McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
3. Special Right Triangles
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Exercise 31 Page 645

In the given figure, you can find a 45^(∘)- 45 ^(∘) - 90 ^(∘) triangle.

x=5
y=10

Practice makes perfect

Let's begin by reviewing the properties of special right triangles.

45^(∘) - 45 ^(∘) - 90 ^(∘) Triangle
In a 45^(∘) - 45 ^(∘) - 90 ^(∘) triangle, the legs l are congruent and the hypotenuse h is sqrt(2) times the length of a leg.
30 ^(∘) - 60 ^(∘) - 90 ^(∘) Triangle
In a 30 ^(∘) - 60 ^(∘) - 90 ^(∘) triangle, the hypotenuse h is 2 times the length of the shorter leg s and the longer leg l is sqrt(3) times the length of the shorter leg.
Let's name the vertices of the given figure. Then we can find y and x, one at a time.

Finding y

We are given that ∠ BEC is a right angle and that EC and EB are congruent. Therefore, △ BCE is a 45^(∘) - 45 ^(∘) - 90 ^(∘) triangle. We also know that the length of the leg EC is 5 sqrt(2). By the 45^(∘) - 45 ^(∘) - 90 ^(∘) Triangle Theorem, l= 5 sqrt(2) and the hypotenuse y is sqrt(2) times the length of a leg, l. y=l sqrt(2) To find y, let's substitute l= 5 sqrt(2) in this equation.
y=l sqrt(2)
y= 5 sqrt(2) * sqrt(2)
Simplify right-hand side
y=5sqrt(4)
y=5*2
y=10

Finding x

We are given that ∠ BCD is a right angle, and through the process of finding y, we know that the measure of ∠ BCE is 45 ^(∘). Since ∠ DCE and ∠ BCE are complementary, the measure of ∠ DCE is also 45^(∘). Therefore, △ ECD is also a 45^(∘) - 45 ^(∘) - 90 ^(∘) triangle.

The length of the leg is x and the length of the hypotenuse is 5 sqrt(2). By the 45^(∘) - 45 ^(∘) - 90 ^(∘) Triangle Theorem, the length of the hypotenuse is sqrt(2) times the length of a leg, x. x sqrt(2) = 5 sqrt(2) ⇔ x=5