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Pearson Geometry Common Core, 2011

PG

Pearson Geometry Common Core, 2011 View details

2. Special Right Triangles

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Exercise 11 Page 503

Notice that the given figure is a 45^(∘)-45^(∘)-90^(∘) triangle.

x=5sqrt(2)

Practice makes perfect

Consider the given right triangle.

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

10= sqrt(2) * x

▼

Solve for x

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

10/sqrt(2)=x

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

10sqrt(2)/sqrt(2)* sqrt(2)=x

sqrt(a)* sqrt(a)= a

10sqrt(2)/2=x

a/b=.a /2./.b /2.

5sqrt(2)=x

Rearrange equation

x=5sqrt(2)

Subchapter links

Lesson Check p.503

Practice and Problem-Solving Exercises p.503-505

Standardized Test Prep p.505

Mixed Review p.505