The resulting solid will be a right circular cone. The total surface area of a right circular cone is the
sum of the area of the circular base plus the lateral area. If we call the radius of the circular base r,
then the area of the circular base is πr^2 . If we call the circumference of the circular base c and the
slant height ℓ, then the lateral area is .
When right triangle PQR is rotated about PR, the resulting right circular cone will have a circular base
with a radius that is leg PQ of right triangle PQR and a slant height that is hypotenuse QR of this
triangle. So in order to find the total surface area of the right circular cone, we need to find the
length of leg PQ and the length of hypotenuse QR of right triangle PQR.
Let’s first find the length of PQ. The area of any triangle is . The area of a right
triangle is , because one leg can be considered to be the base and the other leg can
be considered to be the height.
The area of right triangle PQR is . We know that PR = 4 and that the area of right
triangle PQR is 6. So . Let’s solve this equation for PQ. We have 2 × (PQ) = 6, so PQ
= 3. The radius of the base of the right circular cone is 3. The area of the circular base is π(3^2 ) = 9π.
Now let’s find the hypotenuse QR of right triangle PQR. This triangle has legs of length 3 and 4, so
this is a 3-4-5 right triangle. So the length of QR is 5. Thus, the slant height ℓ of the right circular cone
is 5. The circumference c of the circular base is 2πr = 2π(3) = 6π. Thus, c = 6π and ℓ = 5. The lateral
area is
The area of the circular base is 9π and the lateral area is 15π. The total surface area is 9π + 15π = 24π.
Now 24π ≈ 75.3982236862. To the nearest hundredth, 24π ≈ 75.40. Choice (C) is correct.
A typical SAT Subject Test: Mathematics Level 2 will include one or two questions that entail
visualizing. Here’s another one in Example 5.
Example 5