Topics
5. Modeling in Two & Three Dimensions
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Integrated Math 3

5.02 Modeling with 2D figures

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Lesson

Concept summary

When solving problems in the real world, we can often approximate them with geometric problems by matching the properties of 2D figures to the context.

When modeling with 2D figures, a key property that we want to remember is how to find the area. We have encountered many area formulas already:

Area of triangle

The area of triangle is half the product of the base times height, where b is the base and h is the height.

A = \dfrac{1}{2}bh

A triangle drawn with its base as the horizontal side labelled b and the height drawn as a dashed segment from one of the vertices perpendicular to the base and labelled as h.
Area of rectangle

The area of a rectangle is the product of the base times height, where b is the base and h is the height.

A=bh

A rectangle in which the vertical side on the right is labelled h and the bottom horizontal side at the bottom is labelled b.
Area of parallelogram

The area of a parallelogram is the product of the base times height, where b is the base and h is the height.

A=bh

A parallelogram with its height drawn as a dashed segment from one of the vertices perpendicular to the horizontal side labelled b.
Area of trapezoid

The area of a trapezoid is one half of the product of the height and the sum of the lengths of the bases, where h is the height and a and b are the bases.

A = \dfrac{1}{2}h(a+b)

A trapezoid with the top base side labelled as a and bottom base side labelled as b. A dashed segment from the top base angle perpendicular to the bottom base b is drawn and labelled with h.
Area of a square

The area of a square is the side length squared, where s is the side length.

A=s^2

A square with side length labelled s.
Area of a regular polygon

The area of a regular polygon is one half of the product of the perimeter and apothem, where P is the perimeter and a is the apothem. An apothem is the distance from the center to a side of a regular polygon.

A=\dfrac{1}{2}Pa

A regular six sided polygon with segment called apothem and labelled as a drawn from the center to a side of hte polygon. The sides are marked showing the lengths as distances from one vertex to another and labelled with P for perimeter.
Area of rhombus

The area of a rhombus is the product of the base and height, where b is the base and h is the height.

A=bh

The area of a rhombus can also be determined by one half of the product of the diagonals, where d_1 and d_2 are the diagonals.

A=\dfrac{1}{2}d_1d_2

A rhombus with diagonals drawn as dashed segments intersecting each other. The diagonals are labelled d sub 1 and d sub 2 respectively.

These area formulas are particularly useful for finding the area of 2D figures that are not so simple, like a composite figure. A composite figure is a figure that can be decomposed into smaller figures that have been added together or sometimes subtracted from each other.

A composite shape composed of a triangle on top of a rectangle.
A composite figure formed by adding a rectangle and triangle together.
A composite shape composed of a semicircle cut out of a rectangle.
A composite figure formed by subtracting a semicircle from a rectangle.

We can determine the area of composite figures by breaking them down into simpler shapes. After we find the area of the simpler shapes, we can add or subtract those areas to find the area of the composite shape.

Another use for finding the area of irregular shapes is determining the population density of that region. The population density is calculated as the population in a region divided by the area of that region.

\displaystyle \text{Population density}=\frac{\text{Population}}{\text{Area}}
\bm{\text{Population}}
The number of people or objects
\bm{\text{Area}}
The space available for the population

Worked examples

Example 1

Find the area of a sandbox which has been approximated by this geometric figure. All measurements are in feet.

A quadrilateral with sides of length 11, 8, 4, and 13. Between sides 11 and 8 is an included right angle. The included angle between side lengths 11 and 8 as well as the angle between side lengths 4 and 13 is a right angle.

