Exploring Geometry in Design: Trigonometry and Similarity Unveiled

In this lesson, geometric methods will be used to solve design problems.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Circles
Arcs and arc length
Circle sector and area of a circle sector
Triangles
Area
Perimeter
Trigonometric ratios
Pythagorean Theorem

Try your knowledge on these topics.

Find the perimeter and area of the following figures.

c Calculate the sine, cosine, and tangent of

∠ θ

in the right triangle. Write the answers as fractions in their simplest form.

d By using the Pythagorean Theorem, calculate the missing side length.

e Calculate the length of the arc and the area of the sector. Write the answer correct to one decimal place.

Challenge

Investigating Efficient Solutions Using Geometry

Paulina bought two sprinklers to water her

20 -

meter by

10 -

meter rectangular garden. Each sprinkler waters a circular region, and the radius of each circle measures equally. She does not want any part of her garden to flood. That is, the circles should not overlap. The radii can be adjusted by changing the water pressure. Try to place them in the garden in the most efficient manner!

Where should Paulina place the sprinklers if she wants to water the greatest possible area of her garden?

Example

Optimizing Area Using Geometry

Ali bought

20

meters of fence to construct a playground in his backyard for his dog Rover. He is wavering between the ideas of making the playground's shape into a square, an equilateral triangle, or a circle. Help give Rover the most space to run.

Which of these three shapes has the greatest area?

Hint

Calculate and compare the area of the three shapes.

Solution

The area of the three shapes will be calculated one at a time. Then, they will be compared.

Square

Let

ℓ

be the side length of a square region enclosed by

20

meters of fence. The perimeter of the region

4 ℓ,

will then be equal to

20 .

P = 4 ℓ

Substitute

$P = 20$

20 = 4 ℓ

Solve for

ℓ

DivEqn

$LHS / 4 = RHS / 4$

5 = ℓ

RearrangeEqn

Rearrange equation

ℓ = 5

The side length of the square is

5

meters.

Now, to find its area, the side length can be squared.

Area of the Square : 5^{2} = 25 m^{2}

Equilateral Triangle

The three sides of an equilateral triangle have the same length. Therefore, to find the side length, the perimeter of the triangle, which is equal to the length of the fence, must be divided by

3 .

Side Length : \frac{2 0}{3} meters

To find the area of the triangle, its height

h

must be found first. To do so, the altitude of the triangle will be drawn. Recall that the altitude of an equilateral triangle bisects and is perpendicular to the base.

It can be seen that the right triangle formed at the left of the diagram has hypotenuse

\frac{2 0}{3}

and legs

\frac{1 0}{3}

and

h .

To find the missing value, the Pythagorean Theorem can be used.

a^{2} + b^{2} = c^{2}

SubstituteValues

Substitute values

(\frac{1 0}{3})^{2} + h^{2} = (\frac{2 0}{3})^{2}

Solve for

h

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

\frac{1 0 0}{9} + h^{2} = \frac{4 0 0}{9}

SubEqn

$LHS - \frac{1 0 0}{9} = RHS - \frac{1 0 0}{9}$

h^{2} = \frac{4 0 0}{9} - \frac{1 0 0}{9}

SubFrac

Subtract fractions

h^{2} = \frac{3 0 0}{9}

SqrtEqn

$LHS = RHS$

h = \frac{3 0 0}{9}

SqrtQuot

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

h = \frac{3 0 0}{9}

SplitIntoFactors

Split into factors

h = \frac{1 0 0 ( 3 )}{9}

SqrtProd

$a \cdot b = a \cdot b$

h = \frac{1 0 0 3}{9}

CalcRoot

Calculate root

h = \frac{1 0 3}{3}

Because side lengths are non-negative, when solving the equation only the principal root was considered. Therefore, the length of the leg of the right triangle, which is also the height of the equilateral triangle, is

\frac{1 0 3}{3}

meters. Recall that the side length of the equilateral triangle is

\frac{2 0}{3}

meters. This means that its base is also

\frac{2 0}{3}

meters.

The area of a triangle is half the product of its base and its height. With this information, the area of the triangle can be found.

