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Here you will learn about area, including how to calculate the area of triangles, quadrilaterals, and composite shapes.

Students will first learn about area as part of measurement and data in 3rd grade and 4 th grade. They continue to build upon this topic as a part of geometry in 6th grade and 7th grade.

**Area** is a measure of the amount of space there is inside a 2D shape.

The area of a shape is measured in square units such as \mathrm{cm}^2 ,\mathrm{~mm}^2 ,\mathrm{~m}^2 .

To calculate area, you can use different formulas depending on the shape.

Below are the formulas to find the area of common shapes.

For each of the shapes, there are also pages where you can see more examples and practice questions.

Shape | Formula |
---|---|

\text{Area}=\text{base} \times \text{height}
| |

Triangle | Area=\cfrac{1}{2} \times base \times height \, (perpendicular) |

**Step-by-step guide: **Area of a triangle

**Step-by-step guide:** Area of a right triangle

**Step-by-step guide:** Area of isosceles triangle

**Step-by-step guide: **Area of equilateral triangle

**Step-by-step guide:** Area of obtuse triangle

A **composite shape** can be split into two or more smaller shapes and the area of these shapes can be found. The total area of the composite shape can be found by adding the smaller areas together.

You can also find the area of **composite shapes **by breaking them up into triangles and rectangles and using the formulas.

\hspace{1cm} **Composite Shapes (composed of triangles and rectangles)**

Shape | Solving Strategy/Formula |
---|---|

Break the figure up into rectangles, find the area of each \begin{aligned} Area &= 4 \times 9+6 \times 2 \\ &=36+12 \\ &=48 \, cm^2 \end{aligned} | |

Composite shapes | Break the figure up into rectangles and triangles, find the \begin{aligned} Area &= 11 \times 7+\cfrac{1}{2} \times 7 \times 2 \\ &=77+7\\ &=84 \, m^2 \end{aligned} |

**Step-by-step guide:** Area of composite shapes

Quadrilaterals | Solving Strategy/Formula |
---|---|

Parallelograms are made of a rectangle and two triangles. | |

A rhombus can be broken up into rectangles and triangles. | |

Trapezoid | Trapezoids are made of a rectangle and one or two triangles. |

Use this quiz to check your grade 4 to 6 students’ understanding of area. 15+ questions with answers covering a range of 4th, 5th and 6th grade area topics to identify areas of strength and support!

DOWNLOAD FREEUse this quiz to check your grade 4 to 6 students’ understanding of area. 15+ questions with answers covering a range of 4th, 5th and 6th grade area topics to identify areas of strength and support!

DOWNLOAD FREEHow does this relate to 3rd grade math, 4th grade math, 6th grade math and 7th grade math?

**Grade 3 – Measurement and Data (3.MD.C.7)**Relate area to the operations of multiplication and addition.

a. Find the area of a rectangle with whole-number side lengths by tiling it, and show that the area is the same as would be found by multiplying the side lengths.

b. Multiply side lengths to find areas of rectangles with whole number side lengths in the context of solving real world and mathematical problems, and represent whole-number products as rectangular areas in mathematical reasoning.

c. Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a \times b and a \times c . Use area models to represent the distributive property in mathematical reasoning.

d. Recognize area as additive. Find areas of rectilinear figures by decomposing them into non-overlapping rectangles and adding the areas of the non-overlapping parts, applying this technique to solve real world problems.

**Grade 4 – Measurement and Data (4.MD.A.3)**Apply the area and perimeter formulas for rectangles in real world and mathematical problems. For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor.

**Grade 6 – Geometry (6.G.A.1)**

Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.

**Grade 7 – Geometry (7.G.B.6)**

Solve real-world and mathematical problems involving area, volume and surface area of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms.

In order to calculate the area of a shape with a formula:

**Write down the formula for the area of the shape.****Identify the dimensions required for the formula.****Substitute the dimensions and calculate.****Write down your final answer with units squared.**

Find the area of this rectangle below:

**Write down the formula for the area of the shape.**

Rectangle:

\text { Area }=\text { base } \times \text { height }

2**Identify the dimensions required for the formula.**

\begin{aligned}& \text { Base }=7 \mathrm{~m} \\\\ & \text { Height }=4 \mathrm{~m}\end{aligned}

3**Substitute the dimensions and calculate.**

\begin{aligned}\text { Area } & =\text { base } \times \text { height } \\\\ & =7 \times 4 \\\\ & =28\end{aligned}

4**Write down your final answer with units squared.**

In this case, we are working with meters, so our final answer must be in square meters.

