The Pythagorean Theorem in Three Dimensions Practice

Question

Math 8 Q2 (Pre-Algebra) / Triangles

The pyramid is an arena in Memphis, Tennessee. its height is 321 ft., and its slant height is 436.5 ft. find the length of the edge of the square base. round the answer to the nearest tenth

Answer 1

To find the length of the edge of the square base, we can use the Pythagorean theorem in three dimensions. Here, we are considering a right-angled triangle formed by the height of the pyramid (321 ft), the slant height (436.5 ft), and half the diagonal of the square base.

Let's denote the following:
- \( h \) = height of the pyramid = 321 ft
- \( l \) = slant height of the pyramid = 436.5 ft
- \( d \) = diagonal length of the square base
- \( s \) = side length of the square base

We will use a two-step process to find the side length (\( s \)) of the square base. First, we find the diagonal (\( d \)) by applying the Pythagorean theorem, and then we calculate the side length (\( s \)) by relating the diagonal to the sides of the square.

Step 1: Finding the diagonal (\( d \))
We use the Pythagorean theorem which says that in a right-angled triangle, the square of the length of the hypotenuse (slant height in this case) is equal to the sum of the squares of the other two sides (height and half the diagonal of the base).

The equation is:

\[ l^2 = h^2 + \left(\frac{d}{2}\right)^2 \]

We are given \( l = 436.5 \) ft and \( h = 321 \) ft, so we plug those into our equation:

\[ (436.5)^2 = (321)^2 + \left(\frac{d}{2}\right)^2 \]

\[ 190500.25 = 103041 + \frac{d^2}{4} \]

Now, solve for \( d^2 \):

\[ \frac{d^2}{4} = 190500.25 - 103041 \]

\[ \frac{d^2}{4} = 87459.25 \]

\[ d^2 = 87459.25 \times 4 \]

\[ d^2 = 349837 \]

Now take the square root to find \( d \):

\[ d = \sqrt{349837} \]

\[ d \approx 591.45 \] ft

Step 2: Finding the side length (\( s \))
We know that a square's diagonal and side length are related by the equation \( d = s\sqrt{2} \) because the diagonal forms an isosceles right triangle with two sides of the square.

So we can write:

\[ s = \frac{d}{\sqrt{2}} \]

Using the value we found for \( d \):

\[ s = \frac{591.45}{\sqrt{2}} \]

\[ s = \frac{591.45}{1.4142...} \]

\[ s \approx 418.2 \] ft

So the length of the edge of the square base, rounded to the nearest tenth, is approximately 418.2 feet.