Geometry (B): READ: Geometric Mean Reading

Using Similar Right Triangles

Learning Objectives

Identify similar triangles inscribed in a larger triangle.
Evaluate the geometric mean of various objects.
Identify the length of an altitude using the geometric mean of a separated hypotenuse.
Identify the length of a leg using the geometric mean of a separated hypotenuse.

Introduction

In this lesson, you will study figures inscribed, or drawn within, existing triangles. One of the most important types of lines drawn within a right triangle is called an altitude. Recall that the altitude of a triangle is the perpendicular distance from one vertex to the opposite side. By definition each leg of a right triangle is an altitude. We can find one more altitude in a right triangle by adding an auxiliary line segment that connects the vertex of the right angle with the hypotenuse, forming a new right angle.

You may recall this is the figure that we used to prove the Pythagorean Theorem. In right triangle $ABC$ above, the segment $\overline {CD}$ is an altitude. It begins at angle $C$ , which is a right angle, and it is perpendicular to the hypotenuse $\overline {AB}$ . In the resulting figure, we have three right triangles, and all of them are similar.

Inscribed Similar Triangles

You may recall that if two objects are similar, corresponding angles are congruent and their sides are proportional in length. In other words, similar figures are the same shape, but different sizes. To prove that two triangles are similar, it is sufficient to prove that all angle measures are congruent (note, this is NOT true for other polygons. For example, both squares and “long” rectangles have all $90^\circ$ angles, but they are not similar). Use logic, and the information presented above to complete Example 1.

Example 1

Justify the statement that $\triangle {TQR} \sim \triangle {TSQ} \sim \triangle {QSR}$ .

In the figure above, the big triangle $\triangle {TQR}$ is a right triangle with right angle $\angle {Q}$ and $m \angle {R} = 30^\circ$ and $m \angle {T} = 60^\circ$ . So, if $\triangle {TQR} , \triangle {TSQ}$ , and $\triangle {QSR}$ are similar, they will all have angles of $30^\circ , 60^\circ$ , and $90^\circ$ .

First look at $\triangle {TSQ}$ . $m \angle {QST}=90^\circ$ , and $m \angle {T} = 60^\circ$ . Since the sum of the three angles in a triangle always equals $180^\circ$ , the missing angle, $\angle {TQS}$ , must measure $30^\circ$ , since $30 + 60 + 90 = 180$ . Lining up the congruent angles, we can write $\triangle {TQR} \sim \triangle {TSQ}$ .

Now look at $\triangle{QRS} . \angle{QSR}$ has a measure of $90^\circ$ , and $m\angle{R}=30^\circ$ . Since the sum of the three angles in a triangle always equals $180^\circ$ , the missing angle, $\angle {RQS}$ , must measure $60^\circ$ , since $30 + 60 + 90 = 180.$ Now, since the triangles have congruent corresponding angles, $\triangle {QSR}$ and $\triangle {TQR}$ are similar.

Thus, $\triangle {TQR} \sim \triangle {TSQ} \sim \triangle {QSR}$ . Their angles are congruent and their sides are proportional.

Note that you must be very careful to match up corresponding angles when writing triangle similarity statements. Here we should write $\triangle {TQR} \sim \triangle {TSQ} \sim \triangle {QSR}$ . This example is challenging because the triangles are overlapping.

Geometric Means

When someone asks you to find the average of two numbers, you probably think of the arithmetic mean (average). Chances are good you’ve worked with arithmetic means for many years, but the concept of a geometric mean may be new. An arithmetic mean is found by dividing the sum of a set of numbers by the number of items in the set. Arithmetic means are used to calculate overall grades and many other applications. The big idea behind the arithmetic mean is to find a “measure of center” for a group of numbers.

A geometric mean applies the same principles, but relates specifically to size, length, or measure. For example, you may have two line segments as shown below. Instead of adding and dividing, you find a geometric mean by multiplying the two numbers, then finding the square root of the product.

To find the geometric mean of these two segments, multiply the lengths and find the square root of the product.

$\text{mean}&=\sqrt{8 \cdot 2}\\ &=\sqrt{16} \\ &=4$

So, the geometric mean of the two segments would be a line segment that is $4\;\mathrm{cm}$ in length. Use these concepts and strategies to complete example 2.

Example 2

In $\triangle{BCD}$ below, what is the geometric mean of $BC$ and $CD$ ?

When finding a geometric mean, you first find the product of the items involved. In this case, segment $BC$ is $12\;\mathrm{inches}$ and segment $CD$ is $3\;\mathrm{inches}$ . Then find the square root of this product.

$\text{mean}&=\sqrt{12 \cdot 3}\\ &=\sqrt{36} \\ &=6$

So, the geometric mean of $BC$ and $CD$ in $\triangle{BCD}$ is $6\;\mathrm{inches}$ .

Altitude as Geometric Mean

In a right triangle, the length of the altitude from the right angle to the hypotenuse is the geometric mean of the lengths of the two segments of the hypotenuse. In the diagram below we can use $\triangle{BDC} \sim \triangle{CDA}$ to create the proportion $\frac{d}{f} = \frac{f}{e}$ . Solving for $f , f = \sqrt{d \cdot e}$ .

You can use this relationship to find the length of the altitude if you know the length of the two segments of the divided hypotenuse.

Example 3

What is the length of the altitude $\overline {AD}$ in the triangle below?

To find the altitude of this triangle, find the geometric mean of the two segments of the hypotenuse. In this case, you need to find the geometric mean of $9$ and $3$ . To find the geometric mean, find the product of the two numbers and then take its square root.

$\text{mean} &= \sqrt{9 \cdot 3}\\ &=\sqrt{27}\\ &=3\sqrt{3}$

So $AD=3\sqrt{3}\;\mathrm{feet}$ , or approximately $3 (1.732) = 5.2\;\mathrm{feet}$ .