\text { Area }=28 \mathrm{~m}^{2}

Find the area of the right triangle below:

**Write down the formula for the area of the shape.**

Triangle:

\text { Area }=\cfrac{1}{2} \, \times \text { base } \times \text { height }

**Identify the dimensions required for the formula.**

\begin{aligned}& \text { Base }=8 \mathrm{~cm} \\\\ & \text { Height }=6 \mathrm{~cm}\end{aligned}

**Substitute the dimensions and calculate.**

\begin{aligned}\text { Area } & =\cfrac{1}{2} \, \times \text { base } \, \times \text { height } \\\\ & =\cfrac{1}{2} \, \times 8 \, \times 6 \\\\ & =24\end{aligned}

**Write down your final answer with units squared.**

In this case, we are working with centimeters, so our final answer must be in square centimeters.

\text { Area }=24 \mathrm{~cm}^{2}

Find the area of the isosceles triangle below:

**Write down the formula for the area of the shape.**

Triangle:

\text { Area }=\cfrac{1}{2} \, \times \text { base } \, \times \text { height }

**Identify the dimensions required for the formula.**

\begin{aligned}& \text { Base }=7.7 \mathrm{~m} \\\\ & \text { Height }=20.9 \mathrm{~m}\end{aligned}

**Substitute the dimensions and calculate.**

\begin{aligned}\text { Area } & =\cfrac{1}{2} \, \times \text { base } \, \times \text { height } \\\\ & =\cfrac{1}{2} \, \times 7.7 \, \times 20.9 \\\\ & =80.465\end{aligned}

**Write down your final answer with units squared.**

\text { Area }=80.465 \mathrm{~m}^2

Find the area of the parallelogram below:

**Write down the formula for the area of the shape.**

Parallelogram:

\text { Area }=\text { base } \times \text { height }

**Identify the dimensions required for the formula.**

\text { Base }=6 \mathrm{~cm}

\text {Perpendicular height} =30 \mathrm{~mm}

All units need to be the same to calculate the area.

Let’s change 30 \, mm to 3 \, cm .

\text { Height }=3 \mathrm{~cm}

**Substitute the dimensions and calculate.**

\begin{aligned}\text { Area }&= \text { base } \times \text { height } \\\\ &=6 \times 3 \\\\ &=18\end{aligned}

**Write down your final answer with units squared.**

\text { Area }=18 \mathrm{~cm}^{2}

Find the area of the trapezoid below:

**Write down the formula for the area of the shape.**

This trapezoid is made up of two congruent triangles and a rectangle.

Rectangle:

\text { Area }=\text { base } \times \text { height }

Triangle:

\text { A }=\cfrac{1}{2} \times \text { base } \times \text { height }

**Identify the dimensions required for the formula.**

Rectangle:

\begin{aligned}& \text { Base }=5 \mathrm{~m} \\\\ & \text { Height }=4 \mathrm{~m}\end{aligned}

Triangle:

\begin{aligned}
& \text { Base }=2 \mathrm{~m} \\\\
& \text { Height }=4 \mathrm{~m}
\end{aligned}

**Substitute the dimensions and calculate.**

Rectangle:

\begin{aligned}\text { Area } & =\text { base } \times \text { height } \\\\ & =5 \times 4 \\\\ & =20\end{aligned}

Triangle:

\begin{aligned}\text { Area } & =\cfrac{1}{2} \, \times \text { base } \times \text { height } \\\\
& =\cfrac{1}{2} \, \times 2 \times 4 \\\\
& =4
\end{aligned}

Add the area of the rectangle and the two triangles together:

20 + 4 + 4 = 28

This can also be solved with the formula for trapezoids – since the two bases and height are given.

\begin{aligned}& \text { Area }=\cfrac{1}{2} \, (a+b) h \\\\ & \text { Base } A=5 \mathrm{~cm} \\\\ & \text { Base } B=9 \mathrm{~cm} \\\\ & \text { Height }=4 \mathrm{~cm} \\\\ \end{aligned}

\begin{aligned}\text { Area } & =\cfrac{1}{2} \, (5+9) 4 \\\\
& =\cfrac{1}{2} \,(14) \, 4 \\\\
& =28\end{aligned}

**Write down your final answer with units squared.**

\text {Area }=28 \mathrm{~m}^{2}

Find the area of the composite shape below:

The floor shown below needs to be tiled. Each pack of tile covers a distance of 12 square ft. How many packs of tile will be needed to cover the entire floor?

**Write down the formula for the area of the shape.**

Split the composite shape into three rectangles and label them A, B and C .

Rectangle:

\text { Area }=\text { base } \times \text { height }

**Identify the dimensions required for the formula.**

Notice that rectangle A and C each have a tick mark at the bottom. This means that these sides are congruent, or equal. Therefore, we know the width of rectangle A is 5.5ft .

