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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.

Try your knowledge on these topics.

Find the perimeter and area of the following figures.

a
rectangle
b
triangle
c Calculate the sine, cosine, and tangent of in the right triangle. Write the answers as fractions in their simplest form.
triangle
d By using the Pythagorean Theorem, calculate the missing side length.
triangle
e Calculate the length of the arc and the area of the sector. Write the answer correct to one decimal place.
sector
Challenge

Investigating Efficient Solutions Using Geometry

Paulina bought two sprinklers to water her meter by 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!
Sprinkles
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 From the perspective of Station the satellite is on the horizon. From the perspective of station the satellite is always directly overhead.

earth and satellite

The measure of the angle between the lines from the satellite to the stations is To answer the following questions, assume that the Earth is a sphere with a diameter of kilometers. Write all the answers in kilometer, rounded to the nearest hundred.

a How many kilometers will a signal sent from Station to the satellite and then to Station travel?
b Two astronomers are flying from Station to Station 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 and Station

Hint

a Consider the right triangle formed by the center of the Earth, Station and the satellite.
b The distance the astronomers will fly is the length of the arc whose endpoints are Station and Station
c Consider the isosceles triangle formed by the center of the Earth, Station and Station

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 kilometers, its radius can be calculated.
The radius of the Earth is kilometers. Since the satellite is on the horizon from the perspective of Station 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 and the satellite is a right triangle.
right triangle
The distance traveled by signal from Station to the satellite is equal to This length can be calculated using trigonometric ratios. Note that is given. The length of which is to is also known. The length of the side is desired. Therefore, the tangent ratio can be used.
The above equation can be solved by using inverse operations and Properties of Equality.
Solve for
The distance from Station to the satellite is kilometers.
right triangle
When two legs of a right triangle are known, its hypotenuse can be found. That can be done by using the Pythagorean Theorem.
Solve for
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 kilometers.
right triangle

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

right triangle
The distance from Station to the satellite can be calculated by using the Segment Addition Postulate.
The distance from Station to the satellite is kilometers.
right triangle
Finally, the distance that a signal sent from Station to the satellite and then to Station is the sum of and
This number approximated to the nearest hundred is 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.
right triangle
To find the length of an arc, a proportion that relates it with the measure of the arc must be solved.
Recall that the measure of an arc is the same as the measure of its corresponding central angle. When considering the angle of the corresponding sector, can be found using the Triangle Angle Sum Theorem.
Solve for
The angle of the sector that corresponds to measures Therefore, the measure of is also
right triangle
Recall that the circumference of a sphere, Earth, is equal to its diameter multiplied by
Now the arc length can be calculated by using the formula previously mentioned.
Solve for

Approximate to nearest hundred

The astronomers will travel 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 and are both Therefore, the triangle formed by the Earth's center and the stations is an isosceles triangle. It is also known that
right triangle
The distance that the signal will travel is equal to To find this value, the altitude of through vertex 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 be the point of intersection of the and
right triangle
It can be seen above that is a right triangle whose hypotenuse is Since in this triangle is the side opposite the sine ratio can be used to find Once this length is obtained, it can be used to obtain — the distance that the signal will travel.
Solve for
The length of is For symmetry reasons, the length of is also
right triangle
According to the Segment Addition Postulate, the length of — the distance that the signal will travel — can be found by adding and
Rounded to the nearest hundred the distance that the signal will travel is 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 centimeters. They need to be stored in a toolbox of width centimeters. The pipes can be staggered or non-staggered piles.
pipes
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.
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.
non-staggered pipes

The height of the pile formed by the non-staggered pipes is 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.
Therefore, the triangle is an equilateral triangle with a side length of centimeters.
staggered pipes and triangle

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 centimeters and with side length of centimeters is obtained.

pipes and right triangle
The altitude of the equilateral triangle is the missing leg of the right triangle. It can be found by using the Pythagorean Theorem. Let and be the legs of the right triangle, and its hypotenuse.
Solve for
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 centimeters.
staggered pipes and altitude of triangle

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

staggared pipes in toolbox
The height of the pile can be calculated by using the Segment Addition Postulate.

Difference

Finally, the difference between the heights of the piles can be calculated.
Evaluate
With a difference of 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 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.
playground
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 be the side length of a square region enclosed by meters of fence. The perimeter of the region will then be equal to
Solve for
The side length of the square is meters.
playground
Now, to find its area, the side length can be squared.

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
To find the area of the triangle, its height 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.
playground
It can be seen that the right triangle formed at the left of the diagram has hypotenuse and legs and To find the missing value, the Pythagorean Theorem can be used.
Solve for
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 meters. Recall that the side length of the equilateral triangle is meters. This means that its base is also meters.
playground
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.
Evaluate right-hand side
The area of the equilateral triangle is square meters.

Circle

Just one more major step. Finally, the area of the circle will be calculated. Since Ali bought meters of fence, the circumference is meters. Recall that the circumference of a circle is twice the product of and its radius. With this information, the radius of the circle can be found.
Solve for
The radius of the circle is meters.
playground
The area of a circle is the product of and the square of its radius.
Evaluate right-hand side
The area of the circle is 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

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

equal radii
By using the Pythagorean Theorem, the height of the triangle can be calculated.
Solve for
When solving the equation, only the principal root was considered because a length is always positive. Therefore, the height of the right triangle is centimeters. With this information, its area can be calculated.
Next, note that one of the circles that Magdalena drew is the incircle of this right triangle.
equal radii
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.
Now, the formula provided by the teacher can be used to find the radius of one of the circles that Magdalena drew.
Evaluate right-hand side