Example 4

What is the length of the altitude in the triangle below?

The altitude of this triangle is $\overline{AD}$ . Remember the altitude does not always go “down”! To find $AD$ , find the geometric mean of the two segments of the hypotenuse. Make sure that you fill in missing information in the diagram. You know that the whole hypotenuse, $\overline{CB}$ is $20\;\mathrm{inches}$ long and $BD=4\;\mathrm{inches}$ , but you need to know $CD$ , the length of the longer subsection of $\overline{CB}$ , to find the geometric mean. To do this, subtract.

$CD &= CB-DB\\ &=20 - 4\\ &=16$

So $CD=16\;\mathrm{inches}$ . Write this measurement on the diagram to keep track of your work.

Now find the geometric mean of $16$ and $4$ to identify the length of the altitude.

$AD &= \sqrt{16 \cdot 4}\\ &=\sqrt{64}\\ &=8$

The altitude of the triangle will measure $8\;\mathrm{inches}$ .

Leg as Geometric Mean

Just as we used similar triangles to create a proportion using the altitude, the lengths of the legs in right triangles can also be found with a geometric mean with respect to the hypotenuse. The length of one leg in a right triangle is the geometric mean of the adjacent segment and the entire hypotenuse. The diagram below shows the relationships.

$a & = \sqrt{d \cdot c} \\ b & = \sqrt{e \cdot c}$

You can use this relationship to find the length of the leg if you know the length of the two segments of the divided hypotenuse.

Example 5

What is the length of $x$ in the triangle below?

To find $x$ , the leg of the large right triangle, find the geometric mean of the adjacent segments of the hypotenuse and the entire hypotenuse. In this case, you need to find the geometric mean of $6$ and $12$ . To find the geometric mean, find the product of the two numbers and then take the square root of that product.

$x &= \sqrt{6 \cdot 12}\\ &= \sqrt{72}\\ &= 6 \sqrt{2}$

So, $x=6 \sqrt{2}\;\mathrm{millimeters}$ or approximately $8.49\;\mathrm{millimeters}$ .

Example 6

If $m = AB$ , what is the value $m$ in the triangle below?

To find $m$ in this triangle, find the geometric mean of the adjacent segment of the hypotenuse and the entire hypotenuse. Make sure that you fill in missing information in the diagram. You know that the two shorter sections of the hypotenuse are $15\;\mathrm{inches}$ and $5\;\mathrm{inches}$ , but you need to know the length of the entire hypotenuse to find the geometric mean. To do this, add.

$AD+DC &= AC\\ 5 + 15 &= 20$

So, $AC = 20\;\mathrm{inches}$ . Write this measurement on the diagram to keep track of your work.

Now find the geometric mean of $20$ and $5$ to identify the length of the altitude.

$m &= \sqrt{20 \cdot 5}\\ &= \sqrt{100}\\ &= 10$

So, $m=10\;\mathrm{inches}$ .

Lesson Summary

In this lesson, we explored how to work with different radicals both in theory and in practical situations. Specifically, we have learned:

How to identify similar triangles inscribed in a larger triangle.
How to evaluate the geometric mean of various objects.
How to identify the length of an altitude using the geometric mean of a separated hypotenuse.
How to identify the length of a leg using the geometric mean of a separated hypotenuse.

These skills will help you solve many different types of problems. Always be on the lookout for new and interesting ways to find relationships between sides and angles in triangles.

Points to Consider

How can you use the Pythagorean Theorem to identify other relationships between sides in triangles?

The following questions are for your own review. The answers are listed below for you to check your work and understanding.

Review Questions

Which triangles in the diagram below are similar?
What is the geometric mean of two line segments that are $1$ and $4\;\mathrm{inches}$ , respectively?
What is the geometric mean of two line segments that are $3\;\mathrm{cm}$ each?
Which triangles in the diagram below are similar?
What is the length of the altitude, $h$ , in the triangle below?
What is the length of $d$ in the triangle below?
What is the geometric mean of two line segments that are $4\;\mathrm{yards}$ and $8\;\mathrm{yards}$ , respectively?
What is the length of the altitude in the triangle below?

Use the following diagram from exercises 9-11:
$g =$ ____
$h =$ ____
$k =$ ____ (for an extra challenge, find $k$ in two different ways)
What is the length of the altitude in the triangle below?

Review Answers

Triangles $DEF, EGF,$ and $DGE$ are all similar
$2\;\mathrm{inches}$
$3\;\mathrm{cm}$
Triangles $MNO, PNM,$ and $PMO$ are all similar.
$6\;\mathrm{inches}$
$2\sqrt6\;\mathrm{mm}$ , or approximately $4.9\;\mathrm{mm}$
$4\sqrt2 \;\mathrm{yards}$ , or approximately $5.66\;\mathrm{yards}$
$5\sqrt2 \;\mathrm{feet}$ , or approximately $7.07\;\mathrm{feet}$
$g=\sqrt91 \;\mathrm{inches}$ , or approximately $9.54\;\mathrm{inches}$
$h=\sqrt78 \;\mathrm{inches}$ or approximately $8.83\;\mathrm{inches}$
$k=\sqrt42\;\mathrm{inches}$ or approximately $6.48\;\mathrm{inches}$ . One way to find $k$ is with the geometric mean: $k=\sqrt6.7= \sqrt42\;\mathrm{inches}$ . Alternatively, using the answer from 9 and one of the smaller right triangles, $k = \sqrt{(\sqrt{91})^2- (7)^2}=\sqrt{91-49}= \sqrt{42} \;\mathrm{inches}$

Last modified: Tuesday, June 29, 2010, 10:24 AM