McGraw Hill Glencoe Algebra 2, 2012
MH
McGraw Hill Glencoe Algebra 2, 2012 View details
2. Angles and Angle Measure
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Exercise 56 Page 805

Begin by finding XY. Then use the Pythagorean Theorem.

G

Practice makes perfect

We are told that the area of the following figure is 60 square units. We want to find the length of XZ.

Notice that in this right triangle we are given the length of one of the bases and we are looking for the length of the hypotenuse. Since we are provided with the area of the triangle, this will only help us find XY.

Finding XY

First, we will call the two missing lengths b and c.

The area of a triangle is half the product of the base and its corresponding height, in this case b and 6. Area: 1/2( b)(6) = 6b/2 Since we already know the area of the triangle, we can substitute it into the formula and solve for b.

A = 6b/2
60=6b/2
60=3b
20=b
b=20

The other base of the triangle, XY, is 20 units.

Finding XZ

Now that we have the lengths of both of the bases of the triangle, we can use the Pythagorean Theorem to find the missing length.

Given the theorem, let's write the related equation and solve it for c.

6^2+( 20)^2= c^2
36+400=c^2
436=c^2
sqrt(436)=sqrt(c^2)
sqrt(436)=c
â–¼
Calculate root
sqrt(4* 109)=c
sqrt(4)* sqrt(109)=c
2sqrt(109)=c
c=2sqrt(109)

The hypotenuse XZ of the triangle equals 2sqrt(109) units, which corresponds to answer G.