Video: Calculating the Volume and the Areas of Faces of a Cuboidal Object

The brick shown in the diagram has a length of 15 cm, a height of 6.5 cm, and a width of 5.5 cm. What is the volume of the brick? Answer to the nearest cubic centimeter. What is the area of the largest face of the brick? Answer to the nearest square centimeter. What is the area of the smallest face of the brick? Answer to the nearest square centimeter.

09:20

Video Transcript

The brick shown in the diagram has a length of 15 centimeters, a height of 6.5 centimeters, and a width of 5.5 centimeters. What is the volume of the brick? Answer to the nearest cubic centimeter. What is the area of the largest face of the brick? Answer to the nearest square centimeter. What is the area of the smallest face of the brick? Answer to the nearest square centimeter.

Okay, so in this question, we have been given a brick. Now, this brick seems to be shaped like a cuboid. And we’ve been given information about the length of the brick, the width of the brick, and the height of the brick. Specifically from the very first sentence in the question, we’ve been told that the length of the brick is 15 centimeters, the height of the brick is 6.5 centimeters, and the width is 5.5 centimeters. And we’ve been asked to use this to calculate the volume of the brick, the area of the largest face of the brick, and the area of the smallest face of the brick.

So, let’s first start by recalling that the volume of any object is the amount of space it occupies. And for a cuboid, we can recall that the volume, which we’ll call 𝑉 subscript cuboid — we’ll draw a little nice 3D cuboid here — is equal to the length multiplied by the height multiplied by the width. And luckily for us, we’ve been given the length of the cuboid, the height of the cuboid, and the width of the cuboid because our cuboid, in this case, is the brick, and we’ve been given all three quantities. Hence, we can say that the volume of our brick in particular, we’ll call it 𝑉 subscript 𝑏, is equal to the length of the brick, that’s 15 centimeters, multiplied by the height of the brick, that’s 6.5 centimeters, multiplied by the width of the brick, which happens to be 5.5 centimeters.

And then, to make a bit more progress, we can start by multiplying all the numbers together, so 15 times 6.5 times 5.5. And we can also multiply all the units together, so centimeters multiplied by centimeters multiplied by centimeters, which will give us a unit of cubic centimeters or centimeters cubed. And this is a good thing because we’ve also been asked to give our answer to the nearest cubic centimeter, which confirms for us, firstly, that cubic centimeters is indeed a unit measuring volume. And secondly, that we’ll give our answer in cubic centimeters to the nearest cubic centimeter. So, evaluating the right-hand side of this equation, then, we find that the volume of our brick is 536.25 cubic centimeters.

However, as we’ve already mentioned, we need to answer to the nearest cubic centimeter. And so, we’re going to have to round our value at this position here. We’ll have to keep these first three significant figures. But in order to work out what happens to this last significant figure, we’re going to have to look at the next one. This value is a two and two is less than or equal to five. Therefore, our final significant figure that we’re going to keep is going to stay exactly the same. It’s not going to round up. And hence, we find that the volume of our brick is 536 cubic centimeters to the nearest cubic centimeter, which we can say is the final answer to the first part of our question

moving on then, the second part of our question asks us to find the area of the largest face of the brick. Now, this time we’re not calculating the volume of the entire brick. Instead, we’re gonna have to start by considering the fact that the brick has six different faces. This here is one of the faces, and there’s another face with the exact same dimensions on the other side of the brick, on the backside. If we were to look at the brick from behind, we would see the other face.

The second face of the brick is this one here. This one looks a lot smaller just in the diagram. And there’s an identically shaped face on the brick, which we would see if we looked at it from this side. And finally, out of the three visible faces, there’s also this one to consider. And there’s another shape with the exact same dimensions, which we would see if we looked at the brick from below.

So essentially, we can get away with considering just the three faces that we’ve shaded in here because the three faces that we cannot see have the exact same dimensions as the three faces that we can see. And so, out of these three, we need to find the largest face. Well, just from the diagram, it looks like the orange-shaded face is the largest face. However, it is possible that we haven’t drawn the diagram to scale. So, let’s start by realizing that the three faces that we’ve shaded in are all rectangles.

For example, the orange-shaded face is a rectangle because it’s got four sides. So, here’s side number one, side number two, side number three, and side number four. And those four sides form angles with each other. Now, each angle that is formed in this rectangle is a right angle. In other words, we see four 90-degree angles in this shape. Lastly, not all four sides have the same length, and so that means that this must be a rectangle.

