Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
2. Special Right Triangles
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Exercise 28 Page 504

Divide the given quadrilateral into a rectangle and a right triangle.

a=14, b=6sqrt(2)

Practice makes perfect

Let's divide the given quadrilateral into a rectangle and a right triangle.

We will deal with these two shapes one at a time.

Triangle

Since the right triangle has an acute angle that measures 45^(∘), by the Triangle Angle Sum Theorem the third angle must have a measure of 45^(∘). Let c be the length of the missing leg of the triangle.

Therefore, we have a 45^(∘)-45^(∘)-90^(∘)- triangle. In this type of triangle, the legs are congruent and the length of the hypotenuse is sqrt(2) times the length of a leg. With this information, we can find the length of the hypotenuse and missing leg. b&= sqrt(2) * 6=6sqrt(2) c&= 6 Let's add these newly found side lengths to the triangle.

Rectangle

Let's consider the given shape and the previously obtained information. Let s be the length of the bottom side of the rectangle.

Since opposite sides of a rectangle are congruent, we can conclude that s=8. We can find the value of a using the Segment Addition Postulate. a=8+6 ⇔ a=14