Question
The area of a square-based pyramid is 50 square yards. The volume of the pyramid is 250 cubic yards. What is the height of the pyramid?(1 point) The height of the pyramid is yards.
Answers
GPT-4o mini
To find the height of the square-based pyramid, we can use the formulas for the volume and the surface area of a pyramid.
1. **Volume of a square-based pyramid:**
\[
V = \frac{1}{3} \times B \times h
\]
where \( V \) is the volume, \( B \) is the area of the base, and \( h \) is the height.
2. **Surface area of a square-based pyramid:**
\[
A = B + \frac{1}{2} P l
\]
where \( A \) is the surface area, \( P \) is the perimeter of the base, and \( l \) is the slant height.
Given:
- Area \( A = 50 \) square yards
- Volume \( V = 250 \) cubic yards
Assuming the base is a square, let's denote the side length of the base as \( s \). Thus, the base area \( B \) can be expressed as:
\[
B = s^2
\]
Using the volume formula:
\[
250 = \frac{1}{3} \times s^2 \times h
\]
From this, we can solve for \( h \):
\[
h = \frac{750}{s^2} \quad \text{(1)}
\]
Now, we also know the surface area:
\[
A = s^2 + \frac{1}{2} P l
\]
The perimeter \( P \) of the base is:
\[
P = 4s
\]
The area of the base is already given by \( s^2 \). So:
\[
50 = s^2 + \frac{1}{2} \times 4s \times l
\]
Simplifying the surface area equation:
\[
50 = s^2 + 2s \times l \quad \text{(2)}
\]
Now we need to express \( l \) in terms of \( h \) and \( s \). By the Pythagorean theorem, the relationship between the height \( h \), half of the side length \( \frac{s}{2} \), and the slant height \( l \) is:
\[
l = \sqrt{h^2 + \left(\frac{s}{2}\right)^2}
\]
Substituting this back, we have:
\[
2s \times l = 2s \times \sqrt{h^2 + \left(\frac{s}{2}\right)^2}
\]
Now, we can substitute equation (1) for \( h \) into equation (2). However, solving the entire system could be complex. Instead, let's eliminate \( l \) and express \( s \) in terms of known values.
Consider the volume formula again. If \( s^2 = 50 - 2sl \), substituting gives a single variable equation, which may be computationally intensive.
Let's reconcile:
From \( 250 = \frac{1}{3} s^2 h \) and substituting:
1. Obtain \( h \) directly from \( B \):
If we know \( B = 50 \) (not \( s^2 \)), use \( 250 = \frac{1}{3} \cdot 50 \cdot h \),
then \( 250 = \frac{50}{3}h \),
\[
h = \frac{250 \cdot 3}{50} = 15.
\]
Thus, the calculated height is:
\[
\boxed{15} \text{ yards}.
\]
1. **Volume of a square-based pyramid:**
\[
V = \frac{1}{3} \times B \times h
\]
where \( V \) is the volume, \( B \) is the area of the base, and \( h \) is the height.
2. **Surface area of a square-based pyramid:**
\[
A = B + \frac{1}{2} P l
\]
where \( A \) is the surface area, \( P \) is the perimeter of the base, and \( l \) is the slant height.
Given:
- Area \( A = 50 \) square yards
- Volume \( V = 250 \) cubic yards
Assuming the base is a square, let's denote the side length of the base as \( s \). Thus, the base area \( B \) can be expressed as:
\[
B = s^2
\]
Using the volume formula:
\[
250 = \frac{1}{3} \times s^2 \times h
\]
From this, we can solve for \( h \):
\[
h = \frac{750}{s^2} \quad \text{(1)}
\]
Now, we also know the surface area:
\[
A = s^2 + \frac{1}{2} P l
\]
The perimeter \( P \) of the base is:
\[
P = 4s
\]
The area of the base is already given by \( s^2 \). So:
\[
50 = s^2 + \frac{1}{2} \times 4s \times l
\]
Simplifying the surface area equation:
\[
50 = s^2 + 2s \times l \quad \text{(2)}
\]
Now we need to express \( l \) in terms of \( h \) and \( s \). By the Pythagorean theorem, the relationship between the height \( h \), half of the side length \( \frac{s}{2} \), and the slant height \( l \) is:
\[
l = \sqrt{h^2 + \left(\frac{s}{2}\right)^2}
\]
Substituting this back, we have:
\[
2s \times l = 2s \times \sqrt{h^2 + \left(\frac{s}{2}\right)^2}
\]
Now, we can substitute equation (1) for \( h \) into equation (2). However, solving the entire system could be complex. Instead, let's eliminate \( l \) and express \( s \) in terms of known values.
Consider the volume formula again. If \( s^2 = 50 - 2sl \), substituting gives a single variable equation, which may be computationally intensive.
Let's reconcile:
From \( 250 = \frac{1}{3} s^2 h \) and substituting:
1. Obtain \( h \) directly from \( B \):
If we know \( B = 50 \) (not \( s^2 \)), use \( 250 = \frac{1}{3} \cdot 50 \cdot h \),
then \( 250 = \frac{50}{3}h \),
\[
h = \frac{250 \cdot 3}{50} = 15.
\]
Thus, the calculated height is:
\[
\boxed{15} \text{ yards}.
\]