The same is also true for the other shaded faces because, for example, for the pink-shaded face, we firstly got four sides. And the reason we don’t see 90-degree angles drawn in, specifically in the diagram that we’ve drawn, is because this is a three-dimensional drawing. So, it’s difficult to draw three-dimensional shapes on a two-dimensional surface, that is our screen. However, rest assured that the angles that have been drawn in are in reality all 90 degrees. And so basically, we’re dealing with three different rectangles here, the orange one, the pink one, and the green one.

Now, we can recall that the area of a rectangle, we’ll call this 𝐴 subscript rectangle, is given by multiplying one of the lengths of the rectangle by one of the other lengths that’s perpendicular to that length. In other words, if we were dealing with a rectangle like this, and we’d been told that one of the lengths was 𝑥 and a length perpendicular to that was 𝑦, then the area of our rectangle would be 𝑥 multiplied by 𝑦. So, we could use this equation to find the three areas of the three rectangles that we’ve shaded in and then pick the one with the largest area to find the area of the largest face of the brick.

However, we can be a bit more clever and realize that the length, height, and width of the brick that we’ve been given are going to form the required lengths of the three rectangles for which we’re trying to find the area. For example, for the orange-shaded face, we know that what we’ve been told is the length is one of the lengths of the rectangle. And we need to multiply it by this distance here, which, coincidentally, happens to be the height of the rectangle.

And so, the length multiplied by the height will give us the area of the orange-shaded face. Similarly, the width multiplied by the height will give us the area of the pink-shaded face. And lastly, the length multiplied by the width will give us the area of the green-shaded face. And so, to find the area of the largest face, we need to choose the two largest lengths we’ve been given at a 15 centimeters, 6.5 centimeters, and 5.5 centimeters and multiply them together to give us the area of the largest face.

In this particular case, it so happens that 15 centimeters and 6.5 centimeters are indeed the largest lengths given to us, and they will calculate us the area of the orange-shaded face. So, we were right in earlier, assuming that the orange-shaded face is the largest face of the brick. And the area of the largest face, we’ll call 𝐴 subscript 𝑙, is equal to the length, which is 15 centimeters, multiplied by the height, which is 6.5 centimeters. When we evaluate the right-hand side of this equation, we find that the area of the largest face is 97.5 square centimeters.

However, remember that we’ve been asked to give our answer to the nearest square centimeter, which means that we need to round our value at this position. Now, we’re gonna keep these two values. But in order to figure out what happens to this one here, we need to look at the next number. This is a five and five is indeed greater than or equal to five, which means that the last value that we’re going to keep, this seven, is going to round up to an eight. And when the dust settles, we find that the area of the largest face is 98 square centimeters, which means we found the answer to the second part of our question. The area of the largest face of the brick is 98 square centimeters to the nearest square centimeter. And at this point, we can move on to the final question.

This time we’ve been asked to find the area of the smallest face of the brick. And to do this, we repeat a similar procedure. Except this time, we’re going to multiply the two smallest values out of the three given to us. So, that’s going to be the height and the width because out of 15 centimeters, 6.5 centimeters, and 5.5 centimeters, the two smallest values available to us are 6.5 centimeters and 5.5 centimeters. And multiplying those two together is going to give us the area of the pink-shaded face. Which, from the diagram, we can also see to be the smallest area available to us. And this shows us that we’ve at least roughly drawn the diagram correctly.

But anyway, so we can say that the area of the smallest face, 𝐴 subscript 𝑠, is equal to the two smallest lengths multiplied together, the height, which is 6.5 centimeters, multiplied by the width, which is 5.5 centimeters. And note here that we’ve used 6.5 centimeters in both the calculation for the smallest face and for the largest face as well. However, this is not a problem because we couldn’t have multiplied two values together out of the three that we’ve been given, the length, the height, and the width, in any other way than we did here to give us the largest possible area. And the same is true for the smallest possible area.

So, evaluating the right-hand side of this equation, we find that the smallest possible area is 35.75 square centimeters. But once again, we’ve been asked to give it to the nearest square centimeter. And so, we’re going to have to round at this position here. So, to work out what happens to the last digit that we’re going to keep, once again we look at the next value. This is a seven, and seven is greater than or equal to five, which means that the last digit we’re going to keep is going to round up to a six.

And overall, that gives us 36 square centimeters, which means that at this point we’ve answered all three parts of our question. To recap very quickly, the volume of the brick to the nearest cubic centimeter is 536 centimeters cubed. The area of the largest face of the brick to the nearest square centimeter is 98 square centimeters. And the area of the smallest face of the brick to the nearest square centimeter is 36 square centimeters.

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