A parallelogram is a quadrilateral with two pairs of opposite sides parallel. A rectangle is a special type of parallelogram but parallelograms do not have to have right angles.
Both are examples of parallelograms.
You may recall that we can find the area of a rectangle using the formula $\text{Area }=\text{length }\times\text{width }$Area =length ×width , and we will see that finding the area of a parallelogram is very similar. We will make use of the base and perpendicular height of the parallelogram to find its area.
Notice that a rectangle is a type of parallelogram, but not all parallelograms are rectangles. Why might this be? Think of what each shape has in common and how they differ.
Exploration
Parallelograms can be easily rearranged into rectangles. Explore this using the applet below.
Click and drag the blue circle on the $\text{Slide }$Slide slider. This will rearrange the parallelogram into a rectangle.
Click and drag the blue circle on the $\text{Slant }$Slant slider. Does this change the area of the shape?
Click the button $\editable{\text{Change dimensions}}$Change dimensions. You can now adjust the base and height to make a new parallelogram. This can be done with the $b$b slider and $h$h slider or by dragging the vertices of the parallelogram.
The area of the parallelogram is being calculated with a formula as you change its dimensions. What part of the calculation changes when you change a dimension? Can you work out the formula?
Click the button $\editable{\text{Decompose }}$Decompose to see if the new parallelogram can also rearrange into a rectangle.
Click the button $\editable{\text{Show grid}}$Show grid for a grid. Assume that this is a square centimeter grid. To remove it, click the button $\editable{\text{Hide grid}}$Hide grid.
After using the applet above, we can make the following observations:
Changing the slant of the parallelogram without changing the base and height did not affect its area. This means that the area of a parallelogram depends only upon its base and its perpendicular height, not the slanted height.
The base of the parallelogram is the same as the length of the rectangle it creates.
The perpendicular height of the parallelogram is the same as the width of the rectangle it creates.
As the area of a rectangle can be found with $\text{Area }=\text{length }\times\text{width }$Area =length ×width , then the area of a parallelogram can be found in a similar way.
Formula for the area of a parallelogram
The area of a parallelogram is given by
$\text{Area }=\text{base }\times\text{height }$Area =base ×height , or
$A=b\times h$A=b×h
Unlike a rectangle, there are generally no right angles in a parallelogram. But we should remember that the height and base are at right angles to each other when we work out the area of a parallelogram.
Worked examples
Question 1
Find the area of the parallelogram below.
Think: This parallelogram has a base of $6$6 cm and a height of $4$4 cm. We can rearrange it into a rectangle with length $6$6 cm and width $4$4 cm.
This rectangle has the same area as the parallelogram, which means we can find the area of the parallelogram by calculating the product of its base and height.
Do: We can use the given dimensions in the formula to find the area.
$\text{Area }$Area
$=$=
$\text{base }\times\text{height }$base ×height
(Formula for the area of a parallelogram)
$=$=
$6\times4$6×4
(Substitute the values for the base and height)
$=$=
$24$24
(Perform the multiplication to find the area)
So the parallelogram has an area of $24$24 cm2.
Question 2
What is the area of this parallelogram?
Think: The base always refers to a side of the parallelogram, while the height is the perpendicular distance between two opposite sides. In this parallelogram the base is $12$12 m and the height is $17$17 m.
Do: We can use the given dimensions in the formula to find the area.
$\text{Area }$Area
$=$=
$\text{base }\times\text{height }$base ×height
(Formula for the area of a parallelogram)
$=$=
$12\times19$12×19
(Substitute the values for the base and height)
$=$=
$228$228
(Perform the multiplication to find the area)
So the parallelogram has an area of $228$228 m2.
Reflect: Sometimes the height will be labeled within the parallelogram, and sometimes it will be convenient to indicate the height with a label outside the parallelogram.
Practice questions
Question 3
Consider the following parallelogram.
A parallelogram is depicted with a base labeled as 10cm and a vertical height marked as 6cm, indicated by a dashed line. The height is drawn perpendicular to the base, forming a shaded right triangle, which is denoted by a square at the intersection of the height and the base. There are double arrows on two opposing bases, suggesting congruency.
If the parallelogram was cut and reformed into the shape of a rectangle, what would the length and width of that rectangle be?
Length:
$\editable{}$cm
Width:
$\editable{}$cm
Therefore find the area of the parallelogram.
Question 4
Find the area of the parallelogram shown.
A parallelogram with a base labeled 3mm and height labeled 11mm. The top and bottom sides of the parallelogram are each marked with a single arrowhead indicating that they are parallel to each other. The left and right sides of the parallelogram are each marked with double arrowhead, also indicating that they are parallel to each other.
Question 5
Find the area of a parallelogram whose base is $15$15 cm and height is $7$7 cm.
Area of a trapezoid
A trapezoid is a 2D shape where $1$1 pair of opposite sides that are parallel.
All of these are trapezoids:
Exploration
Let's see if we can figure out a formula for finding the area of a trapezoid by using the applet below.
Drag the point D anywhere you like to create a trapezoid of your choice.
Slide the green slider to the right. What is created?
Can we use a formula we already know to create a formula for finding the area of a trapezoid?
The applet shows that if we rotate a trapezoid $180^\circ$180° and connect it to the original trapezoid then a parallelogram is created. We can see that the trapezoid has half of the area of the parallelogram we created.
Recall that we also know that a parallelogram can be used to create a rectangle. Since $\text{Area of a rectangle }=L\times W$Area of a rectangle =L×W, we can work out the area of any trapezoid.
