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Pearson Geometry Common Core, 2011

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Pearson Geometry Common Core, 2011 View details

7. Similarity Transformations

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Exercise 32 Page 600

In a 45^(∘)-45^(∘)-90^(∘) triangle, the hypotenuse is sqrt(2) times the length of a leg.

38.9 in.

Practice makes perfect

The given triangle is a 45^(∘)-45^(∘)-90^(∘) triangle. Therefore the legs are congruent. Let x be its length.

In a 45^(∘)-45^(∘)-90^(∘) triangle, the hypotenuse is sqrt(2) times the length of a leg. 55 = x sqrt(2) Let's solve the above equation for x.

55 = x sqrt(2)

Rearrange equation

xsqrt(2)=55

.LHS /sqrt(2).=.RHS /sqrt(2).

x=55/sqrt(2)

a/b=a * sqrt(2)/b * sqrt(2)

x=55sqrt(2)/sqrt(2)*sqrt(2)

sqrt(a)* sqrt(a)= a

x=55sqrt(2)/2

Use a calculator

x=38.89087...

Round to 1 decimal place(s)

x≈ 38.9

Subchapter links

Lesson Check p.597

Practice and Problem-Solving Exercises p.598-600

Standardized Test Prep p.600

Mixed Review p.600