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Math 102: College Mathematics14 chapters | 108 lessons

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Lesson Transcript

Instructor:
*DaQuita Hester*

DaQuita has taught high school mathematics for six years and has a master's degree in secondary mathematics education.

Similar triangles are used to solve problems in everyday situations. Learn how to solve with similar triangles here, and then test your understanding with a quiz.

Sarah is standing outside next to a flagpole. The sun casts a 4 ft. shadow of Sarah and a 7 ft. shadow of the flagpole. If Sarah is 5 ft. tall, how tall is the flagpole?

Without a ladder and measuring stick, you may think that solving this problem is impossible. However, finding the solution may be easier than you think. For this problem, and others like it, solving becomes a matter of **similar triangles**.

In another lesson, we learned that similar triangles have congruent corresponding angles and proportional corresponding sides and can be quickly identified by the **Angle-Angle (AA)**, **Side-Angle-Side (SAS)**, and **Side-Side-Side (SSS)** similarity theorems. To solve with similar triangles, we will use their side lengths to set up proportions, meaning that we will create fractions for the corresponding sides and set them equal to each other. Let's begin our practice with some basic examples.

Take a look at triangle ABE and triangle ACD above. From the picture, we see that *Angle B* is congruent to *Angle C* and *Angle E* is congruent to *Angle D*. Therefore, these triangles are similar by AA. Let's set up our proportions.

By going in the same order for each fraction, our beginning proportion is 8/5 = *x*/6. To get *x* by itself, we must cross-multiply, which will give us 5*x* = 48. Next, we will divide both sides by 5 to conclude that *x* = 9.6.

Here's another one: This time, we have triangle MOQ and triangle NOP. Once again, based on the congruent marks in the picture above, we know that these two triangles are similar by AA. So, let's start the problem.

In the figure, we were given the length of side MO, meaning that we also need the length of side QO for the proportion. But, we were not given this information, so we must calculate it.

To get the length of QO, we must add QP + PO since these two pieces form segment QO. By doing this, we add 7 + 8 to see that QO = 15. Now, we are ready for our proportion.

By going in the same order, our proportion is 7/15 = *x*/18. Then, when we cross-multiply, we see that 15*x* = 126. Once we divide both sides by 15, we can conclude that *x* = 8.4.

Now that we've covered some of the basics, let's do some real-world examples, starting with Sarah and the flagpole. Recall that Sarah is 5 ft. tall and has a 4 ft. shadow, and we are looking for the height of the flagpole, which has a 7 ft. shadow.

Do you see the similar triangles above? Sarah and the flagpole are the vertical legs and the shadows are the horizontal legs. Connecting the end of the shadows to the top of Sarah and the flagpole complete the triangles. Since the sun is casting both shadows, and since Sarah is standing right beside the flagpole, the top angles in both triangles are congruent. Additionally, since Sarah and the flagpole are perpendicular with the ground, each triangle has a right angle. Therefore, these triangles are similar by the AA similarity theorem.

Now we can write a proportion. Using *x* to represent the height of the flagpole, we begin with 5/4 = *x*/7. After cross-multiplying, we have 4*x* = 35, and then, by dividing both sides by 4, we see that *x* = 8.75. Therefore, the flagpole has a height of 8.75 ft.

Let's try another: Julie places a mirror on the ground 30 ft. away from the base of her office building and walks backwards until she can see the top of the building in the mirror. If Julie is 5.5 ft. tall and is standing 5ft. away from the mirror, how tall is the building?

For this scenario, Julie and the building serve as the vertical legs of the triangles and the distances from the mirror to Julie and the building are the horizontal legs. The triangles are completed by connecting the mirror to the top of both Julie and the building.

Since Julie can see the top of her building through the mirror, the two angles formed at the mirror are congruent. Also, Julie and her office building form right angles with the ground, showing us that these two triangles above are also similar by AA. Now, it's time for the proportion.

By letting *x* represent the height of the building, our beginning proportion is *x*/30 = 5.5/5. From cross-multiplying, we get 5*x* = 165. Then, by dividing both sides by 5, we can conclude that *x* = 33. Therefore, the height of Julie's office building is 33 ft.

In conclusion, **similar triangles** can be applied to solve everyday, real-world problems. Since the triangles are similar, solving with them requires us to set up proportions that compare their corresponding sides. After going in the same order to set up the proportions, cross-multiplying is necessary to complete and solve the problem.

You'll be able to apply the angle-angle theorem of similar triangles to solve real-world problems after watching this video lesson.

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Math 102: College Mathematics14 chapters | 108 lessons

- Go to Logic

- Go to Sets

- Properties of Shapes: Rectangles, Squares and Rhombuses 5:46
- Properties of Shapes: Triangles 5:09
- Perimeter of Triangles and Rectangles 8:54
- Area of Triangles and Rectangles 5:43
- Circles: Area and Circumference 8:21
- The Pythagorean Theorem: Practice and Application 7:33
- How to Identify Similar Triangles 7:23
- Applications of Similar Triangles 6:23
- Types of Angles: Vertical, Corresponding, Alternate Interior & Others 10:28
- Angles and Triangles: Practice Problems 7:43
- Properties of Shapes: Circles 4:45
- Go to Geometry

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