Sign In
| 18 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
A diagonal of a rectangle divides the rectangle into two right triangles. Because of this, a formula for the area of a right triangle can be derived from the formula for the area of a rectangle. The good news is that the same formula applies to any type of triangle!
The area of a triangle is half the product of its base b and its height h.
A=21bh
The triangle's base can be any of its sides. The height – or altitude – of the triangle is the segment that is perpendicular to the base and connects the base or its extension with its opposite vertex.
A=21bh
On his birthday, Mark's uncle gave him a tangram, a Chinese puzzle made of seven polygons that can be used to create different shapes. The seven individual pieces are called tans.
Mark's uncle warned him that once the pieces are taken out of the box, putting them back is a challenge.
b=8, h=4
Multiply
ca⋅b=ca⋅b
Identity Property of Multiplication
Calculate quotient
Substitute values
ca⋅b=ca⋅b
Calculate quotient
LHS/2=RHS/2
Rearrange equation
Property | Justification |
---|---|
The opposite sides are congruent | Parallelogram Opposite Sides Theorem |
The opposite angles are congruent | Parallelogram Opposite Angles Theorem |
The diagonals bisect each other | Parallelogram Diagonals Theorem |
These properties are illustrated graphically in the next diagram.
The area of a parallelogram is equal to the product of its base b and height h. The base can be any side of the parallelogram and the height is the perpendicular distance to the opposite side.
Parallelograms can be divided into three main types: rectangles, rhombuses, and squares. It is the time to learn about rhombuses.
The area of a rhombus is half the product of the lengths of the diagonals.
Alternatively, since a rhombus is a parallelogram, its area can also be calculated by multiplying its base and height.
Since Mark received the tangram puzzle, he sees polygons everywhere.
d1=127, d2=127
Multiply
ca⋅b=ca⋅b
Calculate quotient
d1=8.7, d2=5
Multiply
ca⋅b=ca⋅b
Calculate quotient
A=21.75
LHS/4.3=RHS/4.3
Calculate quotient
Rearrange equation
Round to 1 decimal place(s)
The following applet shows a general parallelogram or rhombus. Calculate the missing dimension of the given polygon. Round the answer to two decimal places.
A trapezoid is a quadrilateral with exactly one pair of parallel sides. The parallel sides are called the bases of the trapezoid, and the two other sides are called the legs. Two angles that have a base as a common side are called the base angles.
Trapezoids with congruent legs have a special name.
An isosceles trapezoid is a trapezoid whose legs are congruent.
Isosceles trapezoids have two main properties.
Property | Justification |
---|---|
The diagonals are congruent. | Isosceles Trapezoid Diagonals Theorem |
Each pair of base angles is congruent. | Isosceles Trapezoid Base Angles Theorem |
The area of a trapezoid is half the height times the sum of the lengths of the bases. In other words, the area of a trapezoid is the height multiplied by the average of the bases.
A=21h(b1+b2)
Mark is getting ready to go to school. As he eats breakfast with his parents, he looks up and begins to see trapezoids everywhere.
Substitute values
Add terms
Multiply
ca⋅b=ca⋅b
Calculate quotient
Substitute values
Add terms
Commutative Property of Multiplication
ca⋅b=ca⋅b
LHS⋅7.82=RHS⋅7.82
a⋅cb=ca⋅b
Multiply
Calculate quotient
Rearrange equation
Round to 1 decimal place(s)
Mark's futsal team has their final championship game tonight. They have been practicing different strategies for this game. The coach prepared some plays on a whiteboard with a coordinate system. One unit on the board represents 2 meters on the actual court.
Substitute (-6,1) & (2,4)
a−(-b)=a+b
Add and subtract terms
Calculate power
Add terms
Substitute (-3,5) & (-1,0)
a−(-b)=a+b
Add and subtract terms
Calculate power
Add terms
Length on the Whiteboard | Length on the Court |
---|---|
d1=73 | d1=273 |
d2=29 | d2=229 |
d1=273, d2=229
Commutative Property of Multiplication
Multiply
ca⋅b=ca⋅b
Calculate quotient
a⋅b=a⋅b
Multiply
Use a calculator
Multiply
Round to nearest integer
Length on the Whiteboard | Length on the Court |
---|---|
b1=10.2 | b1=20.4 |
b2=5.1 | b2=10.2 |
Substitute values
Add terms
Commutative Property of Multiplication
ca⋅b=ca⋅b
Calculate quotient
LHS/15.3=RHS/15.3
Rearrange equation
Calculate quotient
Round to 1 decimal place(s)
Sometimes a plane figure can be made up of two or more geometric shapes. These figure are called composite figures. The area of a composite figure is the sum of the areas of all the basic figures that make it up. A good example of composite figures are those formed by a tangram puzzle. For instance, consider the following rocket.
The area of the rocket is equal to the sum of the areas of all seven polygons. However, some pieces can be considered together to perform fewer computations. For example, the rocket can be seen as three trapezoids.
Tans Forming the Trapezoid | Dimensions | A=21h(b1+b2) | Area (cm2) |
---|---|---|---|
7 and 3 | b1=8 b2=4 h=2 |
A7,3=21⋅2(8+4) | 12 |
1, 2, and 4 | b1=12 b2=8 h=4 |
A1,2,4=21⋅4(12+8) | 40 |
5 and 6 | b1=28 b2=8 h=8 |
A5,6=21⋅8(28+8) | 12 |
The area of the rocket is 12+40+12=64 square centimeters. In fact, any figure that is formed using the seven tans will also have an area of 64 square centimeters! The following applet shows more examples and allows playing with the tans.
