McGraw Hill Glencoe Geometry, 2012
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McGraw Hill Glencoe Geometry, 2012 View details
2. The Pythagorean Theorem and Its Converse
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Exercise 49 Page 554

We need to use the Pythagorean Theorem a few times.

x≈5.4

Practice makes perfect

We are given a figure that contains at least four right triangles and asked to find the value of x. Let's look at the picture and name all of the missing sides with the consecutive letters.

Our first step will be to evaluate the value of a.

To do this, we will use the Pythagorean Theorem. According to this theorem, the sum of the squared legs of a right triangle is equal to its squared hypotenuse. 7.5^2+ a^2= 11.6^2 Now, we will solve the above equation. Remember that if we are taking a square root of a squared number that represents a side length, we consider only the positive case.
7.5^2+a^2=11.6^2
Solve for a
56.25+a^2=134.56
a^2=78.31
sqrt(a^2)=sqrt(78.31)
a=sqrt(78.31)
a=8.8492...
a≈8.85
The value of a is approximately 8.85. Next we can solve for b.
To do this, we will again use the Pythagorean Theorem. 8.85^2+ 15^2= b^2 Let's solve the above equation.
8.85^2+15^2=b^2
Solve for b
78.31+225=b^2
303.31=b^2
b^2=303.31
sqrt(b^2)=sqrt(303.31)
b=sqrt(303.31)
b=17.4157...
b≈17.42
The value of b is approximately 17.42. Next, notice that we can find the value of c.
We will again create and solve an equation using the Pythagorean Theorem.
2.7^2+c^2= 13^2
Solve for c
7.29+c^2=169
c^2=161.71
sqrt(c^2)=sqrt(161.71)
c=sqrt(161.71)
c=12.7165...
c≈12.72
The value of c is approximately 12.72.

Notice that using this value we can find the value of d. d= 17.42-12.72=4.7 Finally, we can use the value of d to evaluate the value of x.

Let's use the Pythagorean Theorem for the last time.
4.7^2+2.7^2=x^2
Solve for x
22.09+7.29=x^2
29.38=x^2
x^2=29.38
sqrt(x^2)=sqrt(29.38)
x=sqrt(29.38)
x=5.42033...
x≈5.4
The value of x is approximately 5.4.