Rectangle A :

Length=13 \mathrm{ft}

Width =5.5 \mathrm{ft}

Rectangle C is congruent to Rectangle A , so the dimensions are the same.

Rectangle B :

You need to calculate the unknown width. To do this, use the measurements you are given in the question that are parallel to the unknown side:

The entire width of the shape is 13ft .

Subtract 5.5ft and 5.5ft (the width of rectangles A and C ) to find the missing width of rectangle B: 13 - 5.5 - 5.5 = 2ft .

\begin{aligned}& \text { Length }=9 \mathrm{ft} \\\\
& \text { Width }=2 \mathrm{ft}\end{aligned}

**Substitute the dimensions and calculate.**

Area of Rectangle A and C :

\begin{aligned}\text { Area } & =\text { length } \times \text { width } \\\\ & =13 \times 5.5 \\\\ & =71.5\end{aligned}

Area of Rectangle B:

\begin{aligned}\text { Area } & =\text { length } \times \text { width } \\\\
& =9 \times 2 \\\\
& =18\end{aligned}

\begin{aligned}\text { Composite Area }&= \text { Area of } A+\text { Area of } B+\text { Area of } C \\\\ & =71.5+18+71.5 \\\\ & =161\end{aligned}

**Write down your final answer with units squared.**

\text { Composite Area }=161 \mathrm{ft}^2

Since each pack of tiles covers 12 square ft, divide the composite area by 12 to see how many packs are needed.

161 \div 12=13.416

Since a part of a pack cannot be bought, it will take 14 packs to cover the floor.

- Before learning the area formulas, allow students to explore the area by tiling with square tiles or by using shapes that are shown on square grids. This will help students build an understanding of area before they begin calculating it.

- Rather than having students practice finding the area with just worksheets, give them a variety of activities and projects that have a real-world context.

- Be sure to expose students to all types of triangles (acute, right and obtuse) and quadrilaterals in all different orientations when working with area. It is important that students can learn to identify the base and the height of a shape no matter its position. Always look for worksheets or study materials that provide this type of variety.

- For common shapes and composite shapes, allow students opportunities to create their own (either on paper, using manipulatives, or using an interactive website) and have a partner find the area.

**Using the incorrect unit of area or not including units**

Area calculations are always measured in square units.

For example,

Square millimeters (\mathrm{mm}^2) , square centimeters (\mathrm{cm}^2) , square meters (\mathrm{m}^2) , square kilometers (\mathrm{km}^2) , square inches (\mathrm{inch}^2) , square feet (\mathrm{ft}^2) , square yards (\mathrm{yd}^2) , square miles (\mathrm{mi}^2) , etc.

**Using an incorrect area formula**

There are several different area formulas for the different shapes – make sure you use the correct one.

**Confusing perimeter, area and volume**

Two dimensional shapes have a perimeter and an area. Three dimensional figures have a surface area and a volume. It is easy to confuse these terms and their meaning, if you don’t have a deep understanding of what they are.

For example,

**Calculating with different units**

All measurements must be in the same units before calculating the area.

For example,

To find the area of this rectangle, the units need to be all ft or all inches .

Since 24 \text { inches}=2 \mathrm{ft} , the dimensions 5ft and 2ft can be multiplied to calculate the area of this rectangle in feet (10 \mathrm{ft}^2) .

1. Find the area of the rectangle.

11 \mathrm{~m}^2

22 \mathrm{~m}^2

24 \mathrm{~m}^2

12 \mathrm{~m}^2

Use the formula:

\text{Area }= \text{ base } \times \text{ height}

8 \times 3=24\mathrm{~cm}^{2}

2. Find the area of the triangle.

8 \, \cfrac{7}{16} \text { units }^2

16 \, \cfrac{7}{8} \text { units }^2

23 \, \cfrac{5}{20} \text { units }^2

209 \, \cfrac{5}{20} \text { units }^2

The height of the triangle is always perpendicular to the base.

\text { Base }=3 \cfrac{3}{4}

\text { Height } =4 \cfrac{1}{2}

\begin{aligned}A & =\cfrac{1}{2} \, \times \text { base } \times \text { height } \\\\ & =\cfrac{1}{2} \times 3 \cfrac{3}{4} \, \times 4 \cfrac{1}{2} \\\\ & =\cfrac{1}{2} \times \cfrac{15}{4} \, \times \cfrac{9}{2} \\\\ & =\cfrac{135}{16} \\\\ & =8 \cfrac{7}{16} \text { units }^2\end{aligned}

3. Find the area of the triangle.

12 \mathrm{~cm}^2

10 \mathrm{~cm}^2

7.5 \mathrm{~m}^2

6 \mathrm{~cm}^2

The height of the triangle is always perpendicular to the base.

