Work out the perimeter of the trapezium. Give your answer correct to 3 significant figures. (5)
Mark scheme
Answer
Mark
Mark scheme
44.9
P1
for process to find an expression for the area of the trapezium, eg \(\tfrac{1}{2}(12 + CD)8\) or \(8 \times 12 + \tfrac{1}{2} \times 8 \times x\)
or for process to find the area of the triangle, eg \(112 - 8 \times 12\ (= 16)\)
P1
for forming an equation and isolating terms in the unknown length, eg \(4CD = 112 - 48\) or \(\tfrac{1}{2} \times 8 \times x = 112 - 8 \times 12\) or \(\tfrac{1}{2} \times 8 \times x = \text{``}16\text{''}\) or \(CD = 16\) or \(x = 4\)
P1
for start of process to find length of \(BC\), eg \(8^2 + \text{``}4\text{''}^2\ (= 80)\) or \(8^2 + [\text{their } x]^2\) or \(\tan^{-1}\left(\dfrac{\text{``}4\text{''}}{8}\right)\ (= 26.5\ldots)\) oe or \(\tan^{-1}\left(\dfrac{8}{\text{``}4\text{''}}\right)\ (= 63.4\ldots)\) where “4” can be [their \(x\)]
P1
for \(\sqrt{8^2 + \text{``}4\text{''}^2}\) or \(\sqrt{64 + \text{``}16\text{''}}\) or \(\sqrt{80}\) or \(4\sqrt{5}\ (= 8.9\ldots)\) oe or \(\sqrt{8^2 + [\text{their } x]^2}\) or \(\dfrac{\text{``}4\text{''}}{\sin \text{``}26.5\ldots\text{''}}\) or \(\dfrac{8}{\cos \text{``}26.5\ldots\text{''}}\) or \(\dfrac{8}{\sin \text{``}63.4\ldots\text{''}}\) or \(\dfrac{\text{``}4\text{''}}{\cos \text{``}63.4\ldots\text{''}}\) where “4” can be [their \(x\)]
A1
for answer in the range 44.9 to 44.95
Additional guidance
\(x\) is the length of the line from \(C\) to where perpendicular from \(B\) meets \(CD\) Allow use of other letters in place of \(CD\) and \(x\) (do not have to be defined unless otherwise stated) Award P1 for \(112 - 8 \times 12\ (= 16)\) even if not used
Award P2 for \(CD = 16\) or \(x = 4\) even if not used unless clearly from incorrect working eg \(12 - 8\ (= 4)\) Only award P2 for 16 if it is clearly identified as \(CD\)
[their \(x\)] can be any value less than 12 or clearly identified as the length of the line from \(C\) to where perpendicular from \(B\) meets \(CD\) (may be seen on the diagram)
Award P4 for (\(BC =\)) \(\sqrt{80}\) or \(4\sqrt{5}\) or 8.9… unless \(x = 4\) is clearly from incorrect working
If an answer is shown in the range in working and then incorrectly rounded award full marks
Work out the perimeter of the trapezium. Give your answer correct to 3 significant figures. (5)
Mark scheme
Answer
Mark
Mark scheme
44.9
P1
for process to find an expression for the area of the trapezium, eg \(\tfrac{1}{2}(12 + CD)8\) or \(8 \times 12 + \tfrac{1}{2} \times 8 \times x\)
or for process to find the area of the triangle, eg \(112 - 8 \times 12\ (= 16)\)
P1
for forming an equation and isolating terms in the unknown length, eg \(4CD = 112 - 48\) or \(\tfrac{1}{2} \times 8 \times x = 112 - 8 \times 12\) or \(\tfrac{1}{2} \times 8 \times x = \text{``}16\text{''}\) or \(CD = 16\) or \(x = 4\)
P1
for start of process to find length of \(BC\), eg \(8^2 + \text{``}4\text{''}^2\ (= 80)\) or \(8^2 + [\text{their } x]^2\) or \(\tan^{-1}\left(\dfrac{\text{``}4\text{''}}{8}\right)\ (= 26.5\ldots)\) oe or \(\tan^{-1}\left(\dfrac{8}{\text{``}4\text{''}}\right)\ (= 63.4\ldots)\) where “4” can be [their \(x\)]
P1
