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 46 Page 647

Carmen is right, Audrey is not. See solution.

Practice makes perfect

Let's begin with recalling the 30^(∘)-60^(∘)-90^(∘) Triangle Theorem. This theorem tells us that the length of the hypotenuse of this right triangle is 2 times the length of the shorter leg s, and the length of the longer leg is sqrt(3) times the length of the shorter leg.

Now let's take a look at the given picture. Notice that the drawn triangle is equilateral, as it has three congruent angles.

As we can see, the altitude divides the triangle into two 30^(∘)-60^(∘)-90^(∘) triangles. Therefore, according to the theorem we recalled, the shorter leg of each right triangle, s, is half of the length of the hypotenuse, 6. s=1/2( 6)=3 Let's add this information to our picture.

From the theorem, we know that the longer leg is sqrt(3) times the length of the shorter leg, 3. x=3sqrt(3) The value of x is 3sqrt(3). Therefore, Carmen is correct in her reasoning and Audrey is not.