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

Using Geometry in Space

A satellite orbiting the Earth uses radar to communicate with two stations on the surface. The satellite is orbiting in such a way that it is always in line with the center of Earth and Station $B .$ From the perspective of Station $A,$ the satellite is on the horizon. From the perspective of station $B,$ the satellite is always directly overhead.

The measure of the angle between the lines from the satellite to the stations is $1 2^{\circ} .$ To answer the following questions, assume that the Earth is a sphere with a diameter of $12700$ kilometers. Write all the answers in kilometer, rounded to the nearest hundred.

a How many kilometers will a signal sent from Station

A

to the satellite and then to Station

B

travel?

b Two astronomers are flying from Station

A

to Station

B

in a direct path along the surface of the Earth. What is the distance they will fly?

c Suppose a signal could travel through the surface of the Earth between stations. What is the shortest distance the signal would travel between Station

A

and Station

B ?

Hint

a Consider the right triangle formed by the center of the Earth, Station

A,

and the satellite.

b The distance the astronomers will fly is the length of the arc whose endpoints are Station

A

and Station

B .

c Consider the isosceles triangle formed by the center of the Earth, Station

A

and Station

B .

Solution

a The radius of a sphere is half its diameter. When this information is combined with the knowledge that the diameter of the Earth is

12700

kilometers, its radius can be calculated.

r = \frac{d}{2}

Substitute

$d = 12700$

r = \frac{1 2 7 0 0}{2}

CalcQuot

Calculate quotient

r = 6350

The radius of the Earth is

6350

kilometers. Since the satellite is on the horizon from the perspective of Station

A,

the line through this station and the satellite is tangent to the Earth. Therefore, it makes a right angle with the radius. Consequently, the triangle formed by the center of the Earth, Station

A,

and the satellite is a right triangle.

The distance traveled by signal from Station

A

to the satellite is equal to

A S .

This length can be calculated using trigonometric ratios. Note that

m ∠ S

is given. The length of

\overline{A C},

which is

opposite

∠ S,

is also known. The length of the side

adjacent

∠ S

is desired. Therefore, the tangent ratio can be used.

tan ∠ S = \frac{Opposite side to ∠ S}{Adjacent side to ∠ S} ⇓ tan 1 2^{\circ} = \frac{6 3 5 0}{A S}

The above equation can be solved by using inverse operations and Properties of Equality.

tan 1 2^{\circ} = \frac{6 3 5 0}{A S}

▼

Solve for

A S

MultEqn

$LHS \cdot A S = RHS \cdot A S$

tan 1 2^{\circ} \cdot A S = 6350

DivEqn

$LHS / tan 1 2^{\circ} = RHS / tan 1 2^{\circ}$

A S = \frac{6 3 5 0}{tan 1 2 ^{\circ}}

UseCalc

Use a calculator

A S = 29874.401201 \dots

RoundInt

Round to nearest integer

A S \approx 29874

The distance from Station

A

to the satellite is

29874

kilometers.

When two legs of a right triangle are known, its hypotenuse can be found. That can be done by using the Pythagorean Theorem.

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

SubstituteValues

Substitute values

6350^{2} + 29874^{2} = C S^{2}

▼

Solve for

C S

CalcPow

Calculate power

40322500 + 892455876 = C S^{2}

AddTerms

Add terms

932778376 = C S^{2}

SqrtEqn

$LHS = RHS$

932778376 = C S

CalcRoot

Calculate root

30541.42066 \dots = C S

RearrangeEqn

Rearrange equation

C S = 30541.42066 \dots

RoundInt

Round to nearest integer

C S \approx 30541

Note that when solving the equation, only the principal root was considered. This is because side lengths are always positive. Therefore, it can be said that the distance from the center of the Earth to the satellite is

30541

kilometers.

Next, consider Station $B .$ Because it is located on the Earth's surface, its distance from the center is equal to the radius of the Earth, which is $6350$ kilometers.

The distance from Station

B

to the satellite can be calculated by using the Segment Addition Postulate.