A = \frac{1}{2} b h

SubstituteII

$b = \frac{2 0}{3}$ , $h = \frac{1 0 3}{3}$

A = \frac{1}{2} (\frac{2 0}{3}) (\frac{1 0 3}{3})

Evaluate right-hand side

MultFrac

Multiply fractions

A = \frac{2 0}{6} (\frac{1 0 3}{3})

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

A = \frac{1 0}{3} (\frac{1 0 3}{3})

MultFrac

Multiply fractions

A = \frac{1 0 0 3}{9}

The area of the equilateral triangle is

\frac{1 0 0 3}{9}

square meters.

Circle

Just one more major step. Finally, the area of the circle will be calculated. Since Ali bought

20

meters of fence, the circumference is

20

meters. Recall that the circumference of a circle is twice the product of

π

and its radius. With this information, the radius of the circle can be found.

C = 2 π r

Substitute

$C = 20$

20 = 2 π r

Solve for

r

DivEqn

$LHS / 2 π = RHS / 2 π$

\frac{2 0}{2 π} = r

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

\frac{1 0}{π} = r

RearrangeEqn

Rearrange equation

r = \frac{1 0}{π}

The radius of the circle is

\frac{1 0}{π}

meters.

The area of a circle is the product of

π

and the square of its radius.

A = π r^{2}

Substitute

$r = \frac{1 0}{π}$

A = π (\frac{1 0}{π})^{2}

Evaluate right-hand side

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

A = π (\frac{1 0 0}{π ^{2}})

MoveLeftFacToNum

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

A = \frac{π ( 1 0 0 )}{π ^{2}}

CancelCommonFac

Cancel out common factors

SimpQuot

Simplify quotient

A = \frac{1 0 0}{π}

The area of the circle is

\frac{1 0 0}{π}

square meters.

Comparison

Now that the areas of the three figures are known, they can be compared. To do so, the area of the triangle and the area of the circle will be approximated to two decimal places.

Area of the Square	Area of the Equilateral Triangle	Area of the Circle
$25 m^{2}$	$\frac{1 0 0 3}{9}$ $\approx$ $19.25 m^{2}$	$\frac{1 0 0}{π}$ $\approx$ $31.83 m^{2}$

It can be seen above that the circle is the figure with the greatest area. Therefore, Ali should construct Rover's playground in the shape of a circle. Run and feel the wind Rover!

Example

Comparing Areas Using Geometry

Magdalena and Dylan want to build three grain bins in their field in the form of an equilateral triangle. Both sketch the field as an equilateral triangle with a side length of

10

centimeters. Magdalena draws three congruent circles, in contrast to Dylan, who draws the incircle of the triangle and two circles with a radius of

1

centimeter.

For the most efficient use of the field, the total area of the circles should be as large as possible. To find the total area in each case, they ask their teacher for some help. The teacher tells them that the radius of an incircle of a triangle is twice the quotient between the area and the perimeter of the triangle.

Radius of an Incircle r = 2 (\frac{A}{P})

Use the given information to determine who drew their circles with the greatest sum of the circles' areas.

Hint

The altitude of an equilateral triangle divides it into two right triangles. Use the Pythagorean Theorem to find the height of this triangle and then calculate its area. Finally, use the formula provided by the teacher to find the radius of Magdalena's circles and the radius of the incircle drawn by Dylan.

Solution

The area of the circles that Magdalena and Dylan drew will be calculated one at a time. Then, the results will be compared.

Area of Magdalena's Circles

The circles will be ignored for a moment, and the altitude of the triangle will be drawn. The altitude of a triangle is perpendicular to the base. Also, because the triangle is equilateral, the altitude bisects the base. As a result, the length of one leg and the hypotenuse of the obtained right triangle are $5$ and $10$ centimeters, respectively.

By using the Pythagorean Theorem, the height of the triangle can be calculated.

a^{2} + b^{2} = c^{2}

SubstituteValues

Substitute values

5^{2} + h^{2} = 10^{2}

Solve for

h

CalcPow

Calculate power

25 + h^{2} = 100

SubEqn

$LHS - 25 = RHS - 25$

h^{2} = 75

SqrtEqn

$LHS = RHS$

h = 75

SplitIntoFactors

Split into factors

h = 25 (3)

SqrtProd

$a \cdot b = a \cdot b$

h = 253

CalcRoot

Calculate root

h = 53

When solving the equation, only the principal root was considered because a length is always positive. Therefore, the height of the right triangle is

53

centimeters. With this information, its area can be calculated.

