# Video: Finding the Curved Surface Area of a Hemisphere given the Area of Its Great Circl

Find, to the nearest tenth, the curved surface area of a hemisphere, given that the area of the great circle is 441𝜋 mm².

02:41

### Video Transcript

Find, to the nearest tenth, the curved surface area of a hemisphere, given that the area of the great circle is 441𝜋 square millimeters.

Let’s begin by sketching this hemisphere out. Remember that the great circle of a sphere is the circle which divides the sphere up into two hemispheres. So it is the flat circular base of our hemisphere. It’s this face here. We’re told that the area of this great circle is 441𝜋 square millimeters. And we also know that the general formula for calculating the area of a circle is 𝜋𝑟 squared. So we can use these two pieces of information to form an equation. 𝜋𝑟 squared equals 441𝜋. We can solve this equation to find the radius of our hemisphere. First, we can divide through by 𝜋 to give 𝑟 squared equals 441. We then take the square root of each side of the equation. And as 441 is a square number, its square root is 21. So we have that the radius of this hemisphere is 21.

We then recall that the curved surface area of a hemisphere is half the surface area of a full sphere. It’s two 𝜋𝑟 squared. So we can substitute our value for the radius into this formula. That gives two 𝜋 multiplied by 21 squared. We know from our earlier working that 𝑟 squared or 21 squared is 441. So we have two 𝜋 multiplied by 441 which is 882𝜋. The question asks us to give this value to the nearest tenth. So we can evaluate 882𝜋 on a calculator, and it gives 2770.884. The deciding digit in this case is the eight in the hundredths column. So we’re rounding up, which gives 2770.9, and the units for this will be square millimeters.

Now you may’ve spotted that, actually, we didn’t need to calculate the radius of this hemisphere at all. If the area of the great circle is 𝜋𝑟 squared and the curved surface area of the hemisphere is two 𝜋𝑟 squared, then we could have simply doubled the area we were given for the great circle. That would give two multiplied by 441𝜋, which is 882𝜋. And so we’d have arrived at the same stage of calculation as we had here in our previous method. Both of these methods would be equally fine and give the same answer of 2770.9 square millimeters.