### Video Transcript

Find the volume of the following regular pyramid approximating the result to the nearest hundredth.

We’re given a diagram, and note that we’re told that this is a regular pyramid. We remember that a regular pyramid is a right pyramid. That means it’s a pyramid whose apex lies above the centroid of the base. So a regular pyramid is a right pyramid whose base is a regular polygon. The base of this pyramid is a quadrilateral. And if it’s a regular quadrilateral, that means it will be a square. In a regular pyramid, the lateral faces are congruent isosceles triangles. We’re asked in this question to find the volume of the pyramid. We recall that the volume of a pyramid is equal to one-third times the area of the base times ℎ, where ℎ is the height.

If we look at the pyramid, we have two measurements, one of 15 centimeters and one of 17 centimeters. But neither of these is the height of the pyramid. And in fact, we don’t have the area of the base either. That means that we’ll need to do a few calculations before we can attempt to use this volume formula. Let’s consider this triangle, which would form part of the cross section of the pyramid. The height of this triangle will be the same as the height of the pyramid. So let’s define this as ℎ centimeters. We can use the triangle to help us work out the value of ℎ and therefore the height of the pyramid. It may be helpful to draw the two-dimensional triangle alongside the pyramid.

We know that the height is ℎ centimeters and the longest side in this right triangle is 15 centimeters. We don’t however know the value at the base of the triangle. Let’s define this as 𝑥 centimeters. At the minute, however, we have two unknown side lengths, ℎ and 𝑥. But let’s see if we can work out the value of 𝑥 centimeters by considering another triangle. This triangle is formed of half of one of the lateral faces. The height of this triangle will be the same as the slant height in the pyramid; that’s 15 centimeters. The hypotenuse of the triangle will be the same as the length of the lateral edge; that’s 17 centimeters. The base of this triangle will be the same length as the base in the blue triangle. They’ll both be 𝑥 centimeters.

This means that we have a way in which we can calculate 𝑥. Given that we have a right triangle, we can apply the Pythagorean theorem. This theorem states that in every right triangle, the square of the hypotenuse is equal to the sum of the squares on the other two sides. And so we have 17 squared is equal to 15 squared plus 𝑥 squared. Evaluating the squares gives 289 is equal to 225 plus 𝑥 squared. Subtracting 225 from both sides gives us 64 is equal to 𝑥 squared. We then take the square root of both sides. And because 𝑥 is a length, we only need to consider the positive value of the square root. And so 𝑥 is equal to eight.

Now, we can return to the first triangle using the fact that 𝑥 is eight to work out the value of ℎ. Once again, we use the Pythagorean theorem. And be careful when we’re filling in the values; this time the hypotenuse is 15 and not ℎ. Evaluating the squares, we have 225 is equal to 64 plus ℎ squared. Subtracting 64 from both sides, we have 161 is equal to ℎ squared. We then take the square root of both sides of this equation. So we have the square root of 161 is equal to ℎ. We know that 161 is not a square number. And since we haven’t finished with our calculation, let’s keep the value of ℎ as the square root of 161.

We’ll clear some space for the next part of the calculation. So let’s fill in the values that we’ve worked out onto the diagram. We have the height as the square root of 161 centimeters. And half of the side length is eight centimeters. Let’s return to the formula for the volume of a pyramid and see if we now have enough information. Well, we have worked out the height of the pyramid, but we still don’t know the area of the base. As we know that this is a regular pyramid, that means that we have a regular polygon on the base, which is a square.

The area of a square which has a side length of 𝑙 is given as 𝑙 squared. But the length of the square here is not eight centimeters. In fact, we know that this length of eight is half of the side length. That means that the side length will be 16 centimeters. And so, the area of this square will be 16 squared, which is 256 square centimeters. So now, we finally have enough information to work out the volume of the pyramid.

Remember that the area of the base will be the area of the square, which is 256. So the volume of the pyramid is equal to one-third times 256 times the square root of 161. Using a calculator, we can evaluate this as 1,082.7586 and so on cubic centimeters. The final thing to do is to approximate it to the nearest hundredth. Therefore, the final answer is that the volume of this pyramid to the nearest hundredth is 1,082.76 cubic centimeters.