Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
1. Section 7.1
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Exercise 10 Page 401

Practice makes perfect
a Notice that we are given a right triangle with one acute angle measuring 60^(∘).

This means that it is a 30-60-90 triangle and, by the Angle-Angle (AA) Similarity Theorem, it is similar to the following general triangle.

The shorter side in our triangle is 9 units long, so in order to get the lengths x and y, we need to multiply 2 and sqrt(3) by 9, respectively. lx = 9( 2) y = 9( sqrt(3)) ⇒ lx = 18 y = 9sqrt(3)

b This time we are given a right triangle with one acute angle measuring 45^(∘).

This means that it is a 45-45-90 triangle and, by the AA Similarity Theorem, it is similar to the following general triangle.

One of the legs in our triangle is 24 units long. In order to find the lengths x and y, we need to multiply sqrt(2) and 1 by 24, respectively. lx = 24( sqrt(2)) y = 24( 1) ⇒ lx = 24sqrt(2) y = 24

c This time we are given a right triangle with one acute angle measuring 30^(∘).

This means that the triangle is a 30-60-90 triangle and, by the AA Similarity Theorem, it is similar to the following generic triangle.

In a pair of similar triangles, the ratios of corresponding sides are equal. This lets us write the following two equations using the sides of our triangles. 8/sqrt(3) = x/1 = y/2 This gives us two equations, one for x and one for y. 8/sqrt(3) = x/1 8/sqrt(3) = y/2 Let's solve the first equation for x.
8/sqrt(3) = x/1
8/sqrt(3) = x
x = 8/sqrt(3)
x = 8sqrt(3)/sqrt(3)(sqrt(3))
x = 8sqrt(3)/3
We found that x = 8sqrt(3)3. Next, let's solve the equation for y.
8/sqrt(3) = y/2
16/sqrt(3) = y
y = 16/sqrt(3)
y = 16sqrt(3)/sqrt(3)(sqrt(3))
y = 16sqrt(3)/3
We found that y = 16sqrt(3)3.