Exploring Geometry in Design: Trigonometry and Similarity Unveiled

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.

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

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}

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

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}

What is the area of the shaded part of the triangle? Round the answer to two decimal places.

Triangle divided in three sections. One section is shaded

From the diagram, we can identify two isosceles triangles with base angles of 53^(∘) and a base length of 8 and 4 units respectively. The height drawn to the vertex angle in an isosceles triangle bisects the base creating two congruent segments, and it divides the triangle into two congruent right triangles. Let's highlight these in the diagram.

Notice that the shaded area is the area of the larger right triangle minus the area of the smaller right triangle. To determine these areas, we want to know the length of both legs for both triangles. Since we know one leg and one acute angle, we can use the tangent ratio to determine the second leg. tan 53^(∘) = h_1/4 ⇔ h_1= 4tan 53^(∘) [1em] tan 53^(∘) = h_2/2 ⇔ h_2= 2tan 53^(∘) Now we can determine the area of each triangle by multiplying their two legs and dividing by 2. A_1=1/2(4)( 4tan 53^(∘))=8 tan 53^(∘) [1em] A_2=1/2(2)( 2tan 53^(∘))=2 tan 53^(∘) Notice that we are keeping the expressions for each area in exact form to avoid rounding errors. Finally, we will subtract the smaller triangle's area from the larger triangle's area.

The area of the shaded part of the triangle is about 7.96 square inches.

The figure consists of five congruent squares and has an area of $405 cm^{2} .$ What is the perimeter of the figure?

The perimeter consists of 12 congruent sides. Let's label these sides s.

To be able to calculate the perimeter, we first want to determine s. Since the area of a square is the square of its side, we can find s if we determine the area of one of these five squares. Notice that we have been given the area of the entire figure. Therefore, if we divide this number by 5, we get the area of one of the squares. 405/5=81 cm^2 When we know the area of one square, we can use it to determine the side length s.

The side is 9 centimeters. By multiplying this by 12, we find the perimeter of the shape. P=12( 9)=108 cm

What percentage of the rectangle is shaded? Round the answer to the nearest whole percent.

a rectangle with two right triangles in it

Let's determine the area of the rectangle A_R by multiplying its length and width. A_R=(6)(4)=24cm^2 Next, we want to determine the area of the triangles, A_1 and A_2. Since these are right triangles, we can use the formula for the area of a triangle. In A_1, we have been given an acute angle of 75^(∘) and its opposite side of 6 centimeters. In A_2, we have an acute angle of 10^(∘) and its adjacent side of 6 centimeters. Let's label the legs with unknown lengths a and b, respectively.

By using the tangent ratio, we can write two equations. tan 10^(∘) = b/6 &⇔ b= 6tan 10^(∘) [1em] tan 75^(∘) = 6/a &⇔ a= 6/tan 75^(∘) Since we know the base values of each triangle and we have expressions for the height of each triangle, we can substitute them into the formula for the area of a triangle. A_1= 1/2(6)(6tan 10^(∘)) = 18tan 10^(∘) [1em] A_2= 1/2(6)(6/tan 75^(∘))=18/tan 75^(∘) The percent of the entire rectangle that is shaded is the ratio of the sum of the triangle areas A_1 and A_2 to the rectangle area A_R. That can be written as A_1+A_2A_R. Then we will substitute the corresponding values into this equation.

Geometry in Design

Catch-Up and Review

Investigating Efficient Solutions Using Geometry

Using Geometry in Space

Hint

Solution

Optimizing Storage Using Geometry

Hint

Solution

Non-Staggered Pipes

Staggered Pipes

Difference

Optimizing Area Using Geometry

Hint

Solution

Square

Equilateral Triangle

Circle

Comparison

Comparing Areas Using Geometry

Hint

Solution

Area of Magdalena's Circles

Area of Dylan's Circles

Comparison

Optimizing Efficiency Using Geometry

Hint

Solution

Option $1$

Option $2$

Difference

Geometry in Design

Recommended exercises

	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 $2$