Find the required information in each of the following diagrams.
In the following pentagon, the segment connecting B and D is parallel to AE.
The area of the following figure is 17.
We do not know a formula for finding the area of the given pentagon. However, we can divide it into polygons for which we know how to find the areas. For example, let's draw a segment connecting B and D.
We can see that the upper polygon is a triangle. We are told that BD and AE are parallel, which means that ABDE is a trapezoid.
We can find the area of the pentagon by adding the area of the trapezoid and the area of the triangle. A = A_1 + A_2 Let's write the formulas for the area of each polygon.
Area of a Trapezoid | Area of a Triangle |
---|---|
A_1=1/2h(b_1+b_2) | A_2=1/2bh |
The height of the trapezoid is 3 and its bases are 3 and 7. Let's use these values to find its area.
In △ BCD, we can see that the height corresponding to BD is 2, so we will use BD=7 as the base of the triangle.
Finally, we add the two areas to get the area of the pentagon. A = 15+7 ⇒ A = 22 The area of the pentagon is 22 square units.
We need to find the perpendicular distance between AB and CD. We know the area of the whole figure, but we do not know a formula to find it. Let's divide the figure into polygons for which we know how to find the area. For example, let's connect vertices A and D with a segment.
This segment divides the figure into two polygons, ABCD and DEFA. But there is more! Since AB∥DC and DE∥FA, these two polygons are both trapezoids!
Notice that x is the height of trapezoid ABCD. We know the lengths of the bases of this trapezoid. If we knew its area, we could use the formula for the area to find the height. A = 1/2h(b_1+b_2) A = ? h = ? b_1 = 7 b_2 = 4 We do not know the area of either of the trapezoids, but we are given the area of the whole figure — we know that the areas of the trapezoids add up to 17. \begin{gathered} A_\text{Total}=A_{ABCD}+A_{DEFA}\\ \Updownarrow\\ 17=A_{ABCD}+A_{DEFA} \end{gathered} If we find the area of trapezoid DEFA, we will be able to find the area of ABCD, and from there, the height of ABCD. Let's find the area of DEFA.
The bases of trapezoid DEFA are DE and FA and notice that EF is the height. Therefore, let's substitute the corresponding lengths to find the area of this trapezoid.
The area of trapezoid DEFA is 6 square units.
Let's substitute 17 for A_\text{Total} and 6 for A_(DEFA) into the equation for the area of the whole figure. Then, we will solve the equation for A_(ABCD).
The area of trapezoid ABCD is 11 square units.
Now that we know the area of trapezoid ABCD, we can substitute it along with the lengths of the bases into the formula for the area and solve it for the height, x.
We found that the height of the trapezoid, x, is 2 units.
The parallel lines of an average parking are 18 feet long and about 9 feet apart.
Some dimensions of the state of Nevada are shown in the following diagram.
Let's start by noticing that a parking space usually has the shape of a parallelogram.
We are told that the parallel lines of the parking space are 18 feet long and that the distance between the lines, which is also the height of the parallelogram, is 9 feet.
We can find the area of a parallelogram by multiplying the base and height.
The area of the parking space is 162 square feet.
Notice that the state of Nevada forms a rough trapezoid. This means that we can estimate its area with the following formula.
A = 1/2h(b_1+b_2)
The bases are the parallel sides and the height is the perpendicular distance between the bases.
From the diagram, the bases are 505 and 205 miles long and the height is 315 miles. Let's substitute these values into the formula and simplify to find the area of the trapezoid.
The area of Nevada is about 111 825 square miles.
Find the areas of the following polygons.
We need to find the area of quadrilateral ABCD. Notice that AD is parallel to BC and that CD is parallel to AB. This means that ABCD is a parallelogram.
Even more, it seems like ABCD is a rhombus. We can verify this by finding the lengths of two consecutive sides. Let's start by calculating the distance between A and B by using the distance formula.
Let's follow the same procedure to find the distance between A and D. Our computations are summarized in the table below.
Points | d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | Distance |
---|---|---|
A(-3,-2) and D(0,-1) | d=sqrt(( 0-( -3))^2+( -1-( -2))^2) | d=sqrt(10) |
As we can see, AB and AD have the same length. This means that the four sides have the same length, so we know that ABCD is a rhombus. We can then find the area of ABCD by calculating half the product of the lengths of the diagonals. A = 1/2d_1d_2 Let's use the distance formula to find the lengths of the diagonals, starting with AC.
Now let's find the length of BD.
We are ready to find the area of ABCD! Let's do it!
The given polygon is a composite figure. We can find its area by dividing it into polygons that we can find the areas of by using the formulas we know. For instance, let's draw a segment connecting Q and T.
We divided the polygon into a triangle and a rectangle. The area of the polygon is the sum of the areas of these two polygons. A = A_(△) + A_r
The triangle is a right triangle, so its area is half the product of its legs. Let's use the distance formula to find the lengths of the legs, starting with the distance between O and P.
Next, let's find the length of PT.
We are ready to find the area of △ OPT.
The area of rectangle QRST is the product of its dimensions. Let's find the length of QR.
Next, let's find the length of RS.
Now let's multiply the length and the width to find the area of the rectangle.
Let's substitute the area of the triangle and the area of the rectangle into the equation for the area of the composite figure. A = A_(△) + A_r ⇒ A= 12 + 12 Finally, let's add them together to find the area of the composite figure. A=12+12 ⇓ A = 24 The area of the composite figure is 24 square units.