Sign In
| | 18 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Consider a rectangle with length l and width w. The area of the rectangle is the product of its dimensions, A=lw. If a diagonal is drawn across the rectangle, the shape is divided into two triangles.
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=1/2bh
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.
First, consider the particular case of a right triangle. It is always possible to reflect a right triangle across its hypotenuse to form a rectangle.
Note that the area of the rectangle formed is twice the area of the original right triangle. Because of this, the formula for the area of the rectangle, A_r=l w, can be used to find the area of the right triangle. A_r = 2A_t ⇒ l w = 2A_t Furthermore, the height and base of the right triangle have the same measures as the width and length of the rectangle formed by reflecting the triangle. Based on this observation, b and h can be substituted for l and w, respectively, to solve for the area of the original right triangle in terms of its base and height.
This shows that the area of a right triangle can be calculated by using the formula A = 12bh.
To generalize the previous result, it is useful to note that any non-right triangle can be split into two right triangles by drawing one of its heights.
Note that the area of the non-right triangle A is equal to the sum of the individual areas of the smaller right triangles A_1 and A_2. Therefore, it is possible to calculate the area of the non-right triangle by using the previous result for the areas of the smaller right triangles.
Add fractions
Factor out h
a/b=1/b* a
b_1+b_2= b
It has been found that the area of the non-right triangle is half the product of its base b and its height h. This is the same result as the area for a right triangle. Therefore, the area of any triangle is half the product of its base b and its height h.
A = 1/2bh
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.
Find the area of the cat's hind leg, Tan 1.
If the right foot has an area of 4 square centimeters, what is the value of x?
A_(△) = 1/2bh The triangle's base can be any of its sides. The height is the segment that is perpendicular to the base and connects the base with its opposite vertex.
From the diagram, the base of the hind leg is 8 centimeters and the height is 4 centimeters. Substitute these values into the formula to find the area of Tan 1.
b= 8, h= 4
Multiply
a/c* b = a* b/c
Identity Property of Multiplication
Calculate quotient
The area of the cat's hind leg is 16 square centimeters.
A_(△) = 1/2bh The 4-centimeter-long side acts as a base of the triangle, so the x-centimeter-long segment is its corresponding height.
Therefore, substitute 4 for the area, 4 for the base, and x for the height into the formula. Then, solve the equation for x to find the height of the triangle.
Substitute values
a/c* b = a* b/c
Calculate quotient
.LHS /2.=.RHS /2.
Rearrange equation
The height of the triangle is 2 centimeters.
A parallelogram is a quadrilateral with two pairs of parallel sides. Parallelograms can be divided into three main types: rectangles, rhombuses, and squares.
| 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.
Mark continued playing with the tangram and learned to make different animal shapes, including a swan and a rabbit.
The area of a parallelogram is the product of its base and height. The base can be any side and the height is the perpendicular distance to the opposite side. A = bh For Tan 7, the longest sides measure 4 centimeters and the perpendicular distance between them is 2 centimeters. Therefore, to find the area of the piece, substitute 4 for b and 2 for h into the formula and simplify. A = 4* 2 ⇒ A = 8 The area of the swan's neck is 8 square centimeters.
The longest sides have a length of 8 centimeters, and the shape has an area of 32 square centimeters. Substitute these values into the formula for the area of a parallelogram to determine the height, represented by w.
Substitute values
.LHS /8.=.RHS /8.
Rearrange equation
The height of the large parallelogram is 4 centimeters.
Parallelograms can be divided into three main types: rectangles, rhombuses, and squares. It is the time to learn about rhombuses.
A rhombus is a parallelogram with four congruent sides. In other words, a rhombus is a quadrilateral with two pairs of parallel sides, all four of which have the same length.
Rhombuses are symmetric about both diagonals.
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.
Home plate and second base are about 127 feet apart, as are first and third bases. What is the area of the rhombus formed by the bases? Round the answer to the nearest hundred.
The diagonals of each rhombus are about 5 and 8.7 centimeters long. If the perpendicular distance between opposite sides is 4.3 centimeters, what is the side length of each rhombus? Round the answer to one decimal place.
A = 1/2d_1 d_2 The diagonals of the infield are formed by connecting the home plate and the second base, and also the first and third base.
The length of each diagonal is about 127 feet. Substitute this value into the formula for the area.
d_1= 127, d_2= 127
Multiply
a/c* b = a* b/c
Calculate quotient
The area of the rhombus formed by the bases, rounded to the nearest hundred, is 8100 square feet.
A = bh ⇒ A = s* 4.3 The side length can be found by first determining the area of each rhombus. The area of a rhombus is also equal to half the product of the diagonals. A = 1/2d_1 d_2 It is given that the diagonals are 5 and 8.7 centimeters long. The area of each rhombus can be found by substituting these values into the previous formula.
d_1= 8.7, d_2= 5
Multiply
a/c* b = a* b/c
Calculate quotient
Finally, substitute the area into the equation written at the beginning and solve it for s.
A= 21.75
.LHS /4.3.=.RHS /4.3.
