9. Areas of Parallelograms, Triangles, and Trapezoids
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Chapter 9
9.
Areas of Parallelograms, Triangles, and Trapezoids
This lesson explores methods for calculating the area of various geometric shapes, including parallelograms, triangles, trapezoids, and rhombuses. It provides insights into their properties and offers practical guidance to help learners solve geometry problems effectively.
Lesson Settings & Tools
| | 18 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Areas of Parallelograms, Triangles, and Trapezoids
Slide of 18
Area represents the amount of space enclosed within a two-dimensional figure. The method of calculating the area of a figure depends on the type of figure. This lesson will show how the formulas for the area of some figures can be derived from the formula for the area of a rectangle. Buckle up — the journey is about to begin!
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Dividing a Rectangle in Half
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.
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Discussion
Deriving the Formula for the Area of a Triangle
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!
Rule
Area of a 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.
Proof
Proof for Right Triangles
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.
Proof for Non-Right Triangles
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.
A = b_1h/2 + b_2h/2
▼
Simplify right-hand side
AddFrac
Add fractions
A = b_1h + b_2h/2
FactorOut
Factor out h
A = (b_1 + b_2)h/2
MoveNumRight
a/b=1/b* a
A = 1/2(b_1+b_2)h
Substitute
b_1+b_2= b
A = 1/2 bh
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
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Example
Tangram Puzzle
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.
a After trying for a while, Mark managed to form a cat.
Find the area of the cat's hind leg, Tan 1.
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b Mark's uncle showed him how to form a running person.
If the right foot has an area of 4 square centimeters, what is the value of x?
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Hint
a Piece 1 has the shape of a triangle, so use the formula for the area of a triangle.
b Use the formula for the area of a triangle and solve it for the height.
Solution
a The cat's hind leg has the shape of a triangle. The area of a triangle is half the base times the height.
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.
A_(△) = 1/2bh
SubstituteII
b= 8, h= 4
A_(△) = 1/2* 8* 4
Multiply
Multiply
A_(△) = 1/2* 32
MoveRightFacToNum
a/c* b = a* b/c
A_(△) = 1* 32/2
IdPropMult
Identity Property of Multiplication
A_(△) = 32/2
CalcQuot
Calculate quotient
A_(△) = 16
The area of the cat's hind leg is 16 square centimeters.
b The right foot of the running person is another triangle, so begin by recalling the formula for the area of a triangle again.
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.
A_(△) = 1/2bh
SubstituteValues
Substitute values
4 = 1/2* 4* x
MoveRightFacToNum
a/c* b = a* b/c
4 = 4/2* x
CalcQuot
Calculate quotient
4 = 2x
DivEqn
.LHS /2.=.RHS /2.
2 = x
RearrangeEqn
Rearrange equation
x = 2
The height of the triangle is 2 centimeters.
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Pop Quiz
Finding the Missing Dimension
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Discussion
Parallelograms
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.
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Discussion
Area of a Parallelogram
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.
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Example
Tangram Animals
Mark continued playing with the tangram and learned to make different animal shapes, including a swan and a rabbit.
a Find the area of the swan's neck, made up of Tan 7.
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b The rabbit's body is made of Tans 1 and 2 and has an area of 32 square centimeters. Find w.
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Hint
a The swan's neck is a parallelogram.
b The rabbit's body is a parallelogram.
Solution
a The swan's neck is represented by Tan 7, which is a parallelogram.
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.
b The way Tans 1 and 2 are placed forms a large parallelogram.
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.
A = bh
SubstituteValues
Substitute values
32 = 8 w
DivEqn
.LHS /8.=.RHS /8.
4 = w
RearrangeEqn
Rearrange equation
w = 4
The height of the large parallelogram is 4 centimeters.
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Discussion
Parallelograms With Congruent Sides
Parallelograms can be divided into three main types: rectangles, rhombuses, and squares. It is the time to learn about rhombuses.
Concept
Rhombus
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.
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Discussion
Area of a Rhombus
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.
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Example
Rhombuses Everywhere
Since Mark received the tangram puzzle, he sees polygons everywhere.
a Mark went to a baseball game with his family last Sunday. At one point, the game was stopped because second base was not in the right spot. Mark then realized that the bases form a rhombus.
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.
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b While walking to his parent's car in the parking lot, Mark saw a car's logo that was made of three identical rhombuses.
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.
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Hint
a The area of a rhombus the half the product of its diagonals.
b Use the lengths of the diagonals to find the area of each rhombus. Then, use the formula for the area of a parallelogram to determine the length of the base.
Solution
a The bases form a rhombus whose area can be found by using the following formula.
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.
A = 1/2d_1 d_2
SubstituteII
d_1= 127, d_2= 127
A = 1/2* 127* 127
Multiply
Multiply
A = 1/2* 16 129
MoveRightFacToNum
a/c* b = a* b/c
A = 16 129/2
CalcQuot
Calculate quotient
A = 8064.5
The area of the rhombus formed by the bases, rounded to the nearest hundred, is 8100 square feet.
b Let s be the rhombus side length. Notice that the perpendicular distance between opposite sides is the height of the rhombus. Since a rhombus is a parallelogram, its area can be found by multiplying its base and height.
