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Either paper 3 marks
Shape \(A\) and shape \(B\) are mathematically similar triangles.
The base of shape \(A\) is \(3\) cm and the base of shape \(B\) is \(6\) cm.
The area of shape \(A\) is \(21\) cm².
(a)Write down the scale factor of the enlargement from shape \(A\) to shape \(B\).
(b)Calculate the area of shape \(B\).
Show answer
(a) \(2\)
(b) \(84\ \text{cm}^2\)Divide a length on the new shape by the corresponding length on the old one.
\(6 \div 3 = 2\)
Areas do not scale by the length scale factor. Think of the shape as covered in little squares: every square has its side multiplied by \(2\), so its area is multiplied by \(2 \times 2 = 4\).
The area scale factor is the length scale factor squared.
\(21 \times 4 = 84\)
The area of shape \(B\) is \(84\) cm².
- (b) M1area scale factor \(2^2 = 4\)
- (b) A1area of shape \(B\) \(84\) cm²
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Either paper 1 mark\(A\) and \(B\) are similar solids.
Leila says, "The volume of \(B\) is \(64\) times the volume of \(A\) because the length of \(B\) is \(4\) times the length of \(A\)." Is she correct? Tick a box. ☐ Yes ☐ No Give a reason for your answer.Solid Length (cm) \(A\) \(9\) \(B\) \(36\) Show answer
Yes. The lengths are in the ratio \(1 : 4\), so the volume scale factor is \(4^3 = 64\), and the volume of \(B\) is \(64\) times the volume of \(A\), which is what is claimed.The lengths of \(A\) and \(B\) are \(9\) cm and \(36\) cm, so the length scale factor is \(36 \div 9 = 4\).
Volume is three-dimensional, so the volume scale factor is the length scale factor cubed:
\(4^3 = 64\)
The volume of \(B\) is \(64\) times the volume of \(A\), so Leila is correct.
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Either paper 3 marks
Here is solid \(A\), a cone of height \(7\) cm.
Solid \(B\) is a cone that is mathematically similar to solid \(A\). The height of solid \(B\) is \(14\) cm.
The volume of solid \(A\) is \(28\) cm³.
Work out the volume of solid \(B\).
Show answer
\(224\ \text{cm}^3\)The heights are corresponding lengths, so they give the length scale factor.
\(14 \div 7 = 2\)
A volume has three dimensions, and every one of them is multiplied by \(2\), so the volume scale factor is \(2 \times 2 \times 2 = 8\).
\(28 \times 8 = 224\)
The volume of solid \(B\) is \(224\) cm³.
- M1length scale factor \(14 \div 7 = 2\)
- M1volume scale factor \(2^3 = 8\), multiplied by \(28\)
- A1volume of solid \(B\) \(224\) cm³
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Either paper 4 marks
Cuboid \(A\) and cuboid \(B\) are mathematically similar.
The surface area of cuboid \(A\) is \(88\) cm² and the volume of cuboid \(A\) is \(48\) cm³.
(a)Work out the surface area of cuboid \(B\).
(b)Work out the volume of cuboid \(B\).
Show answer
(a) \(792\ \text{cm}^2\)
(b) \(1296\ \text{cm}^3\)The two heights are corresponding lengths, so they give the length scale factor.
\(18 \div 6 = 3\)
A surface area is made of two-dimensional pieces, so it scales by the square of that.
\(88 \times 9 = 792\)
The surface area of cuboid \(B\) is \(792\) cm².
The length scale factor is still \(3\), and a volume scales by its cube.
\(48 \times 27 = 1296\)
The volume of cuboid \(B\) is \(1296\) cm³.
The check worth doing: cuboid \(B\) measures \(6\) cm by \(12\) cm by \(18\) cm, and \(6 \times 12 \times 18 = 1296\) as well.
- (a) M1length scale factor \(3\), so area scale factor \(9\)
- (a) A1surface area of B \(792\) cm²
- (b) M1volume scale factor \(27\) from length scale factor \(3\)
- (b) A1volume of B \(1296\) cm³
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Either paper 4 marks
Shape \(P\) and shape \(Q\) are mathematically similar.
Every length on shape \(Q\) is \(3\) times the corresponding length on shape \(P\).
The area of shape \(P\) is \(8\) cm².
Marcus says:
"The area of shape \(Q\) is \(24\) cm², because every length is \(3\) times as big."
(a)Explain why Marcus is wrong.
(b)Work out the correct area of shape \(Q\).
Show answer
(a) An area has two dimensions and both are multiplied by \(3\), so the area is multiplied by \(3^2 = 9\), not by \(3\).
(b) \(72\ \text{cm}^2\)Picture shape \(P\) covered in \(1\) cm squares. Enlarging by a scale factor of \(3\) turns each of those squares into a square with sides \(3\) cm, and that square has an area of \(3 \times 3 = 9\) cm² — not \(3\) cm².
Marcus has scaled one dimension when there are two of them. Multiplying the area by \(3\) would only be right if the shape were stretched in one direction alone, and that would not leave it similar to shape \(P\).
The area scale factor is the length scale factor squared.
\(3^2 = 9\)
\(8 \times 9 = 72\)
The area of shape \(Q\) is \(72\) cm².
- (a) B1area depends on two dimensions, both multiplied by \(3\)
- (a) B1area is multiplied by \(3^2 = 9\), not by \(3\)
- (b) M1area scale factor \(3^2 = 9\) used to multiply \(8\)
- (b) A1area of shape \(Q\) \(72\) cm²
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Either paper 3 marks
Shape \(A\) and shape \(B\) are mathematically similar hexagons.
The area of shape \(A\) is \(16\) cm² and the area of shape \(B\) is \(36\) cm².
The shortest side of shape \(A\) is \(4\) cm.