C B + B S = C S

SubstituteII

$C S = 30541$ , $C B = 6350$

6350 + B S = 30541

SubEqn

$LHS - 6350 = RHS - 6350$

B S = 24191

The distance from Station

B

to the satellite is

24191

kilometers.

Finally, the distance that a signal sent from Station

A

to the satellite and then to Station

B

is the sum of

A S

and

B S .

29874 + 24191 = 54065 km

This number approximated to the nearest hundred is

54100

kilometers.

b First, note that the astronomers will travel along the surface of the Earth which is spherical. Therefore, their path is not linear, but rather an arc length.

To find the length of an arc, a proportion that relates it with the measure of the arc must be solved.

\frac{Arc length}{Circumference} = \frac{Arc measure}{3 6 0 ^{\circ}}

Recall that the measure of an arc is the same as the measure of its corresponding central angle. When considering

△ A C S,

the angle of the corresponding sector,

∠ C,

can be found using the Triangle Angle Sum Theorem.

m ∠ A + m ∠ C + m ∠ S = 18 0^{\circ}

SubstituteII

$m ∠ A = 9 0^{\circ}$ , $m ∠ S = 1 2^{\circ}$

9 0^{\circ} + m ∠ C + 1 2^{\circ} = 18 0^{\circ}

▼

Solve for

m ∠ C

AddTerms

Add terms

m ∠ C + 10 2^{\circ} = 18 0^{\circ}

SubEqn

$LHS - 10 2^{\circ} = RHS - 10 2^{\circ}$

m ∠ C = 7 8^{\circ}

The angle of the sector that corresponds to

A B

measures

7 8^{\circ} .

Therefore, the measure of

A B

is also

7 8^{\circ} .

Recall that the circumference of a sphere, Earth, is equal to its diameter multiplied by

π .

Circumference : 12700 π

Now the arc length can be calculated by using the formula previously mentioned.

\frac{Arc length}{Circumference} = \frac{Arc measure}{3 6 0 ^{\circ}}

SubstituteII

$Arc measure = 7 8^{\circ}$ , $Circumference = 12700 π$

\frac{Arc length}{1 2 7 0 0 π} = \frac{7 8 ^{\circ}}{3 6 0 ^{\circ}}

▼

Solve for

arc length

MultEqn

$LHS \cdot 12700 π = RHS \cdot 12700 π$

Arc length = \frac{7 8 ^{\circ}}{3 6 0 ^{\circ}} (12700 π)

MoveRightFacToNum

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

arc length = \frac{9 9 0 6 0 0 ^{\circ}}{3 6 0 ^{\circ}} π

ReduceFrac

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

Arc length = \frac{8 2 5 5}{3} π

UseCalc

Use a calculator

Arc length = 8644.615785 \dots

Approximate to nearest hundred

Arc length \approx 8600

The astronomers will travel

8600

kilometers, to the nearest hundred kilometers.

c If a signal can travel through the Earth's surface, the shortest distance it will travel equals the length of the segment that connects the stations. It is already known that

A C

and

B C

are both

6350 .

Therefore, the triangle formed by the Earth's center and the stations is an isosceles triangle. It is also known that

m ∠ C = 7 8^{\circ} .

The distance that the signal will travel is equal to

A B .

To find this value, the altitude of

△ A B C

through vertex

C

will be considered. Keep in mind that the altitude is perpendicular to the base. Also, this altitude bisects the vertex angle of the isosceles triangle. Let

M

be the point of intersection of the

altitude

and

\overline{A B} .

It can be seen above that

△ B C M

is a right triangle whose hypotenuse is

6350 .

Since in this triangle

\overline{M B}

is the side opposite

∠ B C M,

the sine ratio can be used to find

M B .

Once this length is obtained, it can be used to obtain

A B

— the distance that the signal will travel.

sin θ = \frac{Length of the side opposite θ}{Hypotenuse}

SubstituteValues

Substitute values

sin 3 9^{\circ} = \frac{M B}{6 3 5 0}

▼

Solve for

M B

MultEqn

$LHS \cdot 6350 = RHS \cdot 6350$

sin 3 9^{\circ} \cdot 6350 = M B

UseCalc

Use a calculator

3996.184483 \dots = M B

RearrangeEqn

Rearrange equation

M B = 3996.184483 \dots

RoundInt

Round to nearest integer

M B \approx 3996

The length of

\overline{M B}

3996 .