for \(\sqrt{8^2 + \text{``}4\text{''}^2}\) or \(\sqrt{64 + \text{``}16\text{''}}\) or \(\sqrt{80}\) or \(4\sqrt{5}\ (= 8.9\ldots)\) oe or \(\sqrt{8^2 + [\text{their } x]^2}\) or \(\dfrac{\text{``}4\text{''}}{\sin \text{``}26.5\ldots\text{''}}\) or \(\dfrac{8}{\cos \text{``}26.5\ldots\text{''}}\) or \(\dfrac{8}{\sin \text{``}63.4\ldots\text{''}}\) or \(\dfrac{\text{``}4\text{''}}{\cos \text{``}63.4\ldots\text{''}}\) where “4” can be [their \(x\)]
A1
for answer in the range 44.9 to 44.95
Additional guidance
\(x\) is the length of the line from \(C\) to where perpendicular from \(B\) meets \(CD\) Allow use of other letters in place of \(CD\) and \(x\) (do not have to be defined unless otherwise stated) Award P1 for \(112 - 8 \times 12\ (= 16)\) even if not used
Award P2 for \(CD = 16\) or \(x = 4\) even if not used unless clearly from incorrect working eg \(12 - 8\ (= 4)\) Only award P2 for 16 if it is clearly identified as \(CD\)
[their \(x\)] can be any value less than 12 or clearly identified as the length of the line from \(C\) to where perpendicular from \(B\) meets \(CD\) (may be seen on the diagram)
Award P4 for (\(BC =\)) \(\sqrt{80}\) or \(4\sqrt{5}\) or 8.9… unless \(x = 4\) is clearly from incorrect working
If an answer is shown in the range in working and then incorrectly rounded award full marks
In the diagram, all measurements are in centimetres.
\(AC = BC\)
The perimeter of the triangle is 72 cm.
Work out the area of the triangle. (5)
Mark scheme
Answer
Mark
Mark scheme
240
P1
for forming an appropriate equation, eg \(2x + 11 = 4x - 4\) or \(2x + 11 + 4x - 4 + 2x + 5 = 72\) or \(8x + 12 = 72\)
P1
(dep P1) for process to correctly isolate terms in \(x\), eg \(4x - 2x = 11 + 4\) or \(2x + 4x + 2x = 72 - 11 + 4 - 5\) or \(x = 7.5\) oe
P1
for correct application of Pythagoras, eg \((\text{``}26\text{''})^2 - \left(\dfrac{\text{``}20\text{''}}{2}\right)^2\) or \([AC]^2 - \left(\dfrac{[AB]}{2}\right)^2\) or height = 24 or a complete method to find the height
or for a correct trig statement to find \(CAB\) or \(CBA\) or \(ACB\), eg \(\cos CAB = \cos CBA = \dfrac{\text{``}20\text{''} \div 2}{\text{``}26\text{''}}\) or \(\cos CAB = \cos CBA = \dfrac{20^2 + 26^2 - 26^2}{2 \times 20 \times 26}\) or \(\cos ACB = \dfrac{26^2 + 26^2 - 20^2}{2 \times 26 \times 26}\) or \(CAB = 67.3\ldots\) or \(CBA = 67.3\ldots\) or \(ACB = 45.2\ldots\)
P1
for process to find area of triangle, eg \(\text{``}20\text{''} \times \text{``}24\text{''} \div 2\) or \([AB] \times [\text{height}] \div 2\)
or for process to find area of triangle, eg \(\dfrac{1}{2} \times \text{``}26\text{''} \times \text{``}20\text{''} \times \sin \text{``}67.3\ldots\text{''}\) or \(\dfrac{1}{2} \times \text{``}26\text{''} \times \text{``}26\text{''} \times \sin \text{``}45.2\ldots\text{''}\) or \(\dfrac{1}{2} \times [AB] \times [AC] \times \sin [BAC]\) or \(\dfrac{1}{2} \times [BC] \times [AC] \times \sin [ACB]\)
A1
cao
Additional guidance
\(8x = 60\) or \(2x = 15\) implies P2 A correct length stated or shown on diagram implies P2, eg \(AB = 20\), \(AC = 26\), \(CB = 26\) \([AC]\) \([BC]\) \([AB]\) \([ACB]\) \([CAB]\) and \([BAC]\) must be clearly identified if incorrect. May be on diagram. \(AB = 2 \times \text{``}7.5\text{''} + 5\ (= 20)\) \(AC = 2 \times \text{``}7.5\text{''} + 11\ (= 26)\) \(CB = 4 \times \text{``}7.5\text{''} - 4\ (= 26)\) Alternative scheme not expected on Foundation tier but may be seen.