Work out the shortest side of shape \(B\).
Show answer
\(6\ \text{cm}\)Start with the area scale factor, which is the areas divided:
\(36 \div 16 = \frac{9}{4}\)
The area scale factor is the length scale factor squared, so square root it to get back to lengths.
\(\sqrt{ \frac{9}{4} } = \frac{3}{2}\)
Now multiply the shortest side of shape \(A\) by the length scale factor.
\(4 \times \frac{3}{2} = 6\)
The shortest side of shape \(B\) is \(6\) cm.
- M1area scale factor \(36 \div 16 = \frac{9}{4}\)
- M1square root of the area scale factor, length scale factor \(\frac{3}{2}\)
- A1shortest side of shape \(B\) \(6\) cm
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Either paper 3 marks
Two pyramids, \(A\) and \(B\), are mathematically similar.
The volume of \(A\) is \(4\) cm³ and the volume of \(B\) is \(256\) cm³.
The height of \(A\) is \(4\) cm.
Work out the height of \(B\).
Show answer
\(16\ \text{cm}\)Divide the volumes to get the volume scale factor.
\(256 \div 4 = 64\)
The volume scale factor is the length scale factor cubed, so take the cube root to get back to lengths.
\(\sqrt[3]{ 64 } = 4\)
Then multiply the height of \(A\) by the length scale factor.
\(4 \times 4 = 16\)
The height of \(B\) is \(16\) cm.
- M1volume scale factor \(256 \div 4 = 64\)
- M1cube root of the volume scale factor, length scale factor \(4\)
- A1height of \(B\) \(16\) cm
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Either paper 4 marks
Idris makes a scale model of a monument.
The model is \(50\) cm tall and the monument is \(4\) m tall. The model and the monument are mathematically similar, and they are made of the same material.
The model has a mass of \(5\) kg.
Work out the mass of the monument.
Show answer
\(2560\ \text{kg}\)The two heights have to be in the same units before they can be divided, so write the monument's height in centimetres:
\(4 \text{ m} = 400 \text{ cm}\)
\(400 \div 50 = 8\), so the length scale factor is \(8\).
Same material means the mass is proportional to the volume, and a volume scales by the cube of the length scale factor.
\(8 \times 8 \times 8 = 512\)
\(5 \times 512 = 2560\)
The monument has a mass of \(2560\) kg.
- M1both heights in the same units, \(400\) cm
- M1length scale factor \(8\)
- M1volume scale factor \(512\)
- A1mass \(2560\) kg
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Either paper 4 marks
The diagram shows two similar solid triangular prisms, prism \(A\) and prism \(B\).
The volume of prism \(A\) is \(88\) cm³.
The volume of prism \(B\) is \(1375\) cm³.
For prism \(B\), the length of the base of the triangle is \(12.5\) cm and the area of the triangle is \(137.5\) cm².
The height of the triangle for prism \(A\) is \(h\) cm.
Work out the value of \(h\).
Show answer
\(8.8\ \text{cm}\)Prisms \(A\) and \(B\) are similar, so divide the volumes to get the volume scale factor from \(B\) to \(A\).
\(88 \div 1375 = \frac{8}{125}\)
The volume scale factor is the length scale factor cubed, so take the cube root.
\(\sqrt[3]{ \frac{8}{125} } = \frac{2}{5}\)
Next find the height of the triangle in prism \(B\) from its area, using area \(= \frac{1}{2} \times \text{base} \times \text{height}\).
\(137.5 = \frac{1}{2} \times 12.5 \times \text{height}\)
\(\text{height} = 137.5 \times 2 \div 12.5 = 22\) cm
Finally multiply by the length scale factor from \(B\) to \(A\).
\(22 \times \frac{2}{5} = 8.8\)
\(h = 8.8\) cm
- M1volume ratio \(88 \div 1375 = \frac{8}{125}\)
- M1cube root, length scale factor \(\frac{2}{5}\)
- M1height of the triangle in prism B \(22\) cm
- A1h = \(8.8\) cm
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Either paper 3 marks
Triangles \(OPX\), \(OQY\) and \(ORZ\) are similar.
\(OPQR\) and \(OXYZ\) are straight lines.
\(OP : PQ : QR = 1 : 1 : 3\)
Show that
area of triangle \(OPX\) : area of \(PQYX\) : area of \(QRZY\) \(= 1 : 3 : 21\)
Show answer
Triangle \(OPX\) : strip \(PQYX\) : strip \(QRZY\) \(= 1 : 4 - 1 : 25 - 4 = 1 : 3 : 21\)The three triangles are similar, so their areas are in the ratio of the squares of their corresponding lengths. The lengths from \(O\) are cumulative, because \(OQ\) is \(OP\) and \(PQ\) together.
Measuring in the parts of the given ratio: \(OP = 1\), \(OQ = 1 + 1 = 2\) and \(OR = 1 + 1 + 3 = 5\)
\(OP : OQ : OR = 1 : 2 : 5\)
Square them to get the areas of the whole triangles:
triangle \(OPX\) : triangle \(OQY\) : triangle \(ORZ\) \(= 1 : 4 : 25\)
Each strip is a bigger triangle with the next smaller triangle removed, so subtract:
area of \(PQYX\) \(= 4 - 1 = 3\)
area of \(QRZY\) \(= 25 - 4 = 21\)
area of triangle \(OPX\) \(= 1\)
So area of triangle \(OPX\) : area of \(PQYX\) : area of \(QRZY\) \(= 1 : 3 : 21\), as required.
- M1cumulative lengths \(OP : OQ : OR = 1 : 2 : 5\)
- M1areas of the three triangles in the ratio \(1 : 4 : 25\)
- A1subtracting to get the strips, \(3\) and \(21\), ending at \(1 : 3 : 21\)