### 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.