For the square pyramid, the lateral edge length e, is 25 in, and the slant height, s, is 24 in. Find the height of the pyramid, h. Round the answer to the nearest whole number. Hint. Before finding the length of h, you will first need to find the length of r.
To find the height of the square pyramid, you'll first need to determine the length of the semi-diagonal of the base, often labeled as "r" for the radius of the inscribed circle in the square base.
If you have a square pyramid with a lateral edge length "e" of 25 inches and a slant height "s" of 24 inches, you can imagine a right triangle formed by half of the diagonal of the base (r), the height of the pyramid (h), and the slant height (s). This right triangle is formed because the slant height of the pyramid is also the hypotenuse of this right triangle.
The base of the pyramid is a square, so if you draw both diagonals, they will intersect at the center at right angles and thus form four right triangles. The diagonals will be equal in length, and each diagonal can be divided into two equal parts, each part representing r, the semi-diagonal of the square base.
Let "a" represent the side of the base square. The diagonal ("d") of the square can be calculated using the Pythagorean theorem:
\[ d = \sqrt{a^2 + a^2} = \sqrt{2} \times a \]
Since r is half the diagonal (r = d/2), we get:
\[ r = \frac{\sqrt{2}}{2} \times a \]
Now, let's use the slant height s (24 in), and the lateral edge length e (25 in) to find "a". We use another right triangle formed with half of the base as one side, the slant s as the hypotenuse, and the height of the pyramid as another side. The half of the base and the slant height form the right triangle with the lateral edge as the hypotenuse.
Using the Pythagorean theorem, where "a/2" is the half of base side:
\[ \left(\frac{a}{2}\right)^2 + h^2 = e^2 \]
\[ \frac{a^2}{4} + h^2 = 625 \] (because e = 25 in)
Now we know that \( s^2 = r^2 + h^2 \):
\[ 24^2 = \left(\frac{\sqrt{2}}{2} \times a\right)^2 + h^2 \]
\[ 576 = \frac{1}{2} a^2 + h^2 \]
\[ 576 = \frac{1}{2} a^2 + h^2 \]
Let's combine terms, considering the two equations together:
From equation: \(\frac{a^2}{4} + h^2 = 625 \),
We can write \( a^2 \) as \( 2 \times (576 - h^2) \),
\[ a^2 = 2(576 - h^2) \]
\[ a^2 = 1152 - 2h^2 \]
Now substitute this value of \( a^2 \) into the equation \(\frac{a^2}{4} + h^2 = 625 \),
\[ \frac{1152 - 2h^2}{4} + h^2 = 625 \]
Solve for \( h^2 \):
\[ 288 - \frac{1}{2} h^2 + h^2 = 625 \]
\[ 288 + \frac{1}{2} h^2 = 625 \]
\[ \frac{1}{2} h^2 = 625 - 288 \]
\[ \frac{1}{2} h^2 = 337 \]
\[ h^2 = 674 \]
\[ h = \sqrt{674} \]
Therefore,
\[ h \approx \sqrt{674} \approx 25.96 \text{ inches} \]
Rounding this to the nearest whole number, the height of the pyramid, \( h \), is approximately 26 inches.