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.
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.
Parallelograms can be easily rearranged into rectangles. Explore this using the applet below.
After using the applet above, we can make the following observations:
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.
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 cm^{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 m^{2}.
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.
Consider the following parallelogram.
If the parallelogram is formed into a rectangle, what would the length and width of the rectangle be?
Length: | $\editable{}$ cm |
Width: | $\editable{}$ cm |
Hence find the area of the parallelogram.
Find the area of the parallelogram shown.
Find the area of a parallelogram whose base is $15$15 cm and height is $7$7 cm.
A trapezoid is a 2D shape where $1$1 pair of opposite sides that are parallel.
All of these are trapezoids:
Let's see if we can figure out a formula for finding the area of a trapezoid by using the applet below.
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:
$\text{Area of a Trapezoid}=\frac{1}{2}\times\left(\text{Base 1 }+\text{Base 2 }\right)\times\text{Height }$Area of a Trapezoid=12×(Base 1 +Base 2 )×Height
$A=\frac{1}{2}\times\left(b_1+b_2\right)\times h$A=12×(b1+b2)×h
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.
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.
Do: Use the formula for the area of a trapezoid.
$\text{Area of a Trapezoid}$Area of a Trapezoid | $=$= | $\frac{1}{2}\times\left(\text{Base 1 }+\text{Base 2}\right)\times\text{Height }$12×(Base 1 +Base 2)×Height |
$=$= | $\frac{1}{2}\times\left(b_1+b_2\right)\times h$12×(b1+b2)×h | |
$=$= | $\frac{1}{2}\times\left(4+8\right)\times3$12×(4+8)×3 | |
$=$= | $\frac{1}{2}\times12\times3$12×12×3 | |
$=$= | $18$18 cm^{2} |
Now we are able to find the areas of rectangles, parallelograms, and trapezoids.
Consider the trapezoid shown below which has been split into a rectangle and a right triangle.
Find the area of the rectangle.
Find the area of the triangle.
Now find the area of the trapezoid.
Find the area of the trapezoid by first calculating the areas of the triangle and rectangle that comprise it.
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.
Use the applet below in order to develop a formula for finding the 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.
$\text{Area of a Rhombus }=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$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.
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?
Think: We need to identify the two diagonals.
Do: Use the formula for area of a rhombus.
$\text{Area of a Rhombus }$Area of a Rhombus | $=$= | $\frac{1}{2}\times\text{diagonal 1 }\times\text{diagonal 2 }$12×diagonal 1 ×diagonal 2 |
$=$= | $\frac{1}{2xy}$12xy | |
$=$= | $\frac{1}{2}\times16\times4$12×16×4 | |
$=$= | $32$32 cm^{2} |
Consider the rhombus on the left.
If the rhombus is formed into the rectangle on the right, what would the length and the width of the rectangle be?
Length: | $\editable{}$ mm |
Width: | $\editable{}$ mm |
Hence find the area of the rhombus.
Find the area of the rhombus by forming a rectangle.
A kite is a 2D shape with $2$2 pairs of adjacent sides that are equal in length and one pair of opposite angles that are equal in measure.
Of course the kite kids fly around on a windy day is named after the geometric shape it looks like.
Use the applet below to discover the formula for the area of a kite.
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.
$\text{Area of a Kite}=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$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.
Think: I need to identify the long diagonal length and the short diagonal length.Find the area of the kite shown below.
Do:
$\text{Area of a Kite }$Area of a Kite | $=$= | $\frac{1}{2}\times x\times y$12×x×y |
$=$= | $\frac{1}{2}\times4\times\left(2\times0.9\right)$12×4×(2×0.9) | |
$=$= | $\frac{1}{2}\times4\times1.8$12×4×1.8 | |
$=$= | $3.6$3.6 mm^{2} |
The kite on the left can be split into two triangles as shown.
Find the area of one of the triangles.
Now find the area of the kite.
Find the area of the kite by summing the areas of the triangles that make it.
Find 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.