Area of a Right Triangle \frac{1}{2} (5) (53) = \frac{2 5 3}{2} cm^{2}

Next, note that one of the circles that Magdalena drew is the incircle of this right triangle.

Therefore, its radius can be calculated by using the formula given by the teacher. To do this, the perimeter of the right triangle is needed.

Perimeter of a Right Triangle 5 + 53 + 10 = (15 + 53) cm

Now, the formula provided by the teacher can be used to find the radius of one of the circles that Magdalena drew.

r = 2 (\frac{A}{P})

SubstituteII

$P = 15 + 53$ , $A = \frac{2 5 3}{2}$

r = 2 (\frac{\frac{2 5 3}{2}}{1 5 + 5 3})

Evaluate right-hand side

MoveLeftFacToNum

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

r = \frac{2 ( \frac{2 5 3}{2} )}{1 5 + 5 3}

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

r = \frac{2 5 3}{1 5 + 5 3}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot ( 1 5 - 5 3 )}{b \cdot ( 1 5 - 5 3 )}$

r = \frac{2 5 3 ( 1 5 - 5 3 )}{( 1 5 + 5 3 ) ( 1 5 - 5 3 )}

Distr

Distribute $253$

r = \frac{3 7 5 3 - 3 7 5}{( 1 5 + 5 3 ) ( 1 5 - 5 3 )}

ExpandDiffSquares

$(a + b) (a - b) = a^{2} - b^{2}$

r = \frac{3 7 5 3 - 3 7 5}{1 5 ^{2} - ( 5 3 ) ^{2}}

PowProdII

$(a b)^{m} = a^{m} b^{m}$

r = \frac{3 7 5 3 - 3 7 5}{1 5 ^{2} - 5 ^{2} ( 3 ) ^{2}}

CalcPow

Calculate power

r = \frac{3 7 5 3 - 3 7 5}{2 2 5 - 2 5 ( 3 ) ^{2}}

PowSqrt

$(a)^{2} = a$

r = \frac{3 7 5 3 - 3 7 5}{2 2 5 - 2 5 ( 3 )}

Multiply

r = \frac{3 7 5 3 - 3 7 5}{2 2 5 - 7 5}

SubTerm

Subtract term

r = \frac{3 7 5 3 - 3 7 5}{1 5 0}

SplitIntoFactors

Split into factors

r = \frac{7 5 ( 5 ) 3 - 7 5 ( 5 )}{1 5 0}

FactorOut

Factor out $75$

r = \frac{7 5 ( 5 3 - 5 )}{1 5 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 7 5}{b / 7 5}$

r = \frac{5 3 - 5}{2}

The radius of Magdalena's circles is

\frac{5 3 - 5}{2}

centimeters.

Now the area of one of the circles can be calculated.

Area of a Circle = π r^{2}

Substitute

$r = \frac{5 3 - 5}{2}$

Area of a Circle = π (\frac{5 3 - 5}{2})^{2}

Evaluate right-hand side

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

Area of a Circle = π \frac{( 5 3 - 5 ) ^{2}}{4}

CommutativePropMult

\CommutativePropMult

Area of a Circle = \frac{( 5 3 - 5 ) ^{2}}{4} π

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

Area of a Circle = \frac{( 5 3 ) ^{2} - 2 ( 5 3 ) ( 5 ) + 5 ^{2}}{4} π

PowProdII

$(a b)^{m} = a^{m} b^{m}$

Area of a Circle = \frac{5 ^{2} ( 3 ) ^{2} - 2 ( 5 3 ) ( 5 ) + 5 ^{2}}{4} π

CalcPowProd

Calculate power and product

Area of a Circle = \frac{2 5 ( 3 ) ^{2} - 5 0 3 + 2 5}{4} π

PowSqrt

$(a)^{2} = a$

Area of a Circle = \frac{2 5 ( 3 ) - 5 0 3 + 2 5}{4} π

Multiply

Area of a Circle = \frac{7 5 - 5 0 3 + 2 5}{4} π

AddTerms

Add terms

Area of a Circle = \frac{1 0 0 - 5 0 3}{4} π

SplitIntoFactors

Split into factors

Area of a Circle = \frac{2 ( 5 0 ) - 2 ( 2 5 ) 3}{4} π

FactorOut

Factor out $2$

Area of a Circle = \frac{2 ( 5 0 - 2 5 3 )}{4} π

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

Area of a Circle = \frac{5 0 - 2 5 3}{2} π

The area of one of the circles that Magdalena drew is

\frac{5 0 - 2 5 3}{2} π

square centimeters.

