Î´qrs is a Right Triangle. Select the Correct Similarity Statement.

## SIMILAR RIGHT TRIANGLES

Two triangles are similar if two of their corresponding angles are congruent.

Example :

In the diagram shown above,

ΔABC

∼

ΔXYZ

Recall that the corresponding side lengths of similar triangles are in proportion.

## Theorem

If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other.

In the diagram shown above,

ΔCBD

∼

ΔABC

ΔACD∼ΔABC

ΔCBD∼ΔACD

Proof of the above Theorem :

Given :

ΔABC is a right triangle ; altitude CD is drawn to hypotenuse AB.

To Prove :

ΔCBD

∼

ΔABC,

ΔACD

∼

ΔABC,

ΔCBD

∼

ΔACD

Proof :

First proof that ΔCBD∼ΔABC.

Each triangle has a right angle,and each includes∠B.

The triangles are similar by the AA Similarity Postulate.

We can use similar reasoning to show that ΔACD∼ΔABC.

To show that ΔCBD∼ΔACD, begin by showing that

∠ACD

≅

∠B

because they are bothcomplementary to

∠

DCB. Then you can use the AA Similarity Postulate.

## Using a Geometric Mean To Solve Problems

In rightΔABC shown above, altitude CD is drawn to the hypotenuse, forming two smaller right triangles they are similar to

ΔABC. From the theorem given above, we know that

ΔCBD

∼

ΔACD

∼

ΔABC.

Notice that

CD

is longer leg ofΔCBD and the shorter leg

ΔACD.

When we write a proportion comparing the leg lengths of ΔCBD and ΔACD, we can see thatCDis the geometric mean of BD and AD.

Sides CB and ACalso appear in more than one triangle. Their side lengths arealso geometric means, as shown by the proportions below :

These results are expressed in the theorems below.

## Geometric Mean Theorems

Theorem 1 :

In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments.

In the rightΔABC shown above, the length of the altitude CD is the geometric mean of the lengths of the two segments. AD and BD.

That is,

BD/CD = CD/AD

CD^{2}= AD⋅ BD

CD

=

√(

A

D

⋅ BD)

Theorem 2 :

In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments.

The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.

In the rightΔABC shown above,

CB

^{}

=

√(

AB

⋅ DB)

AC

= √(AB ⋅ AD)

## Finding the Height of a Roof

Example :

A roof has a cross section that is a right triangle. The diagram shows the approximate dimensions of this cross section.

a. Identify the similar triangles.

b.Find the heighthof the roof.

Solution (a) :

We may find it helpful to sketch the three similar right triangles so that the corresponding angles and sides have the same orientation. Mark the congruent angles.

Notice that some sides appear in more than one triangle.

For instance, BC is the hypotenuse inΔBCA and the shorter leg in

ΔBDC.

In the diagram shown above,

ΔBCA

∼

ΔCDA

∼

ΔBDC

Solution (b) :

Use the fact thatΔBCA

∼

ΔBDC to write a proportion.

Corresponding side lengths are in proportion :

CA/DC = BC/BD

Substitute.

h/5.5 = 3.1/6.3

Apply cross product property.

6.3h = 3.1 (5.5)

Solve for h.

h

≈ 2.7

Hence, the height of the roof is about 2.7 meters.

## Using a Geometric Mean

Example 1 :

Find the value of x in the diagram shown below.

Solution :

Apply Geometric Mean Theorem 1.

x/3 = 6/x

Apply cross product property.

x^{2} = 18

Solve for x.

x = √18

x =

√(3

⋅ 3

⋅ 2)

x = 3

√

2

Example 2 :

Find the value of y in the diagram shown below.

Solution :

Apply Geometric Mean Theorem 2.

y/2 = (5 + 2)/y

y/2 = 7/y

Apply cross product property.

y^{2} = 14

Solve for x.

x = √14

## Using Indirect Measurement

Example :

To estimate the height of a monorail track, your friend holds a cardboard square at eye level. Your friend lines up the top edge of the square with the track and the bottom edge with the ground. You measure the distance from the ground to your friend’s eye and the distance from your

friend to the track.

In the diagram shown below,XY= h – 5.75 is the difference between the track height h and your friend’s eye level. Find the height of the track.

Solution :

Use Geometric Mean Theorem 2 to write a proportion involving XY. Then you can solve for h.

Geometric Mean Theorem 2 :

XY/WY = WY/ZY

Substitute.

(h – 5.75)/16 = 16/5.75

Apply cross product property.

5.75(h – 5.75) = 16^{2}

Distributive property.

5.75h – 33.0625 = 256

Add 33.0625 to each side.

5.75h = 289.0625

Divide each side by 5.75.

h≈ 50

Hence, the height of the track is about 50 feet.

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### Î´qrs is a Right Triangle. Select the Correct Similarity Statement.

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