Exploring the Area of a Parallelogram, Triangle, Trapezoid, and Rhombus

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

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

Property	Justification
The diagonals are congruent.	Isosceles Trapezoid Diagonals Theorem
Each pair of base angles is congruent.	Isosceles Trapezoid Base Angles Theorem

Property

Justification

The diagonals are congruent.

Isosceles Trapezoid Diagonals Theorem

Each pair of base angles is congruent.

Isosceles Trapezoid Base Angles Theorem

a The area of a rhombus is half the product of its diagonals.

A = \frac{1}{2} d_{1} d_{2}

The length of the diagonals can be found with the help of the distance formula.

d = (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 = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(- 6, 1)$ & $(2, 4)$

d_{1} = (2 - (- 6))^{2} + (4 - 1)^{2}

▼

Evaluate right-hand side

SubNeg

$a - (- b) = a + b$

d_{1} = (2 + 6)^{2} + (4 - 1)^{2}

AddSubTerms

Add and subtract terms

d_{1} = 8^{2} + 3^{2}

CalcPow

Calculate power

d_{1} = 64 + 9

AddTerms

Add terms

d_{1} = 73

Next, find the length of the diagonal connecting the vertices

(- 3, 5)

and

(- 1, 0) .

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(- 3, 5)$ & $(- 1, 0)$

d_{2} = (- 1 - (- 3))^{2} + (0 - 5)^{2}

▼

Evaluate right-hand side

SubNeg

$a - (- b) = a + b$

d_{2} = (- 1 + 3)^{2} + (0 - 5)^{2}

AddSubTerms

Add and subtract terms

d_{2} = 2^{2} + (- 5)^{2}

CalcPow

Calculate power

d_{2} = 4 + 25

AddTerms

Add terms

d_{2} = 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}

2

to find the actual lengths.

Length on the Whiteboard	Length on the Court
$d_{1} = 73$	$d_{1} = 273$
$d_{2} = 29$	$d_{2} = 229$

Finally, substitute the actual lengths into the formula for the area of a rhombus.

A = \frac{1}{2} d_{1} d_{2}

SubstituteII

$d_{1} = 273$ , $d_{2} = 229$

A = \frac{1}{2} \cdot 273 \cdot 229

▼

Simplify right-hand side

CommutativePropMult

Commutative Property of Multiplication

A = \frac{1}{2} \cdot 2 \cdot 2 \cdot 73 \cdot 29

Multiply

A = \frac{1}{2} \cdot 4 \cdot 73 \cdot 29

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

A = \frac{4}{2} \cdot 73 \cdot 29

CalcQuot

Calculate quotient

A = 273 \cdot 29

ProdSqrt

$a \cdot b = a \cdot b$

A = 2 73 \cdot 29

Multiply

A = 22117

UseCalc

Use a calculator

A = 2 \cdot 46.010868 \dots

Multiply

A = 92.021736 \dots

RoundInt

Round to nearest integer

A \approx 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 = \frac{1}{2} h (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 = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(4, 5)$ & $(6, - 5)$

b_{1} = (6 - 4)^{2} + (- 5 - 5)^{2}

▼

Simplify right-hand side

SubTerms

Subtract terms

b_{1} = (2)^{2} + (- 10)^{2}

CalcPow

Calculate power

b_{1} = 4 + 100

AddTerms

Add terms

b_{1} = 104

UseCalc

Use a calculator

b_{1} = 10.198039 \dots

RoundDec

Round to $1$ decimal place(s)

b_{1} \approx 10.2

Next, find the length of the shorter base that connects

(- 6, 1)

and

(- 5, - 4) .

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(- 6, 1)$ & $(- 5, - 4)$

b_{2} = (- 5 - (- 6))^{2} + (- 4 - 1)^{2}

▼

Simplify right-hand side

SubNeg

$a - (- b) = a + b$

b_{2} = (- 5 + 6)^{2} + (- 4 - 1)^{2}

AddSubTerms

Add and subtract terms

b_{2} = (1)^{2} + (- 5)^{2}

CalcPow

Calculate power

b_{2} = 1 + 25

AddTerms

Add terms

b_{2} = 26

UseCalc

Use a calculator

b_{2} = 5.099019 \dots

RoundDec

Round to $1$ decimal place(s)

b_{2} \approx 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 = \frac{1}{2} h (b_{1} + b_{2})

SubstituteValues

Substitute values

324.4 = \frac{1}{2} \cdot h (20.4 + 10.2)

▼

Solve for

h

AddTerms

Add terms

324.4 = \frac{1}{2} \cdot h \cdot 30.6

CommutativePropMult

Commutative Property of Multiplication

324.4 = \frac{1}{2} \cdot 30.6 \cdot h

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

324.4 = \frac{3 0 . 6}{2} h

CalcQuot

Calculate quotient

324.4 = 15.3 h

DivEqn

$LHS / 15.3 = RHS / 15.3$

\frac{3 2 4 . 4}{1 5 . 3} = h

RearrangeEqn

Rearrange equation

h = \frac{3 2 4 . 4}{1 5 . 3}

CalcQuot

Calculate quotient

h = 21.202614 \dots

RoundDec

Round to $1$ decimal place(s)

h \approx 21.2

The height of the trapezoid formed by the players on the court is about

21.2

meters.

Tans Forming the Trapezoid	Dimensions	$A = \frac{1}{2} h (b_{1} + b_{2})$	Area $(cm^{2})$
$7$ and $3$	$b_{1} = 8$ $b_{2} = 4$ $h = 2$	$A_{7, 3} = \frac{1}{2} \cdot 2 (8 + 4)$	$12$
$1,$ $2,$ and $4$	$b_{1} = 12$ $b_{2} = 8$ $h = 4$	$A_{1, 2, 4} = \frac{1}{2} \cdot 4 (12 + 8)$	$40$
$5$ and $6$	$b_{1} = 28$ $b_{2} = 8$ $h = 8$	$A_{5, 6} = \frac{1}{2} \cdot 8 (28 + 8)$	$12$

Tans Forming the Trapezoid

Dimensions

A = \frac{1}{2} h (b_{1} + b_{2})

Area

(cm^{2})

7

and

3

b_{1} = 8

b_{2} = 4

h = 2

A_{7, 3} = \frac{1}{2} \cdot 2 (8 + 4)

12

1,

2,

and

4

b_{1} = 12

b_{2} = 8

h = 4

A_{1, 2, 4} = \frac{1}{2} \cdot 4 (12 + 8)

40

5

and

6

b_{1} = 28

b_{2} = 8

h = 8

A_{5, 6} = \frac{1}{2} \cdot 8 (28 + 8)

12

Areas of Parallelograms, Triangles, and Trapezoids

Catch-Up and Review

Dividing a Rectangle in Half

Deriving the Formula for the Area of a Triangle

Area of a Triangle

Proof

Proof for Right Triangles

Proof for Non-Right Triangles

Tangram Puzzle

Hint

Solution

Finding the Missing Dimension

Parallelograms

Area of a Parallelogram

Tangram Animals

Hint

Solution

Parallelograms With Congruent Sides

Rhombus

Area of a Rhombus

Rhombuses Everywhere

Hint

Solution

Solving Parallelograms and Rhombuses

Trapezoids

Isosceles Trapezoid

Area of a Trapezoid

Extra

A Head Full of Polygons

Hint

Solution

Solving Trapezoids

Futsal Formations

Hint

Solution

Composite Figures

Areas of Parallelograms, Triangles, and Trapezoids

Recommended exercises

	18 Theory slides
	8 Exercises - Grade E - A
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