# Question Video: Solving Area Problems in a Real-World Context Mathematics • 8th Grade

A water feature can be modeled as a hemisphere with its base set onto a square patio. If the diameter of the hemisphere is 4 feet and the patio has a side of length 6 feet, what would the visible area of the patio be? Give your answer accurate to two decimal places.

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### Video Transcript

A water feature can be modeled as a hemisphere with its base set onto a square patio. If the diameter of the hemisphere is four feet and the patio has a side of length six feet, what would the visible area of the patio be? Give your answer accurate to two decimal places.

Let’s begin with a sketch of this water feature. It is a hemisphere, so that’s half a sphere, with a diameter of four feet. And it’s sitting on a square patio, which has a side length of six feet. The visible area of the patio will be all of the patio’s area which isn’t covered by the circular base of this hemisphere. It’s the area of the square minus the area of the circle, which is in fact the great circle of this hemisphere. We know how to find the areas of each of these two-dimensional shapes. The area of a square is simply a side length squared; that’s six squared. And the area of a circle is 𝜋𝑟 squared.

Now, we need to be a little careful here because we were given the diameter of the hemisphere, which is the diameter of this circle. It’s four feet. So the radius is half of this; it is two feet. We have then six squared minus 𝜋 multiplied by two squared. That simplifies to 36 minus four 𝜋. And we could leave our answer in this form if we wanted. But this question actually asked us to give our answer accurate to two decimal places. Evaluating on a calculator gives 23.43362, and then rounding to the required two decimal places gives 23.43. As the units for the length in this question were given in feet, the units for area will be square feet. And so we have our answer to the problem. The visible area of the patio to two decimal places is 23.43 square feet.