Since the three circles are congruent, they have the same area. Therefore, to calculate the sum of the areas, it is enough to multiply the area of one of the circle's by

3 .

Area of Magdalena ’ s Circles 3 (\frac{5 0 - 2 5 3}{2} π) = \frac{1 5 0 - 7 5 3}{2} π cm^{2}

Area of Dylan's Circles

Calculating the sum of the areas of Dylan's circles takes less steps than calculating the sum of the areas of Magdalena's circles.

The formula given by the teacher can be used to calculate the radius of the incircle that Dylan drew. First, recall that the height and base of this equilateral triangle are

53

and

10

centimeters, respectively. This information will be used to calculate the area of the triangle.

Area of a Triangle = \frac{1}{2} b h

SubstituteII

$b = 10$ , $h = 53$

Area of a Triangle = \frac{1}{2} (10) (53)

Evaluate right-hand side

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

Area of a Triangle = \frac{1 0}{2} (53)

CalcQuot

Calculate quotient

Area of a Triangle = 5 (53)

Multiply

Area of a Triangle = 253

The area of the equilateral triangle is

253

square centimeters. Its perimeter is the sum of the sides. Therefore, its perimeter is

10 + 10 + 10 = 30

centimeters. With this information, the formula given by the teacher can be used to find the radius of the incircle.

r = 2 (\frac{A}{P})

SubstituteII

$P = 30$ , $A = 253$

r = 2 (\frac{2 5 3}{3 0})

Evaluate right-hand side

MoveLeftFacToNum

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

r = \frac{5 0 3}{3 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 0}{b / 1 0}$

r = \frac{5 3}{3}

The radius of the incircle is

\frac{5 3}{3}

centimeters.

Therefore, the area of the incircle can be found.

Area of a Circle = π r^{2}

Substitute

$r = \frac{5 3}{3}$

Area of a Circle = π (\frac{5 3}{3})^{2}

Evaluate right-hand side

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

Area of a Circle = π (\frac{( 5 3 ) ^{2}}{3 ^{2}})

PowProdII

$(a b)^{m} = a^{m} b^{m}$

Area of a Circle = π (\frac{5 ^{2} ( 3 ) ^{2}}{3 ^{2}})

CalcPow

Calculate power

Area of a Circle = π (\frac{2 5 ( 3 ) ^{2}}{9})

PowSqrt

$(a)^{2} = a$

Area of a Circle = π (\frac{2 5 ( 3 )}{9})

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

Area of a Circle = π (\frac{2 5}{3})

CommutativePropMult

\CommutativePropMult

Area of a Circle = \frac{2 5}{3} π

The area of the incircle that Dylan drew is

\frac{2 5}{3} π

square centimeters.

Next, the area of a circle of radius

1

can be calculated.

Area of One Circle Area = π r^{2} \Rightarrow Area Area = π (1^{2}) = π cm^{2}

Finally, the areas of both circles of radius

1

and the area of the incircle will be added.

\frac{2 5}{3} π + π + π

Simplify

IdPropMult

\IdPropMult

\frac{2 5}{3} π + 1 π + 1 π

OneToFrac

Rewrite $1$ as $\frac{3}{3}$

\frac{2 5}{3} π + \frac{3}{3} π + \frac{3}{3} π

FactorOut

Factor out $π$

(\frac{2 5}{3} + \frac{3}{3} + \frac{3}{3}) π

AddFrac

Add fractions

\frac{3 1}{3} π

The sum of the areas of Dylan's circles is

\frac{3 1}{3} π

square centimeters.

Comparison

The sum of the areas of the circles that Magdalena and Dylan drew are known. For simplicity in the comparison, they will be approximated to one decimal place.