For symmetry reasons, the length of

\overline{A M}

is also

3996 .

According to the Segment Addition Postulate, the length of

\overline{A B}

— the distance that the signal will travel — can be found by adding

M B

and

A M .

A B = 3996 + 3996 \Leftrightarrow A B = 7992

Rounded to the nearest hundred the distance that the signal will travel is

8000

kilometers.

Example

Optimizing Storage Using Geometry

Mark's favorite subject is Geometry. He can never get enough! While organizing his storage space, he came across a situation he calls the staggered pipes situation. There are six pipes, each with a radius of

3

centimeters. They need to be stored in a toolbox of width

18

centimeters. The pipes can be staggered or non-staggered piles.

Mark would like to calculate the difference in heights of the two piles. Help him find that value! Round the answer to three significant figures.

Hint

For the staggered pipes, consider the triangle formed by the centers of two pipes next to each other and the pipe on top of them.

Solution

The heights of each pile will be calculated one at a time. Their difference can then be calculated.

Non-Staggered Pipes

The diameter of the pipes is twice their radius.

\underline{Diameter} 2 \times 3 = 6 centimeters

Since the pipes are not staggered, they are directly on stacked on top of each other without a gap. Therefore, the height of the pile is the sum of the diameters of two vertically stacked pipes.

The height of the pile formed by the non-staggered pipes is $12$ centimeters.

Staggered Pipes

To find the height of this pile, the triangle formed by the centers of two pipes next to each other and the pipe on top of them will be considered. Note that the length of each side of this triangle is equal to the sum of two radii.

\underline{Side Length} 3 + 3 = 6 centimeters

Therefore, the triangle is an equilateral triangle with a side length of

6

centimeters.

Consider now the altitude of the above triangle. Note that the altitude of an equilateral triangle bisects the base. Recall also that the altitude of a triangle is perpendicular to the base. Therefore, a right triangle with hypotenuse $6$ centimeters and with side length of $3$ centimeters is obtained.

The altitude of the equilateral triangle is the missing leg of the right triangle. It can be found by using the Pythagorean Theorem. Let

a

and

b

be the legs of the right triangle, and

c

its hypotenuse.

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

SubstituteII

$a = 3$ , $c = 6$

3^{2} + b^{2} = 6^{2}

▼

Solve for

b

CalcPow

Calculate power

9 + b^{2} = 36

SubEqn

$LHS - 9 = RHS - 9$

b^{2} = 27

SqrtEqn

$LHS = RHS$

b = 27

SplitIntoFactors

Split into factors

b = 9 (3)

SqrtProd

$a \cdot b = a \cdot b$

b = 93

CalcRoot

Calculate root

b = 33

Be aware that, when solving the equation, only the principal root was considered. That is because side lengths are always positive. Therefore, the length of the leg of the right triangle, which is the altitude of the equilateral triangle, is

33

centimeters.

This information can be added to the diagram of the staggered pipes.

The height of the pile can be calculated by using the Segment Addition Postulate.

\underline{Height of the Pile} 3 + 33 + 33 + 3 = (6 + 63) cm

Difference

Finally, the difference between the heights of the piles can be calculated.

(6 + 63) - 12

▼

Evaluate

RemovePar

Remove parentheses

6 + 63 - 12

SubTerm

Subtract term

63 - 6

UseCalc

Use a calculator

4.392304 \dots

RoundSigDig

Round to $3$ significant digit(s)

4.39

With a difference of

4.39

inches, Mark will be able to fit a variety of other objects into the toolbox based on his preferred layout.

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

Commutative Property of Multiplication

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

Commutative Property of Multiplication

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

Identity Property of Multiplication

\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

Commutative Property of Multiplication

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

Commutative Property of Multiplication

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.

	7 Theory slides
	10 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Geometry in Design

Catch-Up and Review

Hint

Solution

Hint

Solution

Non-Staggered Pipes

Staggered Pipes

Difference

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

Geometry in Design

Recommended exercises

Option $1$

Option $2$