Glencoe Math: Course 3, Volume 2
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Glencoe Math: Course 3, Volume 2 View details
6. Use the Pythagorean Theorem
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Exercise 16 Page 429

about 13.9mm

Practice makes perfect

We are given the following rectangular prism.

The segment with the length of x millimeters is the diagonal of the rectangular prism. We are asked to find the value of x. Let's start by taking a closer look at the diagram!

Notice that the interior diagonal makes a right triangle with the height and the diagonal of the base as its legs. If we know these values, we will be able to use the Pythagorean Theorem to find the length of the diagonal. We know that the height of the prism is 5 millimeters, but we do not know the length of the diagonal of the base. Let's consider the base of our prism.

We have another right triangle with an unknown hypotenuse. We can use the Pythagorean Theorem to find it. a^2+ b^2= c^2 In the formula, a and b are the lengths of the legs and c is the length of the hypotenuse of a right triangle. We can see that in the triangle, the lengths of the legs are 5 millimeters and 12 millimeters.

Let's apply the Pythagorean Theorem to this triangle. a^2+ b^2= c^2 ⇕ 5^2+ 12^2= c^2 Now we can solve the equation that we got to find the value of c.

5^2+12^2=c^2
25+144=c^2
169=c^2
sqrt(169)=sqrt(c^2)
13=sqrt(c^2)
13=c
c=13

Since a negative side length does not make sense, we will only consider the positive solution. Therefore, we have that the length of diagonal of the base is 13 millimeters. Returning to our prism, let's consider the larger right triangle made of the height of our prism, its interior diagonal, and the diagonal of the base.

Let's we use the Pythagorean Theorem again to find the length of the diagonal of the prism. We are given a triangle with a= 5, b= 13, and c= x.

We can apply the Pythagorean Theorem to this triangle. a^2+ b^2= c^2 ⇕ 5^2+ 13^2= x^2 Next, we will solve the equation that we got to find the value of x.

5^2+13^2=x^2
25+169=x^2
194=x^2
sqrt(194)=sqrt(x^2)
13.928388... =sqrt(x^2)
13.928388... =x
x=13.928388...
x≈ 13.9

We found that the missing length is about 13.9 millimeters.