# Video: Pack 5 • Paper 3 • Question 19

Pack 5 • Paper 3 • Question 19

04:01

### Video Transcript

The volume, 𝑉, of a pyramid is related to the area of its base, 𝐴, and its height, ℎ, by the formula 𝐴 is equal to three multiplied by 𝑉 all over ℎ. Lucy measured the volume of a pyramid to be 115 centimetres cubed correct to three significant figures. She measured its height to be 7.2 centimetres correct to two significant figures. Calculate the upper and lower bounds on the area of the base of the pyramid.

Let’s start by calculating the upper and lower bounds of both the given volume and the height of the pyramid. To find the bounds for the volume, let’s consider what it might have been rounded to had it gone to the number below. It’s rounded to three significant figures. So the next one down from this would have been 114.

Similarly, had the number rounded to the next one up, that would have been 116. We can find the lower bound by working out the midpoint of 114 and 115, which is 114.5. The smallest possible number it could have been is 114.5.

Similarly, the upper bound is halfway between 115 and 116, which is 115.5. Remember, we call this the upper bound as whilst it does get extremely close to 115.5, it can never actually be 115.5, because this would round up to 116. We can represent this situation by using inequalities. 𝑉 is greater than or equal to 114.5. However, it is smaller than 115.5.

Now let’s repeat this process for the height ℎ. 7.2 has been rounded to two significant figures. Had this number actually rounded to the one below, that would have been 7.1. And the number above would have been 7.3. The lower bound is the midpoint of 7.1 and 7.2, which is 7.15. The upper bound is the midpoint of 7.2 and 7.3, which is 7.25. Once again, we can represent these bounds using inequalities. ℎ must be greater than or equal to 7.15 but less than 7.25.

Now that we’ve got both the upper and lower bound for the volume and the height, we can substitute these into the formula for the area of the base of the pyramid. First, let’s consider the lower bound of the area. When dividing, if we are trying to create the smallest number possible, we will need to divide the smallest number by the largest number.

Here that means we will need to divide the lower bound for the volume by the upper bound for the height. The lower bound of 𝐴 is given by three multiplied by 114.5 over 7.25. Correct to three decimal places, that’s 47.379.

For the upper bound of 𝐴, we want to create the largest number possible. To do this, we’ll divide the biggest possible number by the smallest possible number. That’s the upper bound of the volume divided by the lower bound for the height: three multiplied by 115.5 all over 7.15. That’s 48.462 correct to three decimal places.

The lower bound for the area of the base of the pyramid is 47.379 centimetres squared and the upper bound is 48.462 centimetres squared, all rounded to three decimal places.