ft incorrect figures providing at least one previous P1 awarded. [height] is what they clearly think is the height of the triangle but not 26 or 20 or 10
The radius of the base of the cone is \(\dfrac{3}{4}\) of the height of the cone. The total surface area of the cone is \(54\pi\) cm2
Work out the height of the cone. (4)
Mark scheme
Answer
Mark
Mark scheme
6
P1
for starting process, by defining height, radius and using Pythagoras to form an equation for the slant height \(l\) eg height \(= h\), radius \(= \frac{3}{4}h\) and \(l^2 = h^2 + \left(\dfrac{3h}{4}\right)^2 \left(= \dfrac{25}{16}h^2\right)\) or \((l =) \sqrt{h^2 + \left(\dfrac{3h}{4}\right)^2} \left(= \dfrac{5}{4}h\right)\) oe eg \(r = 3x\) and \(h = 4x\) and \(l^2 = (3x)^2 + (4x)^2\)
P1
(dep P1) for process to form a correct expression for the curved surface area in terms of a single variable, eg \(\pi \times \dfrac{3}{4}h \times \text{``}\dfrac{5}{4}h\text{''} \left(= \dfrac{15}{16}\pi h^2\right)\) or \(\pi \times 3x \times \text{``}5x\text{''}\) where \(h = 4x\)
P1
for forming and simplifying a correct equation to find height, eg \(\dfrac{24\pi}{16}h^2 = 54\pi\)
A1
cao
Additional guidance
Can use any other letter than \(h\) provided it is defined eg height \(= x\)
May include area of circle eg \(\pi \times \left(\dfrac{3}{4}h\right)^2 + \pi \times \dfrac{3}{4}h \times \text{``}\dfrac{5}{4}h\text{''} \left(= \dfrac{24}{16}\pi h^2\right)\)
In the diagram, all measurements are in centimetres.
\(AC = BC\)
The perimeter of the triangle is 72 cm.