Area of Magdalena's Circles	Area of Dylan's Circles
$\frac{1 5 0 - 7 5 3}{2} π$ $\approx$ $31.6 cm^{2}$	$\frac{3 1}{3} π$ $\approx$ $32.5 cm^{2}$

It can be concluded that the circles drawn by Dylan have a greater area than the circles drawn by Magdalena.

Closure

Optimizing Efficiency Using Geometry

In this lesson, different geometric methods were used to solve design problems. Here, the challenge presented at the beginning of the lesson will be examined in detail.

Paulina bought two sprinklers to water her

20 -

meter by

10 -

meter rectangular garden. Each sprinkler waters a circular region, and the radius of each circle has the same measurement. Paulina can adjust the radius, which will affect both sprinklers equally. Recall that the sprinklers should not water any same region of the garden.

Paulina is considering two options. For the first option, she is thinking of placing the sprinklers in such a way that the circles formed by the water are tangent between them. Additionally, each circle is tangent to three sides of the garden.

For the second option, Paulina is thinking of placing the sprinklers at the midpoints of the shorter sides of the garden. Here, the circles formed by the water are also tangent to each other.

By paying close attention to the diagrams, it can be seen that the sprinklers water a greater area of the garden in option

2 .

How much more area is watered in Option

2

than in Option

1 ?

Approximate the answer to the nearest integer.

Hint

For the second option, consider dividing the area watered by one sprinkler into two right triangles and one circle sector.

Solution

The area of the garden watered in each of the options will be calculated one at a time. Then, their difference will be found.

Option $1$

Here, the diameter of each circle that is formed by the water coming out of the sprinklers is equal to the width of the garden, which is

10

meters.

To calculate the area of one of the circles, its radius is needed. Recall that the radius is half the diameter.

Radius : \frac{1 0}{2} = 5 meters

Now the formula for the area of a circle can be used.

A = π r^{2}

Substitute

$r = 5$

A = π (5^{2})

Evaluate right-hand side

CalcPow

Calculate power

A = π (25)

CommutativePropMult

\CommutativePropMult

A = 25 π

The area of one circle is

25 π

square meters. The area of the two circles combined is twice the area of one circle.

Area Watered in Option 1 (m^{2}) 25 π \times 2 = 50 π

Option $2$

For this option, the area watered by one sprinkler will be calculated. Then, to obtain the total area, that value will be multiplied by

2 .

Keep in mind that the radius of one circle is half the length of the garden. Additionally, since the sprinkler is located at the midpoint of the width, its distance to the endpoints of the width can also be calculated.

Radius : Distance to Endpoints : \frac{2 0}{2} = 10 m \frac{1 0}{2} = 5 m

With this information, the area watered by one sprinkler can be divided into a circle sector with a radius of

10

meters and two right triangles with a hypotenuse and leg of

10

and

5

meters, respectively. Let

h

be the height of these right triangles.

To find the height

h

of these right triangles, the Pythagorean Theorem can be used.

a^{2} + b^{2} = c^{2}

SubstituteValues

Substitute values

h^{2} + 5^{2} = 10^{2}

Solve for

h

CalcPow

Calculate power

h^{2} + 25 = 100

SubEqn

$LHS - 25 = RHS - 25$

h^{2} = 75

SqrtEqn

$LHS = RHS$

h = 75

SplitIntoFactors

Split into factors

h = 25 (3)

SqrtProd

$a \cdot b = a \cdot b$

h = 253

CalcRoot

Calculate root

h = 53

Since lengths are always positive, only the principal root was considered in the solution. Therefore, the height of the right triangles is

53

meters.

Now using the formula for the area of a triangle, the area of each triangle can be calculated.

A = \frac{1}{2} b h

SubstituteII

$b = 5$ , $h = 53$

A = \frac{1}{2} (5) (53)

Evaluate right-hand side

Multiply

A = \frac{1}{2} (253)

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

A = \frac{2 5 3}{2}

The area of each right triangle is

\frac{2 5 3}{2}

square meters.

It is now time to find the area of the circle sector. This area depends on the angle of the sector. To find the angle of the sector, one of the acute angles of the triangles should be found first. Consider the triangle located in the bottom-right corner of the diagram, and let

θ

be the desired angle.

