triangle is one half of the area of the rectangle! Since the areas of two congruent triangles are identical to the area of a rectangle, we may have the formula A = ^{1}⁄_{2} b • h if we want to determine the area of a triangle.

When we’re using the formula for triangles, if we want to find its area, it’s key to identify the base and the corresponding height (perpendicular to its base).

**For example:**

The problem: Find the area of a triangle that has a height of 4 (four) inches and a base of 10 (ten) inches.

Area (A) = ^{1}⁄_{2} b • h Begin with our formula for a triangle’s area.

A = ^{1}⁄_{2} • 10 • 4 Substitute 10 (ten) for the base and 4 (four) for the height.

A = ^{1}⁄_{2} • 40 Now, multiply.

A = 20

The answer is: A (area) = 20 in^{2}

Now let’s take a look at a trapezoid. If we want to discover the area of any trapezoid, we must take average lengths of the two (2) parallel bases, and then multiply the average length by its height:

Look at the example provided below. Do you notice that a trapezoid’s height is always perpendicular to its bases (just as when we find a parallelogram’s height)?

**For example:**

Problem: We must find the area of this trapezoid:

We begin with the formula for a trapezoid’s area.

We substitute 4 (four) and 7 (seven) for the bases and 2 (two) for the height, and then find A.

The answer is: The area of our trapezoid is 11 cm^{2}.

**Area Formulas**

We can use the following formulas if we need to find the areas of different shapes:

For a square: A = s^{2}

For a rectangle: A = l • w

For a parallelogram: A = b • h

For a triangle: A = ^{1}⁄_{2} b • h

For a trapezoid:

**Working with Perimeter & Area**

Often, we are asked to find the perimeter or area of shapes which are not standard polygons. Architects and artists, for example, generally are dealing with complex shapes. However, these complex shapes may be thought of as being the composition of less complicated, smaller shapes such as rectangles, triangles, or trapezoids.

If we want to find the perimeter of a non-standard shape, we still can determine the distance around that shape if we add together the lengths of all sides.

To find the area of a non-standard shape, we need to do it a bit different. We need to create different regions within that shape for which we then can find the areas that we then can add together. Just take a look at how we’ve done this below.

**The example:**

Our problem is: Find the perimeter and area of this polygon.

The perimeter (P) = 18 + 6 + 3 + 11 + 9.5 + 6 + 6

P = 59.5 cm

To find our perimeter, lust add together all lengths of the shape’s sides. Begin at the top to work clockwise around our shape.

To find our shape’s area, we can divide our polygon into two (2) separate and simpler regions. Now, the area of the whole shape, the entire polygon, will be equal to the sum of the areas of these two regions.

Area of our polygon = (Area of region A) + (Area of region B)

The area of region A

Area (A) = l • w Region A is a rectangle.

A =18 • 6 To find our area, we multiply length (18) by width (6).

A = 108 Region A’s area is: 108 cm^{2}.

Area of Region B

Area (A) = ^{1}⁄_{2} b • h Region B is a triangle. To find our area, we use this formula: ^{1}⁄_{2} bh

A = ^{1}⁄_{2} • 9 • 9 where the base is 9 (nine) and the height is 9 (nine).

A = ^{1}⁄_{2} • 81

A = 40.5 Region B’s area is: 40.5 cm^{2}.

Now add both regions together: 108 cm^{2} plus 40.5 cm2 = 148.5 cm^{2}.

The answer is:

The perimeter = 59.5 cm

The area = 148.5 cm^{2}

We may also use what we know about perimeters and areas to help us solve problems in situations where we are buying paint or fencing, or to determine how big a rug should be for our living room. Here’s an example of fencing.

**The example**

The problem: Rosie is planting her garden that has the dimensions as shown below. Now, she wants to put an even, thin layer of mulch across the whole surface of her garden. The price if the mulch is $3 per square foot. What’s the amount of money she’ll need to spend on the mulch?

Well, this shape combines two simpler shapes, a trapezoid and a rectangle. First, we need to find the area of each shape.

A = l • w

A = 8 • 4 So first, we find the rectangle’s area.

A = 32 ft^{2}

Then we find the trapezoid’s area.

32 ft^{2} plus 44 ft^{2} = 76 ft^{2} Add the two measurements.

76 ft^{2} • $3 = $228 We are multiplying by $3 to see how much Rosie will be spending.

The answer is: Rosie will be spending $228 for covering the entire garden with mulch.

Now, find the area of this shape shown below:

A) 11 ft^{2}

B) 18 ft^{2}

C) 20.3 ft

D) 262.8 ft^{2}

**Which is the correct answer?**

A) 11 ft^{2}

This is correct. Our shape is a trapezoid, so we can use this formula: for finding the area:

B) 18 ft^{2}

Incorrect. It seems like you’ve multiplied 2 by 9 to come up with 18 ft^{2. }This would have worked if our shape would have been a rectangle but our shape is a trapezoid. So we need to use this formula: . The correct answer would be 11 ft^{2}.

C) 20.3 ft

Incorrect. It seems like you’ve added all of the dimensions together. Well, that gives the perimeter. For finding the trapezoid’s area, we use this formula: . Our correct answer would be: 11 ft^{2}.

D) 262.8 ft^{2}

Incorrect. It seems like you’ve multiplied all dimensions together. Our shape is a trapezoid, which means we have to use this formula: . The correct answer would be: 11 ft^{2}.

**Summary**

A 2-dimensional shape’s perimeter is the distance around that shape. We can fond it by adding up all of the sides (they all have to be the same measure units, though). The area of a 2-dimensional shape can be found by counting the number of all squares that are covering the shape. There are quite a few formulas for quickly finding the area of some standard polygons, like parallelograms and triangles.

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