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Higher June 2025 Paper 2 Q10
10 \(ABD\) and \(ABC\) are right-angled triangles.
Diagram NOT accurately drawn
\(AB = 16\) cm \(BC = 12\) cm \(AD = 1.5 \times AC\)
Find the length of \(CD\)
(5)
Mark scheme
Mark scheme Scheme Marks \(\left(AC^2 =\right)12^2 + 16^2 \;(= 144 + 256 = 400)\) or \((BAC =)\tan^{-1}\left(\dfrac{12}{16}\right)\left(= 36.8(698\ldots)\right)\) or 36.9 or
\((BCA =)\tan^{-1}\left(\dfrac{16}{12}\right)\left(= 53.1(301\ldots)\right)\)
M1 \((AC =)\sqrt{12^2 + 16^2}\left(= \sqrt{144 + 256} = \sqrt{400} = 20\right)\) or \((AC =)\frac{16}{\cos\text{``}{36.8}\text{''}}(= 20)\) or
\((AC =)\frac{12}{\sin\text{``}{36.8}\text{''}}(= 20)\) or \((AC =)\frac{16}{\sin\text{``}{53.1}\text{''}}(= 20)\) or \((AC =)\frac{12}{\cos\text{``}{53.1}\text{''}}(= 20)\)
M1 \(\left(BD^2 =\right)\left(1.5 \times \text{``}{20}\text{''}\right)^2 - 16^2 \;(= 644)\) or \(\left(BD^2 =\right)30^2 - 16^2 \;(= 900 - 256 = 644)\) or
\((BAD =)\cos^{-1}\left(\dfrac{16}{\text{``}{30}\text{''}}\right)\left(= 57.7(690\ldots)\right)\) or 57.8 or \((BDA =)\sin^{-1}\left(\dfrac{16}{\text{``}{30}\text{''}}\right)\left(= 32.2(309\ldots)\right)\)
or \((BCA =)\sin^{-1}\left(\dfrac{16}{\text{``}{20}\text{''}}\right)\left(= 53.1(301\ldots)\right)\) and \(\dfrac{(CD)}{\sin(180 - 126.9 - 32.2)} = \dfrac{\text{``}{30}\text{''}}{\sin(180 - 53.1)}\) oe
M1 \((BD =)\sqrt{\left(1.5 \times \text{``}{20}\text{''}\right)^2 - 16^2}\) (= 25.3(771…)) or
\((BD =)\sqrt{30^2 - 16^2}\left(= \sqrt{900 - 256} = \sqrt{644} = 2\sqrt{161} = 25.3(771\ldots)\right)\) or
\((BD =)16 \times \tan\text{``}{57.7}\text{''}\left(= 25.3(771\ldots)\right)\) or \((BD =)\text{``}{30}\text{''} \times \sin\text{``}{57.7}\text{''}\left(= 25.3(771\ldots)\right)\) or
\((BD =)\sqrt{16^2 + \text{``}{30}\text{''}^2 - 2 \times 16 \times \text{``}{30}\text{''} \times \cos\text{``}{57.7}\text{''}}\left(= 25.3(771\ldots)\right)\) or
\((BD =)30 \times \cos\text{``}{32.2}\text{''}\left(= 25.3(771\ldots)\right)\) or \((BD =)\dfrac{16}{\tan\text{``}{32.2}\text{''}}\left(= 25.3(771\ldots)\right)\) or
\((BD =)\dfrac{16}{\sin\text{``}{32.2}\text{''}} \times \sin\text{``}{57.7}\text{''}\left(= 25.3(771\ldots)\right)\) or \((CD =)\dfrac{\text{``}{30}\text{''}}{\sin\text{``}{126.9}\text{''}} \times \sin\text{``}{20.9}\text{''}\) oe
M1 Working not required, so correct answer scores full marks (unless from obvious incorrect working) A1 (5) (5 marks)
Notes M1: for a correct method using triangle \(ABC\)
M1: for a correct method to find \(AC\)
M1: for a correct method using triangle to find \(BD^2\) or angle \(BAD\) or angle \(BDA\) or for a correct equation for side \(CD\)
M1: for a correct method to find \(BD\) or \(CD\)
A1: awrt 13.4
Higher June 2025 Paper 1 Q9
9
Diagram NOT accurately drawn
\(ABCD\) is a trapezium with one line of symmetry.
angle \(ADC = 60^\circ\) \(AD = 12\) cm \(DC = 47\) cm
Work out the area of the trapezium.