Work out the area of the triangle. (5)
Mark scheme
Answer
Mark
Mark scheme
240
P1
for forming an appropriate equation, eg \(2x + 11 = 4x - 4\) or \(2x + 11 + 4x - 4 + 2x + 5 = 72\) or \(8x + 12 = 72\)
P1
(dep P1) for process to isolate terms in \(x\) for their equation, eg \(4x - 2x = 11 + 4\) or \(2x + 4x + 2x = 72 - 11 + 4 - 5\) or \(x = 7.5\) oe
P1
for correct application of Pythagoras, eg \((\text{``}26\text{''})^2 - \left(\dfrac{\text{``}20\text{''}}{2}\right)^2\) or \([AC]^2 - \left(\dfrac{[AB]}{2}\right)^2\) or height = 24 or a complete method to find the height
or for a correct trig statement to find \(CAB\) or \(CBA\) or \(ACB\), eg \(\cos CAB = \cos CBA = \dfrac{\text{``}20\text{''} \div 2}{\text{``}26\text{''}}\) or \(\cos CAB = \cos CBA = \dfrac{20^2 + 26^2 - 26^2}{2 \times 20 \times 26}\) or \(\cos ACB = \dfrac{26^2 + 26^2 - 20^2}{2 \times 26 \times 26}\) or \(CAB = 67.3\ldots\) or \(CBA = 67.3\ldots\) or \(ACB = 45.2\ldots\)
P1
for process to find area of triangle, eg \(\text{``}20\text{''} \times \text{``}24\text{''} \div 2\) or \([AB] \times [\text{height}] \div 2\)
or for process to find area of triangle, eg \(\dfrac{1}{2} \times \text{``}26\text{''} \times \text{``}20\text{''} \times \sin \text{``}67.3\ldots\text{''}\) or \(\dfrac{1}{2} \times \text{``}26\text{''} \times \text{``}26\text{''} \times \sin \text{``}45.2\ldots\text{''}\) or \(\dfrac{1}{2} \times [AB] \times [AC] \times \sin [BAC]\) or \(\dfrac{1}{2} \times [BC] \times [AC] \times \sin [ACB]\)
A1
cao
Additional guidance
\(8x = 60\) or \(2x = 15\) implies P2 A correct length stated or shown on diagram implies P2 Eg \(AB = 20\), \(AC = 26\), \(CB = 26\) \([AC]\) \([BC]\) \([AB]\) \([ACB]\) \([CAB]\) and \([BAC]\) must be clearly identified if incorrect. May be on diagram. \(AB = 2 \times \text{``}7.5\text{''} + 5\ (= 20)\) \(AC = 2 \times \text{``}7.5\text{''} + 11\ (= 26)\) \(CB = 4 \times \text{``}7.5\text{''} - 4\ (= 26)\) Alternative scheme not expected on Foundation tier but may be seen.
ft incorrect figures providing at least one previous P1 awarded. [height] is what they clearly think is the height of the triangle but not 26 or 20 or 10
Use Pythagoras’ theorem to show that the value of \(x\) is between 10 and 11 [4 marks]
Mark scheme
Answer
Mark
Comments
\(12^2\) or \(13^2\)
M1
oe 144 or 169 implied by 313 or \(\sqrt{313}\) or [17.6, 17.7]
\(13^2 - 12^2\) or \(169 - 144\) or 25 or \(\sqrt{13^2 - 12^2}\) or \(\sqrt{169 - 144}\) or \(\sqrt{25}\) or 5
M1dep
oe 5 may be in correct position on the diagram
\(9^2 + (\text{their } 5)^2\) or \(81 +\) their 25 or 106 or \(\sqrt{9^2 + (\text{their } 5)^2}\) or \(\sqrt{81 + \text{their } 25}\) or \(\sqrt{106}\) or [10.2, 10.3]
M1dep
oe their 5 or their 25 must be from correct working
\(9^2 + 5^2 = 106\) and \(\sqrt{106}\) and [10.2, 10.3] or \(\sqrt{9^2 + 5^2}\) and [10.2, 10.3] or \(\sqrt{81 + 25}\) and [10.2, 10.3] or \(9^2 + 5^2 = 106\) and \(10^2 = 100\) and \(11^2 = 121\)