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.
A = 1/2d_1 d_2
SubstituteII
d_1= 8.7, d_2= 5
A = 1/2* 8.7 * 5
Multiply
Multiply
A = 1/2* 43.5
MoveRightFacToNum
a/c* b = a* b/c
A = 43.5/2
CalcQuot
Calculate quotient
A = 21.75
Finally, substitute the area into the equation written at the beginning and solve it for s.
A = s* 4.3
Substitute
A= 21.75
21.75 = s* 4.3
DivEqn
.LHS /4.3.=.RHS /4.3.
21.75/4.3 = s
CalcQuot
Calculate quotient
5.058139... = s
RearrangeEqn
Rearrange equation
s = 5.058139...
RoundDec
Round to 1 decimal place(s)
s ≈ 5.1
The side length of each rhombus is about 5.1 centimeters.
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Pop Quiz
Solving Parallelograms and Rhombuses
The following applet shows a general parallelogram or rhombus. Calculate the missing dimension of the given polygon. Round the answer to two decimal places.
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Discussion
Trapezoids
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.
Concept
Isosceles Trapezoid
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 |
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Discussion
Area of a Trapezoid
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)
Extra
Graphical DerivationThe formula for the area of a trapezoid with bases b_1 and b_2 and height h can be derived by transforming the trapezoid into a triangle with base b_1+b_2 and height h.
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Example
A Head Full of Polygons
Mark is getting ready to go to school. As he eats breakfast with his parents, he looks up and begins to see trapezoids everywhere.
a The lampshade has bases that are 12 and 6 inches long and it is 10 inches tall.
What is the area of the lampshade?
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b When Mark saw Donatello, his pet turtle, he got the idea to make a turtle out of his tangram.
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.
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Hint
a The lampshade is a trapezoid.
b The area of a trapezoid is one-half the product of the height and the sum of the bases. Solve this formula for the height.
Solution
a The shade of the lamp is a trapezoid. Therefore, its area is half the height times the sum of the length of the bases.
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.
A = 1/2h(b_1+b_2)
SubstituteValues
Substitute values
A = 1/2* 10( 12+ 6)
▼
Simplify right-hand side
AddTerms
Add terms
A = 1/2* 10* 18
Multiply
Multiply
A = 1/2* 180
MoveRightFacToNum
a/c* b = a* b/c
A = 180/2
CalcQuot
Calculate quotient
A = 90
The lampshade has an area of 90 square inches.
b The bases and the area of the trapezoid that makes up the turtle's shell are given.
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.
A = 1/2h(b_1+b_2)
SubstituteValues
Substitute values
6.2 = 1/2* h( 4.7+ 3.1)
▼
Solve for h
AddTerms
Add terms
6.2 = 1/2* h* 7.8
CommutativePropMult
Commutative Property of Multiplication
6.2 = 1/2* 7.8 * h
MoveRightFacToNum
a/c* b = a* b/c
6.2 = 7.8/2 * h
MultEqn
LHS * 2/7.8=RHS* 2/7.8
6.2 * 2/7.8= h
▼
Simplify
MoveLeftFacToNum
a*b/c= a* b/c
6.2* 2/7.8= h
Multiply
Multiply
12.4/7.8= h
CalcQuot
Calculate quotient
1.589743... = h
RearrangeEqn
Rearrange equation
h = 1.589743...
RoundDec
Round to 1 decimal place(s)
h≈ 1.6
The turtle's shell is about 1.6 inches tall.
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Pop Quiz
Solving Trapezoids
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Example
Futsal Formations
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.
a When Mark's team is defending, they arrange themselves to form a parallelogram, where the player closest to the ball will try to recover it while the rest of the team covers the spaces where the opponent could attack.
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.
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b When Mark's team is attacking and the opponent closes the central part of the court, they attack the flanks using a trapezoidal formation.
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.
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Hint
a Use the distance formula to find the lengths of the diagonals, then substitute them into the formula for the area of a rhombus. Remember to convert each distance to meters.
b Use the distance formula to find the lengths of the bases. Then, use the formula for the area of a trapezoid to determine the height of the trapezoid. Remember to convert each distance to meters.
Solution
a The area of a rhombus is half the product of its diagonals.
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).
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstitutePoints
Substitute ( -6,1) & ( 2,4)
d_1 = sqrt(( 2-( -6))^2+( 4- 1)^2)
▼
Evaluate right-hand side
SubNeg
a-(- b)=a+b
d_1 = sqrt((2+6)^2+(4-1)^2)
AddSubTerms
Add and subtract terms
d_1 = sqrt(8^2+3^2)
CalcPow
Calculate power
d_1 = sqrt(64+9)
AddTerms
Add terms
d_1 = sqrt(73)
Next, find the length of the diagonal connecting the vertices (-3,5) and (-1,0).