(5)
Mark scheme
Mark scheme Scheme Marks eg \(12\sin 60\left(= 6\sqrt{3} = 10.3(9\ldots)\right)\) or \(\sqrt{12^2 - \text{``}{6}\text{''}^2}\left(= 6\sqrt{3} = 10.3(9\ldots)\right)\)
or (Area \(ADC\) =) \(\dfrac{1}{2} \times 12 \times 47 \times \sin 60 \;(= 244.2\ldots)\)
M1 eg \(12\cos 60 (= 6)\) or \(\sqrt{12^2 - \left(\text{``}{6\sqrt{3}}\text{''}\right)^2}\;(= 6)\) M1 eg (\(AB\) =) 47 – “6” – “6” (= 35) M1 eg (Trapezium =) \(\dfrac{1}{2} \times \left(47 + \text{``}{35}\text{''}\right) \times \text{``}{10.3(9\ldots)}\text{''}\)
or (Rectangle + 2 × Triangle =) \(\text{``}{35}\text{''} \times \text{``}{10.3(9\ldots)}\text{''} + 2 \times \dfrac{1}{2} \times \text{``}{6}\text{''} \times \text{``}{10.3(9\ldots)}\text{''}\)
or (Rectangle + 2 × Triangle =) \(\text{``}{35}\text{''} \times \text{``}{10.3(9\ldots)}\text{''} + 2 \times \dfrac{1}{2} \times \text{``}{6}\text{''} \times 12 \times \sin 60\)
or (Triangle \(ADC\) + Triangle \(ABC\) =) \(\text{``}{244.2\ldots}\text{''} + \dfrac{1}{2} \times 12 \times \text{``}{35}\text{''} \times \sin 120\)
oe eg \(\left(47 - \text{``}{6}\text{''}\right) \times \text{``}{10.3(9\ldots)}\text{''}\)
M1 Working required A1 (5) (5 marks)
Notes M1: for a method find the height of the trapeziumor the area of triangle \(ADC\)
M1: (indep)
for a method find the base of the triangle, condone missing brackets around \(\text{``}{6\sqrt{3}}\text{''}\)
The first two M1 marks can be awarded in either order
M1: (dep on previous M1)
M1: for a complete method
A1: (dep on M1) allow 420 – 427 from correct working
Higher June 2025 Paper 1R Q9
9 The diagram shows triangle \(PQR\)
Diagram NOT accurately drawn
Work out the length of \(QR\)
(3)
Mark scheme
Mark scheme Scheme Marks eg \(\tan 24 = \dfrac{6.5}{QR}\) or \(\dfrac{6.5}{\sin 24} = \dfrac{QR}{\sin(180 - 90 - 24)}\) oe
or \(\tan(180 - 90 - 24) = \dfrac{QR}{6.5}\) or
\((PR =)\dfrac{6.5}{\sin 24}(= 15.9\ldots)\) and \(6.5^2 + QR^2 = \text{``}{15.9}\text{''}^2\)
M1 eg \((QR =)\dfrac{6.5}{\tan 24}\) or \((QR =)\dfrac{6.5}{\sin 24} \times \sin 66\)
or \((QR =)\;6.5\tan 66\) [where 66 = 180 – 90 – 24]
or \((QR =)\sqrt{\text{``}{15.9}\text{''}^2 - 6.5^2}\)
M1 Correct answer scores full marks (unless from obvious incorrect working) A1 (3) (3 marks)
Notes M1: for setting up a trig equation in \(QR\) or and then setting up Pythagoras or trig equation for \(QR\)
M1: for a complete method
A1: accept 14.5 – 14.61
Higher November 2024 Paper 2 Q25
25 The diagram shows an equilateral triangle \(ABC\) and a circle with centre \(O\)
Diagram NOT accurately drawn
\(AB\), \(BC\) and \(CA\) are tangents to the circle.
