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A cuboid has been removed from a larger cuboid as shown in the diagram.
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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.
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When two shapes are similar, we can say that one is an enlargement of the other.
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And when we enlarge a shape, we enlarge it by a scale factor.
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So how do we find the value of the scale factor?
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Well, to find the scale factor of enlargement, we take a new length and we divide it by the corresponding old length.
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Let’s see what that looks like in this example.
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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.
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Let’s say that the new length then is 10 centimetres and the old corresponding length is 15 centimetres.
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The scale factor of enlargement — remember, enlarging a shape can make it larger or smaller — is therefore 10 divided by 15.
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Both 10 and 15 have a factor of five, so we can divide both the numerator and the denominator of this fraction by five.
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And the scale factor for enlargement simplifies to two-thirds.
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Now this value represents the scale factor of the length of one of its sides.
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Since we’re trying to find the relationship between the volume, we need to find the scale factor for the volume.
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Volume is measured in this case in centimetres cubed.
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That’s a good way to remember that, to find the scale factor for the volume, we cube the scale factor for the length.
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Two cubed is two multiplied by two multiplied by two, which is eight.
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And three cubed is three multiplied by three multiplied by three, which is 27.
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So then the scale factor for the volume of these cuboids is eight over 27.
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Let’s find the volume of the large cuboid then.
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The formula for volume of a cuboid is width multiplied by height multiplied by length.
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The width and the height of this cuboid is 15 centimetres, and its length is 20 centimetres.
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15 multiplied by 15 multiplied by 20 is 4500.
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So the volume of the larger cuboid is 4500 centimetres cubed.
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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.
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That’s 4500 multiplied by eight over 27.
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And that gives us a smaller volume of 4000 over three centimetres cubed.
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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.
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That’s 4500 centimetres cubed minus 4000 over three centimetres cubed.
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And if we type that into our calculator, we get 9500 over three.
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To convert an improper fraction into a mixed number on our calculator, we need to find the button that looks like this.
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And if we press that, we get that 9500 over three is equivalent to 3166 and two-thirds.
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And we have shown, as required, that the volume of the remaining solid is 3166 and two-thirds centimetres cubed.
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Part b) The shaded face of the solid has exactly four lines of symmetry.
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The shaded face has been drawn out again below.
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Work out the size of angle 𝑥 as shown on the diagram.
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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.
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That tells us we’re going to need to use right angle trigonometry.
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Before we can do that though, we need to calculate at least two of the sizes of its sides.
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We know that this shaded face has exactly four lines of symmetry.
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This tells us that the larger and smaller squares share a centre, as shown.
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We can therefore find the values of the two lengths shown by subtracting 10 centimetres from 15 centimetres.
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Halving five will give us the width of the right-angled triangle that we’re interested in.
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Five divided by two is 2.5, so our triangle is 2.5 centimetres wide.
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We also know that this part here is 2.5 centimetres and that the side of the larger square is 15 centimetres in length.
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So we can subtract 2.5 from 15 to tell us the height of our right-angled triangle.
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That’s 12.5 centimetres.
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Now we have a right-angled triangle for which we know the length of two of the sides.
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Let’s begin by labelling the sides of the triangle.
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The hypotenuse is the longest side.
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It’s the side opposite the right angle.
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The side opposite the included angle 𝑥 is known as the opposite.
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And the final side is the adjacent.
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It’s the one left over but is also next to the included angle.
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We know the lengths of the opposite and adjacent sides of this triangle.
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That means we use the tan ratio.
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Tan of 𝜃 is equal to opposite over adjacent.
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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.
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12.5 divided by 2.5 is five, so we get tan of 𝑥 is equal to five.
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To solve this equation, we want to do the opposite of tan.
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That’s the inverse tan.
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So 𝑥 is equal to the inverse tan of five.
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Typing this into our calculator and we get that 𝑥 is equal to 78.690 and so on degrees.
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The question doesn’t specify what we should round our answer to.
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So here I’ve chosen two decimal places.
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𝑥 is equal to 78.69 degrees.