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstitutePoints
Substitute ( -3,5) & ( -1,0)
d_2 = sqrt(( -1-( -3))^2+( 0- 5)^2)
▼
Evaluate right-hand side
SubNeg
a-(- b)=a+b
d_2 = sqrt((-1+3)^2+(0-5)^2)
AddSubTerms
Add and subtract terms
d_2 = sqrt(2^2+(-5)^2)
CalcPow
Calculate power
d_2 = sqrt(4+25)
AddTerms
Add terms
d_2 = sqrt(29)
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.
A = 1/2d_1d_2
SubstituteII
d_1= 2sqrt(73), d_2= 2sqrt(29)
A = 1/2* 2sqrt(73)* 2sqrt(29)
▼
Simplify right-hand side
CommutativePropMult
Commutative Property of Multiplication
A = 1/2* 2* 2 * sqrt(73)*sqrt(29)
Multiply
Multiply
A = 1/2* 4 * sqrt(73)*sqrt(29)
MoveRightFacToNum
a/c* b = a* b/c
A = 4/2* sqrt(73)*sqrt(29)
CalcQuot
Calculate quotient
A = 2sqrt(73)*sqrt(29)
ProdSqrt
sqrt(a)*sqrt(b)=sqrt(a* b)
A = 2sqrt(73* 29)
Multiply
Multiply
A = 2sqrt(2117)
UseCalc
Use a calculator
A = 2* 46.010868...
Multiply
Multiply
A = 92.021736...
RoundInt
Round to nearest integer
A ≈ 92
The area of the rhombus formed by the players on the court is about 92 square meters.
b The area of a trapezoid is half the height times the sum of the lengths of the bases.
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).
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstitutePoints
Substitute ( 4,5) & ( 6,-5)
b_1 = sqrt(( 6- 4)^2 + ( -5- 5)^2)
b_1 ≈ 10.2
Next, find the length of the shorter base that connects (-6,1) and (-5,-4).
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstitutePoints
Substitute ( -6,1) & ( -5,-4)
b_2 = sqrt(( -5-( -6))^2 + ( -4- 1)^2)
b_2 ≈ 5.1
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.
A = 1/2h(b_1+b_2)
SubstituteValues
Substitute values
324.4 = 1/2* h( 20.4+ 10.2)
▼
Solve for h
AddTerms
Add terms
324.4 = 1/2* h* 30.6
CommutativePropMult
Commutative Property of Multiplication
324.4 = 1/2* 30.6 * h
MoveRightFacToNum
a/c* b = a* b/c
324.4 = 30.6/2h
CalcQuot
Calculate quotient
324.4 = 15.3h
DivEqn
.LHS /15.3.=.RHS /15.3.
324.4/15.3 = h
RearrangeEqn
Rearrange equation
h = 324.4/15.3
CalcQuot
Calculate quotient
h = 21.202614...
RoundDec
Round to 1 decimal place(s)
h ≈ 21.2
The height of the trapezoid formed by the players on the court is about 21.2 meters.
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Closure
Composite Figures
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.
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Areas of Parallelograms, Triangles, and Trapezoids
Exercise 3.1
What is the area of the letter A?
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There is no formula that directly gives us the area of this letter A, but we can try breaking it down into some parts that we can find the area of. For example, this letter A can be made by cutting a triangle and small trapezoid out from a large trapezoid.
The area of the letter is the area of the large trapezoid minus the areas of the cut pieces. Unfortunately, we do not know the heights of the small pieces, so we cannot find their areas. However, we can also form the letter A by cutting a large triangle out of the original large trapezoid, then adding back in the middle trapezoid.
We do have all the information to find the area of each of these pieces. Let's calculate the area of the letter A by finding the area of the large trapezoid minus the area of the triangle plus the area of the middle trapezoid. A_(Letter A) = A_1 - A_2 + A_3 Let's find each area one by one.
Finding the Area of the Large Trapezoid A1
The large trapezoid has bases that are 80 and 10 centimeters long and is 61 centimeters high.
The area of a trapezoid is half the height times the sum of the lengths of the bases. Let's substitute these values into the formula and solve for the area of the large trapezoid A_1.
The area of the large trapezoid is 2745 square centimeters.
Finding the Area of the Triangle A_2
The triangle cutout has a base that is 20 centimeters long and is 51 centimeters high.
The area of a triangle is half the product of the base and height. Let's find the area of the triangle A_2!
The area of the triangle cutout is 510 square centimeters.
Finding the Area of the Middle Trapezoid A_3
The middle trapezoid has bases that are 15 and 8 centimeters long and is 20 centimeters high.
Let's substitute these values into the formula for the area of a trapezoid and simplify to find the area of the middle trapezoid A_3.
The area of the middle trapezoid is 230 square centimeters.
Finding the Area of the Letter A
Finally, we can substitute the areas we found into the equation we wrote for the area of the letter A. A_(Letter A) = A_1 - A_2 + A_3 Now let's find the total area of the letter. A_(Letter A)= 2745 - 510 + 230 ⇓ A_(Letter A) = 2465 The area of the letter A is 2465 square centimeters.
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Solution
Areas of Parallelograms, Triangles, and Trapezoids
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