A1
oe
Additional guidance
Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
Correct use of Pythagoras’ theorem using a square root
Alternative method 2
\(\sin E = \dfrac{8}{17}\;\) or \(\;\cos A = \dfrac{8}{17}\) or \(E = 28.(\ldots)\) or \(A = 61.9(\ldots)\) or 62 and \(\cos 28.(\ldots) = \dfrac{EM}{17}\) or \(\tan 28.(\ldots) = \dfrac{8}{EM}\) or \(\sin 61.9(\ldots) = \dfrac{EM}{17}\) or \(\tan 61.9(\ldots) = \dfrac{EM}{8}\)
M1
\(17 \cos 28.(\ldots)\) or \(8 \div \tan 28.(\ldots)\) or \(17 \sin 61.9(\ldots)\) or \(8 \tan 61.9(\ldots)\)
A1
Additional guidance
8, 15, 17 on their own
M0A0
\(EM^2 = 289 - 64 = 225,\ EM = 15\)
M1A0
Mark scheme (b)
Answer
Mark
Comments
Alternative method 1
\(30^2 + (16 \div 2)^2\) or \(\;30^2 + 8^2\) or 964
M1
oe
\(\sqrt{\text{their } 964}\) or \(2\sqrt{241}\) or [31, 31.1]
M1dep
oe CM
\(\tan x = \dfrac{15}{\text{their } [31,\ 31.1]}\)
M1dep
oe \(\;\) eg \(\;90 - \tan^{-1} \dfrac{\text{their } [31,\ 31.1]}{15}\) dep on M1 M1
[25.7, 26]
A1
Alternative method 2
\(30^2 + 17^2\) \(\;\) or \(\;\) 1189
M1
oe
\(\sqrt{\text{their } 1189}\) or [34.4, 34.5]
M1dep
oe CE
\(\sin x = \dfrac{15}{\text{their } [34.4,\ 34.5]}\)
M1dep
oe \(\;\) eg \(\;90 - \cos^{-1} \dfrac{15}{\text{their } [34.4,\ 34.5]}\) or \(\dfrac{\sin x}{15} = \dfrac{\sin 90}{\text{their } [34.4,\ 34.5]}\) dep on M1 M1
[25.7, 26]
A1
Alternative method 3
\(30^2 + (16 \div 2)^2\) or 964 or \(\;30^2 + 17^2\) or 1189
M1
oe
\(\sqrt{\text{their } 964}\) or \(2\sqrt{241}\) or [31, 31.1] or \(\sqrt{\text{their } 1189}\) or [34.4, 34.5]
M1dep
oe CM CE
\(\cos x = \dfrac{\text{their } [31,\ 31.1]}{\text{their } [34.4,\ 34.5]}\)
M1dep
oe \(\;\) eg \(\;90 - \sin^{-1} \dfrac{\text{their } [31,\ 31.1]}{\text{their } [34.4,\ 34.5]}\) dep on M1 M1
[25.7, 26]
A1
Alternative method 4
\(17^2 - (16 \div 2)^2\) or 225 or \(30^2 + (16 \div 2)^2\) or 964 or \(30^2 + 17^2\) or 1189
Work out the value of \(x\) as a decimal. [3 marks]
Mark scheme
Answer
Mark
Comments
\(8^2\) and \(3^2\) seen or \(8 \times 8\) and \(3 \times 3\) seen or 64 and 9 seen or 55
M1
M2 for \(\sin^{-1}\left(\dfrac{3}{8}\right) = 22.(\ldots)\) and \(8\cos(\text{their } 22.(\ldots))\) or \(\cos^{-1}\left(\dfrac{3}{8}\right) = 67.(\ldots)\) or 68 and \(8\sin(\text{their } 67.(\ldots))\)
\(\sqrt{8^2 - 3^2}\) or \(\sqrt{64 - 9}\) or \(\sqrt{55}\)
M1dep
[7.4, 7.42]
A1
Additional guidance
\(\sqrt{8^2 + 3^2}\) or \(\sqrt{64 + 9}\) or \(8^2 + 3^2\) or \(64 + 9\)
M1M0depA0
Only \(\sqrt{73}\) or only 73 or only 8.5...
M0
If trigonometry used it must be a fully correct method that would lead to the correct value of \(x\)
Partial method using trigonometry
M0
Ignore units given
8 cm\(^2\) is not \(8^2\) unless recovered
Correct answer in range seen, ignore further work if truncates or rounds
M2A1
\(8^2 = 16\) and \(3^2 = 6,\ \sqrt{16 - 6}\)
M1M1depA0
Scale drawing with answer in range [7.4, 7.42]
M2A1
Scale drawing with answer not in range [7.4, 7.42]
