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Either paper 4 marks
Here is a rectangle.
(a)Find the perimeter of the rectangle.
(b)Find the area of the rectangle.
Show answer
(a) \(46\ \text{cm}\)
(b) \(112\ \text{cm}^2\)The two unlabelled sides are the same as the two labelled ones.
\(16 + 7 + 16 + 7 = 46\)
The perimeter is \(46\) cm.
\(\text{Area} = \text{length} \times \text{width}\)
\(16 \times 7 = 112\)
The area is \(112\) cm².
- (a) M1adding all four sides, \(16 + 7 + 16 + 7\), or \(2 \times (16 + 7)\)
- (a) A1perimeter \(46\) cm
- (b) M1\(16 \times 7\)
- (b) A1area \(112\) cm²
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Either paper 2 marks
Here is a triangle. The dashed line is the perpendicular height.
Work out the area of the triangle.
Show answer
\(100\ \text{cm}^2\)\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{perpendicular height}\)
\(= \frac{1}{2} \times 20 \times 10 = \frac{ 200 }{2} = 100\)
The area is \(100\) cm².
- M1uses \(\frac{1}{2} \times 20 \times 10\) (or \(200\) halved)
- A1area \(100\) cm²
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Either paper 1 mark
A regular pentagon has sides of length \(2\) cm. Every side is made longer, so that each side is now \(6\) cm.
Work out what the perimeter of the shape is multiplied by.
Show answer
\(\times 3\)Scale factor for each side: \(6 \div 2 = 3\)
The perimeter is the sum of the sides, and every side is multiplied by the scale factor, so the perimeter is multiplied by the scale factor.
Multiplier for the perimeter: \(3^{ 1 } = 3\)
So the perimeter becomes \(3\) times as large.
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Either paper 4 marks
Here is a rectangle with an area of \(54\) cm².
(a)Work out the length of the side marked \(x\).
(b)Work out the perimeter of the rectangle.
Show answer
(a) \(6\ \text{cm}\)
(b) \(30\ \text{cm}\)The area of a rectangle is length times width, so divide the area by the side you know.
\(x = 54 \div 9 = 6\)
\(x = 6\) cm
The rectangle is \(6\) cm by \(9\) cm.
\(2 \times (6 + 9) = 30\)
The perimeter is \(30\) cm.
- (a) M1\(54 \div 9\)
- (a) A1\(x = 6\) cm
- (b) M1\(2 \times (6 + 9)\), or all four sides added
- (b) A1perimeter \(30\) cm
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Either paper 4 marks
The diagram shows a right-angled triangle.
(a)Find the perimeter of the triangle.
(b)Find the area of the triangle.
Show answer
(a) \(24\ \text{cm}\)
(b) \(24\ \text{cm}^2\)The perimeter is the total distance round the outside.
\(6 + 8 + 10 = 24\)
The perimeter is \(24\) cm.
The two sides that meet at the right angle are the base and the height.
\(\frac{1}{2} \times 6 \times 8 = \frac{ 48 }{2} = 24\)
The area is \(24\) cm².
- (a) M1adding the three sides \(6 + 8 + 10\)
- (a) A1perimeter \(24\) cm
- (b) M1\(\frac{1}{2} \times 6 \times 8\)
- (b) A1area \(24\) cm²
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Either paper 2 marks
The diagram shows an isosceles triangle.
Find the perimeter of the triangle.
Show answer
\(41\ \text{cm}\)The dashes show that the two sloping sides are equal, so the unlabelled side is also \(14\) cm.
\(14 + 14 = 28\)
\(28 + 13 = 41\)
The perimeter is \(41\) cm.
- M1third side equal to the marked side, \(14\) cm, and adding all three sides
- A1perimeter \(41\) cm
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Either paper 4 marks
The diagram shows a parallelogram.
(a)Work out the area of the parallelogram.
(b)Work out the perimeter of the parallelogram.
Show answer
(a) \(156\ \text{cm}^2\)
(b) \(56\ \text{cm}\)\(\text{Area} = \text{base} \times \text{perpendicular height}\)
The perpendicular height is the dashed line, \(12\) cm. The \(15\) cm side is sloping, so it is not the height.
\(13 \times 12 = 156\)
The area is \(156\) cm².
Opposite sides of a parallelogram are equal, so the two unlabelled sides are \(13\) cm and \(15\) cm.
\(2 \times (13 + 15) = 2 \times 28 = 56\)
The perimeter is \(56\) cm.
- (a) M1\(13 \times 12\) using the perpendicular height
- (a) A1area \(156\) cm²
- (b) M1\(2 \times (13 + 15)\) or \(2 \times 28\)
- (b) A1perimeter \(56\) cm
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Either paper 3 marks
The square below has an area of \(196\) cm².
(a)Find the length of the side marked \(x\).
(b)Find the perimeter of the square.
Show answer
(a) \(14\ \text{cm}\)
(b) \(56\ \text{cm}\)The area of a square is the side length multiplied by itself, so the side is the square root of the area.
\(x = \sqrt{ 196 } = 14\)
\(x = 14\) cm
All four sides of a square are equal.
\(4 \times 14 = 56\)
The perimeter is \(56\) cm.
- (a) M1square root of the area \(\sqrt{ 196 }\)
- (a) A1side length \(14\) cm
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Either paper 4 marks
Here is a regular octagon.
(a)Calculate the perimeter of the octagon.
(b)A different regular octagon has a perimeter of \(64\) cm.
Calculate the length of one of its sides.
Show answer
(a) \(96\ \text{cm}\)
(b) \(8\ \text{cm}\)Every side of a regular polygon is the same length, and a octagon has \(8\) of them.
\(8 \times 12 = 96\)
The perimeter is \(96\) cm.
The \(64\) cm is shared equally between \(8\) sides.
\(64 \div 8 = 8\)
Each side is \(8\) cm.
- (a) M1\(8 \times 12\)
- (a) A1perimeter \(96\) cm
- (b) M1\(64 \div 8\)
- (b) A1side \(8\) cm
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Either paper 2 marks
Here is a rectangular carpet that measures 8 m by 5 m.
Part of the carpet is covered by a square rug of side 2 m.
Show that \(\frac{ 1 }{ 10 }\) of the carpet is covered by the rug.
Show answer
\(\frac{ 2^2 }{ 8 \times 5 } = \frac{ 4 }{ 40 } = \frac{ 1 }{ 10 }\)Area of the rug: \(2 \times 2 = 4\) m²
Area of the carpet: \(8 \times 5 = 40\) m²
The fraction of the carpet covered is the rug area over the carpet area:
\(\frac{ 4 }{ 40 }\)
Divide the top and bottom by 4:
\(\frac{ 4 \div 4 }{ 40 \div 4 } = \frac{ 1 }{ 10 }\)
So \(\frac{ 1 }{ 10 }\) of the carpet is covered by the rug.
- M1rug area \(4\) m² and carpet area \(40\) m²
- A1\(\frac{ 4 }{ 40 }\) divided through by \(4\) to reach \(\frac{ 1 }{ 10 }\)