The total area, in cm², of the regions shown shaded in the diagram is \(nx^2\)
Find the value of \(n\)
(5)
Mark scheme Mark scheme (using a value for the radius)
Mark scheme Scheme Marks eg \(\tan 30 = \dfrac{x}{0.5AB}\) or \(\tan 60 = \dfrac{0.5AB}{x}\) oe or
\(0.5AB = \dfrac{x}{\tan 30}\) or \(0.5AB = x\tan 60\) oe or \(\dfrac{1}{2}AB = \sqrt{3}x\)
M1 eg \((AB =)\;2x\tan 60\) or \(\dfrac{2x}{\tan 30}\) or \(2\sqrt{3}x\) or \(3.46..x\) oe
OR
area small triangle \(\left(\dfrac{1}{6}\textit{shape}\right) = \dfrac{1}{2} \times x\tan 60 \times x\) or \(\dfrac{1}{2} \times \sqrt{3}x \times x\) or \(\dfrac{\sqrt{3}}{2}x^2\)
M1 eg \(6 \times \dfrac{1}{2} \times x\tan 60 \times x\) or \(\dfrac{1}{2} \times (2x\tan 60)^2 \times \sin 60\) or \(\dfrac{1}{2}\left(\dfrac{2x}{\tan 30}\right)^2 \sin 60\)
or \(\dfrac{1}{2} \times 2\sqrt{3}x \times \sqrt{\left(2\sqrt{3}x\right)^2 - \left(\sqrt{3}x\right)^2}\) or \(0.5 \times 2\sqrt{3}x \times 3x\) or \(3\sqrt{3}x^2\) or \(5.19\ldots x^2\)
M1 \(6 \times \dfrac{\sqrt{3}}{2}x^2 - \pi x^2\) or \(3\sqrt{3}x^2 - \pi x^2\) or \(\text{``}{5.19\ldots}\text{''}x^2 - \pi x^2\)
or \(6 \times \dfrac{1}{2} \times \tan 60 \times x^2 - \pi x^2\) or \(\dfrac{1}{2}\left(\dfrac{2x}{\tan 30}\right)^2 \sin 60 - \pi x^2\) oe
M1 Correct answer scores full marks (unless from obvious incorrect working) Look for values written on the diagram A1 (5) (5 marks)
Notes M1: (off spec but a student who knows without calc that height of triangle = \(3x\): \(\tan 60 = \dfrac{3x}{0.5AB}\) or \(\sin 60 = \dfrac{3x}{BC}\) )
M1: Expression for side of triangle (\(AB\) or \(BC\) or \(AC\))OR
M1: a correct expression for the area of triangle \(ABC\)
M1: a correct expression for the area of the shaded parts of the diagram or a correct equation for the shaded parts of the diagram
A1: 2.05 – 2.06
Mark scheme (using a value for the radius) Scheme Marks eg \(\tan 30 = \dfrac{100}{0.5AB}\) or \(\tan 60 = \dfrac{0.5AB}{100}\) oe or
\(0.5AB = \dfrac{100}{\tan 30}\) or \(0.5AB = 100\tan 60\) oe or
\(\dfrac{1}{2}AB = \sqrt{3} \times 100\)
M1 eg \((AB =)\;200\tan 60\) or \(\dfrac{200}{\tan 30}\) or \(200\sqrt{3}x\) or \(346\ldots x\) )
or
area small triangle \(\left(\dfrac{1}{6}\textit{shape}\right) = \dfrac{1}{2} \times \text{``}{173\ldots}\text{''} \times 100\;(= 8660\ldots)\)
M1 eg \(\dfrac{1}{2} \times 200\sqrt{3} \times 200\sqrt{3} \times \sin 60\;(= 51961.52\ldots)\) or
or
\(6 \times \dfrac{1}{2} \times \text{``}{173\ldots}\text{''} \times 100\;(= 6 \times \text{``}{8660\ldots}\text{''} = 51961.52\ldots)\)
M1 eg \(\text{``}{51961.52}\text{''} - \pi \times 100^2 = 100^2 n\) M1 Correct answer scores full marks (unless from obvious incorrect working) A1
Notes M1: eg using radius = 100 ANY VALUE CAN BE USED CONSISTENTLY FOR AWARD OF MARKS
M1: Expression for side of triangle
M1: a correct expression for the area of the triangle
M1: a correct equation for the area of the shaded parts of the diagram.