We tend to call the top side of the trapezoid base $1$1 or $b_1$b1 and we call the bottom side base $2$2 or $b_2$b2. When we flip it both of those sides make up the base of the new parallelogram. We also need the height, h, of the trapezoid.
This gives us the following formula:
Formula for the area of a trapezoid
$\text{Area of a Trapezoid}=\frac{1}{2}\times\left(\text{Base 1 }+\text{Base 2 }\right)\times\text{Height }$Area of a Trapezoid=12×(Base 1 +Base 2 )×Height
You might wonder why we multiply by $\frac{1}{2}$12 in the formula. Remember, we flipped the trapezoid over to turn it into a parallelogram. That means that $2$2 trapezoids create the parallelogram. Since we only want the area of $1$1 trapezoid we must multiply by $\frac{1}{2}$12.
Let's look at an example using the new formula.
Worked example
Question 6
A new chocolate bar is being made in the shape of a trapezoid. The graphic designer needs to know the area of the trapezoid to begin working on a wrapping design. The diagram below shows the dimensions of the new chocolate bar. Find its area.
Think: We need to identify the base $1$1, base $2$2, and the height. Find these values on the diagram.
Now we are able to find the areas of rectangles, parallelograms, and trapezoids.
Practice questions
QUESTION 7
Consider the trapezoid shown below which has been split into a rectangle and a right triangle.
A right-angled trapezoid is shown, shaded in purple. The trapezoid consists of a rectangle and a right triangle on its right side. The trapezoid has a bottom base measured 19m and a top base measured 10m. The height of the trapezoid, hence also that of the rectangle, is measured 9m.
Find the area of the rectangle.
Find the area of the triangle.
Now find the area of the trapezoid.
QUESTION 8
Find the area of the trapezoid by first calculating the areas of the triangle and rectangle that comprise it.
A trapezoid with the horizontal top and bottom sides, which are also its bases, is shaded purple. The top and bottom sides are both perpendicular to the left side as indicated by a small square. The right side is slanted. The trapezoid can be divided into a rectangle and a triangle. The top side measures $6$6mm and the bottom side measures $10$10mm. The right side, which represents the height, measures $4$4mm.
Area of a rhombus
A rhombus is a 2D shape with $4$4 equal length sides, diagonals that bisect each other, and opposite sides that are equal in length and parallel.
Exploration
Use the applet below in order to develop a formula for finding the area of a rhombus.
Check the box to show the labels on the diagonals.
Slowly drag the slider all the way to the right.
What new shape has been created? Notice the formula that is shown. Why is this the formula for area of a rhombus?
In the applet above we saw that the inner triangles of the rhombus can be rearranged to create a rectangle. Of course, like the trapezoid, the original rhombus is only $\frac{1}{2}$12 of the new rectangle.
Because we know the $\text{Area of a rectangle }=L\times W$Area of a rectangle =L×W, we can work out the area of any rhombus.
Lets call the diagonals $x$x and $y$y. These give us the length and width of the rectangle that the rhombus fits inside.
Formula for the area of a rhombus
$\text{Area of a Rhombus }=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$Area of a Rhombus =12×diagonal 1×diagonal 2
$A=\frac{1}{2xy}$A=12xy
Let's look at an example using this new formula.
Worked example
Question 9
A packing box with a square opening is squeezed together creating the rhombus shown below. What is the area of the opening of the box?
In the applet above we saw that if we copy the inner triangles created by the diagonals and rearrange, a rectangle is created. Of course, like the trapezoid, our original shape is only 1/2 of this rectangle.
Because we know the $\text{Area of a rectangle }=L\times W$Area of a rectangle =L×W, we can work out the area of any kite.
We tend to call the long diagonal $x$x and we call the short diagonal of the kite $y$y. These give us the length and width of the rectangle that the kite fits inside.
Formula for the area of a kite
$\text{Area of a Kite}=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$Area of a Kite=12×diagonal 1×diagonal 2
$A=\frac{1}{2}\times x\times y$A=12×x×y
Let's try an example using this new formula.
Worked example
Question 12
Think: I need to identify the long diagonal length and the short diagonal length.Find the area of the kite shown below.
The kite on the left can be split into two triangles as shown.
The kite on the left is split into two triangles with its longer diagonal measuring $7$7m as the bases of the two triangles. The shorter diagonal of the kite measures $4$4m. The longer diagonal perpendicularly bisects the shorter diagonal into two equal segments forming the height of each triangle. The shorter adjacent sides of the kite are both marked with double tick marks, and the longer adjacent sides are both marked with single tick mark. Half of the kite to the left of the longer diagonal is shaded purple, and similarly, the triangle on the left is shaded purple.
Find the area of one of the triangles.
Now find the area of the kite.
QUESTION 14
Find the area of the kite by summing the areas of the triangles that make it.
A four sided shape resembling a kite is depicted. Its diagonals are drawn as a central vertical line and a horizontal line, intersecting at a right angle as indicated by a small square. The vertical line is labeled as 7m, and the horizontal line is marked as 4m. The left side of the kite is shaded. Single tick marks on two longer sides indicating congruent lengths. Double tick marks on the shorter sides also indicates congruence. The vertical diagonal diagonal divides the kite into 2 triangles: a shaded and unshaded.
Outcomes
6.G.1
Find area of right triangles, other triangles, special quadrilaterals, and polygons by composing and decomposing into rectangles, triangles and/or other shapes; apply these techniques in the context of solving real-world and mathematical problems.