Approach

We don't have a formula for the area of this type of quadrilateral. Instead, we can decompose the shape into two right-triangles as follows:

A quadrilateral with sides of length 11, 8, 4, and 13. The included angle between side lengths 11 and 8 as well as the angle between side lengths 4 and 13 is a right angle. A dashed segment from the angle between sides of length 11 and 13 and the angle between sides of length 8 and 4 serves as the hypotenuse of the right triangles. The triangle with side lengths 11 and 8 is labelled 1 and the the triangle with side lengths 4 and 13 is labelled 2

Solution

Area of triangle 1:

\displaystyle A_1\displaystyle =\displaystyle \dfrac{1}{2}\cdot b\cdot hFormula
\displaystyle {}\displaystyle =\displaystyle \dfrac{1}{2}\cdot 8\cdot11Substitution
\displaystyle {}\displaystyle =\displaystyle \dfrac{1}{2} \cdot 88Simplify
\displaystyle {}\displaystyle =\displaystyle 44Simplify

Area of triangle 2:

\displaystyle A_2\displaystyle =\displaystyle \dfrac{1}{2}\cdot b\cdot hFormula
\displaystyle {}\displaystyle =\displaystyle \dfrac{1}{2}\cdot 4\cdot13Substitution
\displaystyle {}\displaystyle =\displaystyle \dfrac{1}{2} \cdot 52Simplify
\displaystyle {}\displaystyle =\displaystyle 26Simplify

Area of the composite shape:

\displaystyle A\displaystyle =\displaystyle A_1 + A_2
\displaystyle {}\displaystyle =\displaystyle 44+26Substitution
\displaystyle {}\displaystyle =\displaystyle 70Simplify

Area of the sandbox = 70 \text{ ft}^2

Reflection

Consider that the sandbox was not necessarily an exact geometric figure. When modeling with 2D figures, it is reasonable to approximate the shape of an area to a geometric figure that makes our calculations easier.

Example 2

An amphitheatre is designed with a semicircular viewing section and a rectangular stage. The viewing section is designed so that the furthest audience member is 10 meters from the middle of the front of the stage. The stage is 20 meters by 6 meters.

a

Find the approximate area covered by the amphitheatre to the nearest square meter.

Approach

Using the description of the amphitheatre, we can approximate the area to that of a composite figure formed by a rectangle and a semicircle.

A composite figure composed of a semicircle on top of a rectangle. The rectangle is 20 by 6 meters. The semicircle has a radius of 10 meters.

Solution

Area of the semicircle:

\displaystyle A_{\text{semicircle}}\displaystyle =\displaystyle \frac{1}{2}\pi r^2Formula
\displaystyle =\displaystyle \frac{1}{2}\pi\cdot 10^2Substitution
\displaystyle =\displaystyle \frac{1}{2}\pi\cdot 100Simplify
\displaystyle =\displaystyle 50\piSimplify

Area of the rectangle:

\displaystyle A_{\text{rectangle}}\displaystyle =\displaystyle bhFormula
\displaystyle =\displaystyle 20\cdot6Substitution
\displaystyle =\displaystyle 120Simplify

Area of the amphitheatre:

\displaystyle A\displaystyle =\displaystyle A_{\text{semicircle}}+A_{\text{rectangle}}
\displaystyle =\displaystyle 50\pi +120Substitution

If we evaluate the expression for the area of the amphitheatre and round to the nearest square meter, we get that:A\approx 277\text{ m}^2

b

If there are 300 people in the audience and 20 actors on stage, find the population density of the amphitheatre.

Solution

\displaystyle \text{Population density}\displaystyle =\displaystyle \frac{\text{Population}}{\text{Area}}Formula
\displaystyle =\displaystyle \frac{300+20}{277}Substitution
\displaystyle =\displaystyle \frac{320}{277}Simplify

Rounding to two decimal places, the population density of the amphitheatre is 1.16 people per square meter.

Outcomes

M3.N.Q.A.1.B

Use appropriate quantities in formulas, converting units as necessary.

M3.N.Q.A.1.C

Define and justify appropriate quantities within a context for the purpose of modeling.

M3.G.MG.A.1

Use geometric shapes, their measures, and their properties to model objects found in a real-world context for the purpose of approximating solutions to problems.*

M3.MP1

Make sense of problems and persevere in solving them.

M3.MP2

Reason abstractly and quantitatively.

M3.MP3

Construct viable arguments and critique the reasoning of others.

M3.MP4

Model with mathematics.

M3.MP5

Use appropriate tools strategically.

M3.MP6

Attend to precision.

M3.MP7

Look for and make use of structure.

M3.MP8

Look for and express regularity in repeated reasoning.

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