Since the three side lengths are known, any of the trigonometric ratios can be used to find the measure of

∠ θ .

Here, the cosine ratio will be arbitrarily used.

cos θ = \frac{adj}{hyp}

SubstituteII

$adj = 5$ , $hyp = 10$

cos θ = \frac{5}{1 0}

Solve for

θ

ReduceFrac

$\frac{a}{b} = \frac{a / 5}{b / 5}$

cos θ = \frac{1}{2}

$cos^{- 1} LHS = cos^{- 1} RHS$

θ = cos^{- 1} \frac{1}{2}

UseCalc

Use a calculator

θ = 6 0^{\circ}

The angle of the triangle measures

6 0^{\circ} .

By following the same procedure, it can be found that the corresponding angle on the other triangle also measures

6 0^{\circ} .

In the diagram, it can be seen that these two

60 - degree

angles and the angle of the circle sector form a straight angle. Therefore, the angle of the sector can be found.

Angle of the Circle Sector 18 0^{\circ} - 6 0^{\circ} - 6 0^{\circ} = 6 0^{\circ}

The sector has a radius of

10

meters and an angle whose measure is

6 0^{\circ} .

The area of a circle sector is the quotient between the angle of the sector and

36 0^{\circ},

multiplied by the product of

π

and the square of the radius.

A = \frac{θ}{3 6 0 ^{\circ}} \cdot π r^{2}

In the above formula,

6 0^{\circ}

and

10

can be substituted for

θ

and

r,

respectively.

A = \frac{θ}{3 6 0 ^{\circ}} \cdot π r^{2}

SubstituteII

$θ = 6 0^{\circ}$ , $r = 10$

A = \frac{6 0 ^{\circ}}{3 6 0 ^{\circ}} \cdot π (10^{2})

Evaluate right-hand side

CalcPow

Calculate power

A = \frac{6 0 ^{\circ}}{3 6 0 ^{\circ}} \cdot π (100)

CommutativePropMult

\CommutativePropMult

A = \frac{6 0 ^{\circ}}{3 6 0 ^{\circ}} (100) π

MoveRightFacToNum

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

A = \frac{6 0 0 0 ^{\circ}}{3 6 0 ^{\circ}} π

ReduceFrac

$\frac{a}{b} = \frac{a / 1 2 0 ^{\circ}}{b / 1 2 0 ^{\circ}}$

A = \frac{5 0}{3} π

The area of the sector is

\frac{5 0}{3} π

square meters. The area watered by one sprinkler is the sum of the areas of the triangles and the sector.

Area Watered by One Sprinkler (m^{2}) \frac{2 5 3}{2} + \frac{2 5 3}{2} + \frac{5 0}{3} π = 253 + \frac{5 0}{3} π

Finally, the area watered by both sprinklers is twice the above result.

Area Watered in Option 2 (m^{2}) (253 + \frac{5 0}{3} π) \times 2 = 503 + \frac{1 0 0}{3} π

Difference

Now that the area of the garden watered in both options is known, the difference can be found.

difference = (503 + \frac{1 0 0}{3} π) - 50 π

Evaluate right-hand side

RemovePar

Remove parentheses

difference = 503 + \frac{1 0 0}{3} π - 50 π

NumberToFrac

$a = \frac{3 \cdot a}{3}$

difference = 503 + \frac{1 0 0}{3} π - \frac{1 5 0}{3} π

FactorOut

Factor out $π$

difference = 503 + (\frac{1 0 0}{3} - \frac{1 5 0}{3}) π

SubFrac

Subtract fractions

difference = 503 - \frac{5 0}{3} π

UseCalc

Use a calculator

difference = 34.242662 \dots

RoundInt

Round to nearest integer

difference \approx 34

The difference is about

34

square meters. Go ahead and make a recommendation for Paulina's sprinkling system.

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Catch-Up and Review

Hint

Solution

Square

Equilateral Triangle

Circle

Comparison

Hint

Solution

Area of Magdalena's Circles

Area of Dylan's Circles

Comparison

Hint

Solution

Option 1

Option 2

Difference

Option $1$

Option $2$