# Video: AQA GCSE Mathematics Higher Tier Pack 1 • Paper 2 • Question 14

AQA GCSE Mathematics Higher Tier Pack 1 • Paper 2 • Question 14

05:55

### Video Transcript

A cuboid has been removed from a larger cuboid as shown in the diagram.

Part a) Given that the cuboid removed is similar to the larger cuboid, show that the volume of the remaining solid is 3166 and two-thirds centimetres cubed.

When two shapes are similar, we can say that one is an enlargement of the other. And when we enlarge a shape, we enlarge it by a scale factor. So how do we find the value of the scale factor?

Well, to find the scale factor of enlargement, we take a new length and we divide it by the corresponding old length. Let’s see what that looks like in this example. We can see that the smaller cuboid has a side length of 10 centimetres, and the larger cuboid has a corresponding side length of 15 centimetres. Let’s say that the new length then is 10 centimetres and the old corresponding length is 15 centimetres.

The scale factor of enlargement — remember, enlarging a shape can make it larger or smaller — is therefore 10 divided by 15. Both 10 and 15 have a factor of five, so we can divide both the numerator and the denominator of this fraction by five. And the scale factor for enlargement simplifies to two-thirds.

Now this value represents the scale factor of the length of one of its sides. Since we’re trying to find the relationship between the volume, we need to find the scale factor for the volume. Volume is measured in this case in centimetres cubed. That’s a good way to remember that, to find the scale factor for the volume, we cube the scale factor for the length. Two cubed is two multiplied by two multiplied by two, which is eight. And three cubed is three multiplied by three multiplied by three, which is 27. So then the scale factor for the volume of these cuboids is eight over 27.

Let’s find the volume of the large cuboid then. The formula for volume of a cuboid is width multiplied by height multiplied by length. The width and the height of this cuboid is 15 centimetres, and its length is 20 centimetres. 15 multiplied by 15 multiplied by 20 is 4500. So the volume of the larger cuboid is 4500 centimetres cubed.

To find the volume of the small cuboid, the cuboid that’s been taken out of the large cuboid, we’re going to multiply this by the scale factor. That’s 4500 multiplied by eight over 27. And that gives us a smaller volume of 4000 over three centimetres cubed.

To find the volume of the remaining solid, we’re going to subtract the volume of the smaller solid from the volume of the larger solid. That’s 4500 centimetres cubed minus 4000 over three centimetres cubed. And if we type that into our calculator, we get 9500 over three.

To convert an improper fraction into a mixed number on our calculator, we need to find the button that looks like this. And if we press that, we get that 9500 over three is equivalent to 3166 and two-thirds. And we have shown, as required, that the volume of the remaining solid is 3166 and two-thirds centimetres cubed.

Part b) The shaded face of the solid has exactly four lines of symmetry. The shaded face has been drawn out again below. Work out the size of angle 𝑥 as shown on the diagram.

Notice how what we actually have here is a right-angled triangle for which we’re trying to find the size of one of the angles. That tells us we’re going to need to use right angle trigonometry.

Before we can do that though, we need to calculate at least two of the sizes of its sides. We know that this shaded face has exactly four lines of symmetry. This tells us that the larger and smaller squares share a centre, as shown.

We can therefore find the values of the two lengths shown by subtracting 10 centimetres from 15 centimetres. Halving five will give us the width of the right-angled triangle that we’re interested in. Five divided by two is 2.5, so our triangle is 2.5 centimetres wide.

We also know that this part here is 2.5 centimetres and that the side of the larger square is 15 centimetres in length. So we can subtract 2.5 from 15 to tell us the height of our right-angled triangle. That’s 12.5 centimetres.

Now we have a right-angled triangle for which we know the length of two of the sides. Let’s begin by labelling the sides of the triangle. The hypotenuse is the longest side. It’s the side opposite the right angle. The side opposite the included angle 𝑥 is known as the opposite. And the final side is the adjacent. It’s the one left over but is also next to the included angle.

We know the lengths of the opposite and adjacent sides of this triangle. That means we use the tan ratio. Tan of 𝜃 is equal to opposite over adjacent. Substituting what we know about our right-angled triangle into this formula and we get tan 𝑥 is equal to 12.5 divided by 2.5. 12.5 divided by 2.5 is five, so we get tan of 𝑥 is equal to five.

To solve this equation, we want to do the opposite of tan. That’s the inverse tan. So 𝑥 is equal to the inverse tan of five. Typing this into our calculator and we get that 𝑥 is equal to 78.690 and so on degrees. The question doesn’t specify what we should round our answer to. So here I’ve chosen two decimal places. 𝑥 is equal to 78.69 degrees.