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 19 Page 430

168 inches

Practice makes perfect

We are given a stained glass window in the shape of the following figure.

We are asked to find the perimeter of the window. Let's start by marking all the missing lengths in the diagram.

Notice that we need to find the value of x to calculate the perimeter of the window. Let's find the value of y first. Take a closer look at the diagram.

The segment with the length of y is also a leg of a right triangle. This means that we can use the Pythagorean Theorem to find the missing length. 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 are given a triangle with a= 27, b= y, and c= 45.

Let's substitute these values into the formula. a^2+ b^2= c^2 ⇕ 27^2+ y^2= 45^2 Now we can solve the equation that we got to find the value of y.

a^2+b^2=c^2
27^2+ y^2= 45^2
â–¼
Solve for y
729+y^2=2025
729+y^2-729=2025-729
y^2=1296
sqrt(y^2)=sqrt(1296)
y=sqrt(1296)
y=36

Since a negative side length does not make sense, we only need to consider positive solutions. Therefore, we got that y=36. Now we can find the value of x.

The segment with the length of x is also the hypotenuse of a right triangle. To find the value of x, we can use the Pythagorean Theorem again. In this case, we are given a triangle with a= 15, b= 36, and c= x.

Let's substitute these values into the formula. a^2+ b^2= c^2 ⇕ 15^2+ 36^2= x^2 Now we can solve the equation that we got to find the value of x.

a^2+b^2=c^2
15^2+ 36^2= x^2
â–¼
Solve for x
225+1296=x^2
1521=x^2
sqrt(1521)=sqrt(x^2)
39 =sqrt(x^2)
39 =x
x= 39

We found that x=39. Finally, we have all the side lengths of the given figure!

Let's add up all the side lengths of the window to find its perimeter. 45+ 45+ 39+ 39=168 We got that the perimeter of the window is 168 inches.