Rotate the triangle to better see the height or look for the square right angle symbol.

Height =4 \mathrm{~cm}

Base =3 \mathrm{~cm}

\begin{aligned}A&=\cfrac{1}{2} \, \times \text{base} \times {height} \\\\ &=\cfrac{1}{2} \,\times \, 4\times 3\\\\ &=6 \mathrm{~cm}^{2}\\\end{aligned}

4. Find the area of the parallelogram in \mathrm{m}^2.

36 \mathrm{~m}^2

41.76 \mathrm{~m}^2

29 \mathrm{~m}^2

3,600 \mathrm{~m}^2

720 \mathrm{~cm}=7.2 \mathrm{~m}

\begin{aligned} & \text { Base }=7.2 \mathrm{~m} \\\\ & \text { Height }=5 \mathrm{~m}\end{aligned}

*Be careful not to confuse the perpendicular height with the diagonal side.

\begin{aligned}\text { Area } & =\text { base } \times \text { height } \\\\ & =7.2 \times 5 \\\\ & =36 \mathrm{~m}^2\end{aligned}

5. Find the area of the trapezoid below in \mathrm{cm}^2.

6,020 \mathrm{~cm}^2

5,250 \mathrm{~cm}^2

4,480 \mathrm{~cm}^2

5,762 \mathrm{~cm}^2

Convert all units to centimeters. Convert 0.64 \, m into 64 \, cm and 860 \, mm into 86 \, cm .

The trapezoid is made up of a rectangle and a triangle.

For the rectangle, use the formula:

\begin{aligned}\text { Area } & =\text { base } \times \text { height } \\\\ & =64 \times 70 \\\\ & =4,480\end{aligned}

To find the area of the triangle, you need to find the base length. To do this, use the measurements you are given in the question that are parallel to the unknown side.

The opposite side of the square is also 64 \, cm . If the entire side is 86 \, cm , subtract 64 from 86 to find the missing base length.

86 \, – \, 64 = 22

For the triangle, use the formula:

\begin{aligned}\text { Area } & =\cfrac{1}{2} \, \times \text { base } \times \text { height } \\\\ & =\cfrac{1}{2} \, \times 22 \times 70 \\\\ & =770\end{aligned}

Add the area of the rectangle and triangle together, for the total area of the trapezoid:

4,480+770=5,250 \mathrm{~cm}^2

This can also be solved with the formula.

\begin{aligned}\text{Area }&=\cfrac{1}{2}(a+b)h\\\\ &=\cfrac{1}{2} \, \times (64+86) \times 70\\\\ &=\cfrac{1}{2} \, \times 150 \times 70\\\\ &=5250 \mathrm{~cm}^{2}\end{aligned}

6. Below is the front side of a piece of wood that needs to be covered in paint. Joey has 45.34 {~cm}^2 of paint. How much more does he need to cover the front side?

88 \mathrm{~cm}^2

110 \mathrm{~cm}^2

53.5 \mathrm{~cm}^2

42.66 \mathrm{~cm}^2

The pentagon is made up of a rectangle and a triangle.

\begin{aligned}\text{Area of rectangle }&=\text{ base } \times \text{ height}\\\\ &=11 \times 6\\\\ &=66 \mathrm{~cm}^{2}\end{aligned}

\begin{aligned}\text{Area of triangle }&=\cfrac{1}{2} \, \times \text{ base } \times \text{ height }\\\\ &=\cfrac{1}{2} \, \times 11 \times 4\\\\ &=22 \mathrm{~cm}^{2}\end{aligned}

Total area: 66+22=88 \mathrm{~cm}^{2}

To find how much more paint Joey needs, subtract what he has from the total amount needed.

88 \, – \, 45.34 = 42.66

Joey needs 42.66 \mathrm{~cm}^{2} more of paint to cover the front side

Area is the measurement of the space inside of a 2D closed figure.

Yes. Some irregular shapes have formulas, like rectangles and triangles. However, many do not. If an irregular shape does not have a formula, it can be broken up into simple shapes (like rectangles or triangles), finding the area of each part and then adding them all together to find the total area of the irregular shape.**Step-by-step guide: **Area of irregular shapes

Yes, the area of a circle is found with the formula \Pi r^2 . This is taught in 7th grade, per the Common Core State Standards.

They are also referred to as ‘closed figures’ or ‘plane figures.’

- 3D shapes
- Volume
- Area of a quadrilateral
- Area of a hexagon
- Area of a pentagon

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