Calculate quotient
Rearrange equation
Round to 1 decimal place(s)
The side length of each rhombus is about 5.1 centimeters.
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=1/2h(b_1+b_2)
Mark is getting ready to go to school. As he eats breakfast with his parents, he looks up and begins to see trapezoids everywhere.
What is the area of the lampshade?
The bases of the shell are about 4.7 and 3.1 inches long and it has an area of 6.2 square inches. How tall is the turtle's shell? Round the answer to one decimal place.
A = 1/2h(b_1+b_2) It is said that the bases of the lampshade are 12 and 6 inches long and that it is 10 inches tall. Substitute these values into the formula for the area.
Substitute values
Add terms
Multiply
a/c* b = a* b/c
Calculate quotient
The lampshade has an area of 90 square inches.
b_1 = 4.7 b_2 = 3.1 A = 6.2 The shell's height can be found by using the formula for the area of a trapezoid, which is half the height times the sum of the lengths of the bases. A = 1/2h(b_1+b_2) Substitute the given information into the formula and solve it for h.
Substitute values
Add terms
Commutative Property of Multiplication
a/c* b = a* b/c
LHS * 2/7.8=RHS* 2/7.8
a*b/c= a* b/c
Multiply
Calculate quotient
Rearrange equation
Round to 1 decimal place(s)
The turtle's shell is about 1.6 inches tall.
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.
In this play, Mark's team forms a rhombus. What is the area of the rhombus the players form on the actual court? Round the answer to the nearest integer.
The area of the trapezoid the players form on the court is about 324.4 square meters. Find the height of the trapezoid. Round the answer to one decimal place.
A = 1/2d_1d_2 The length of the diagonals can be found with the help of the distance formula. d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) First, find the length of the diagonal connecting the vertices (-6,1) and (2,4).
Substitute ( -6,1) & ( 2,4)
a-(- b)=a+b
Add and subtract terms
Calculate power
Add terms
Next, find the length of the diagonal connecting the vertices (-3,5) and (-1,0).
Substitute ( -3,5) & ( -1,0)
a-(- b)=a+b
Add and subtract terms
Calculate power
Add terms
The lengths of the diagonals correspond to the rhombus on the whiteboard. Since the units on the whiteboard represent 2 meters on the court, multiply d_1 and d_2 by 2 to find the actual lengths.
| Length on the Whiteboard | Length on the Court |
|---|---|
| d_1 = sqrt(73) | d_1 = 2sqrt(73) |
| d_2 = sqrt(29) | d_2 = 2sqrt(29) |
Finally, substitute the actual lengths into the formula for the area of a rhombus.
d_1= 2sqrt(73), d_2= 2sqrt(29)
Commutative Property of Multiplication
Multiply
a/c* b = a* b/c
Calculate quotient
sqrt(a)*sqrt(b)=sqrt(a* b)
Multiply
Use a calculator
Multiply
Round to nearest integer
The area of the rhombus formed by the players on the court is about 92 square meters.
A = 1/2h(b_1+b_2) It is given that the area of the trapezoid is about 324.4 square meters. Notice that the length of the bases can be found by using the distance formula. Start with the longest base, the one connecting (4,5) and (6,-5).
Substitute ( 4,5) & ( 6,-5)
Next, find the length of the shorter base that connects (-6,1) and (-5,-4).
Substitute ( -6,1) & ( -5,-4)
As in Part A, notice that these lengths correspond to the trapezoid on the whiteboard. Multiply these lengths by 2 to determine the actual lengths on the court.
| Length on the Whiteboard | Length on the Court |
|---|---|
| b_1 = 10.2 | b_1 = 20.4 |
| b_2 = 5.1 | b_2 = 10.2 |
Finally, substitute the bases and the area into the formula for the area of a trapezoid and solve the resulting equation for the height.
Substitute values
Add terms
Commutative Property of Multiplication
a/c* b = a* b/c
Calculate quotient
.LHS /15.3.=.RHS /15.3.
Rearrange equation
Calculate quotient
Round to 1 decimal place(s)
The height of the trapezoid formed by the players on the court is about 21.2 meters.
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.
The area of the rocket is the sum of the areas of the three trapezoids that make it up. A_(Rocket) = A_(7,3) + A_(1,2,4) + A_(5,6) The three areas can all be found with the same formula. All the computations are summarized in the table below.
| Tans Forming the Trapezoid | Dimensions | A=1/2h(b_1+b_2) | Area (cm^2) |
|---|---|---|---|
| 7 and 3 | b_1 = 8 b_2=4 h=2 |
A_(7,3) = 1/2* 2(8+4) | 12 |
| 1, 2, and 4 | b_1 = 12 b_2=8 h=4 |
A_(1,2,4) = 1/2* 4(12+8) | 40 |
| 5 and 6 | b_1 = 2sqrt(8) b_2=sqrt(8) h=sqrt(8) |
A_(5,6) = 1/2* sqrt(8)(2sqrt(8)+sqrt(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.
What is the area of the pentagon?
What is the value of x?
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.
What is the area of the parking space?
Use the given dimensions to estimate the area of Nevada.
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.
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.