A1: 2.05 – 2.06
Higher November 2024 Paper 1 Q12
12 \(ABC\) is a right-angled triangle.
Diagram NOT accurately drawn
\(AB = 4250\) m angle \(BAD = 47^\circ\) angle \(BCA = 24^\circ\)
Work out the length of \(DC\)
(4)
Mark scheme
Mark scheme Scheme Marks eg
\(\tan 47 = \dfrac{(BD)}{4250}\) or \(\tan 24 = \dfrac{4250}{(BC)}\) or \(\tan(47 + \text{“}19\text{”}) = \dfrac{(BC)}{4250}\) or \(\dfrac{(BD)}{\sin 47} = \dfrac{4250}{\sin 43}\)
or \((AD =)\;\dfrac{4250}{\cos 47}\;(= 6231.686\ldots)\) or \((AD =)\;\dfrac{4250}{\sin 43}\;(= 6231.686\ldots)\)
or \((AC =)\;\dfrac{4250}{\sin 24}\;(= 10\,449.021\ldots)\) or \((AC =)\;\dfrac{4250}{\cos 66}\;(= 10\,449.021\ldots)\)
M1 eg
\((BD =)\;4250\tan 47\;(= 4557.567\ldots)\) or \((BC =)\;\dfrac{4250}{\tan 24}\;(= 9545.656\ldots)\)
or \((BD =)\;\dfrac{4250}{\sin 43} \times \sin 47\;(= 4557.567\ldots)\) or \(\dfrac{(DC)}{\sin\text{“}19\text{”}} = \dfrac{\text{“}10449\ldots\ldots\text{”}}{\sin\text{“}137\text{”}}\)
or \(\dfrac{(DC)}{\sin\text{“}19\text{”}} = \dfrac{\text{“}6231.686\text{”}}{\sin 24}\) or \((BC =)\;4250 \times \tan(47 + \text{“}19\text{”})\;(= 9545.656\ldots)\)
or \((DC^2 =)\;\text{“}6231\text{”}^2 + \text{“}10449\text{”}^2 - 2 \times \text{“}6231\text{”} \times \text{“}10449\text{”} \times \cos 19\)
M1 eg
“9545.656” – “4557.567” (= 4988.089)
or \((DC =)\;\dfrac{\text{“}6231.686\text{”}}{\sin 24} \times \sin 19\) or \((DC =)\;\dfrac{\text{“}10449\ldots\ldots\text{”}}{\sin\text{“}137\text{”}} \times \sin\text{“}19\text{”}\)
or \((DC =)\sqrt{\text{“}6231\text{”}^2 + \text{“}10449\text{”}^2 - 2 \times \text{“}6231\text{”} \times \text{“}10449\text{”} \times \cos 19}\)
M1 Correct answer scores full marks (unless from obvious incorrect working) A1 (4) (4 marks)
Notes M1: for a complete method
A1: allow in the range 4932 – 4990
Higher November 2024 Paper 2 Q3
3 The diagram shows triangle \(PQR\)
Diagram NOT accurately drawn
Work out the value of \(x\)
(3)
Mark scheme
Mark scheme Scheme Marks \(\cos 43 = \dfrac{x}{8.6}\) or
\(\tan 43 = \dfrac{8.6\sin 43}{x}\) or
\(\sin(90 - 43) = \dfrac{x}{8.6}\) or
\(\dfrac{x}{\sin(90 - 43)} = \dfrac{8.6}{\sin 90}\) or
\((x^2 =)\;8.6^2 - (8.6\sin 43)^2\) or \((x^2 =)\;8.6^2 - 5.8(65\ldots)^2\)
M1 \((x =)\;8.6\cos 43\) or
\((x =)\;\dfrac{8.6\sin 43}{\tan 43}\left(= \dfrac{\text{“}5.8(65\ldots)\text{”}}{\tan 43}\right)\) or
\((x =)\;8.6\sin(90 - 43)\) or
\((x =)\;\dfrac{8.6\sin 47}{\sin 90}\) or
\((x =)\sqrt{8.6^2 - \text{“}5.8(65\ldots)\text{”}^2}\)
M1 Correct answer scores full marks (unless from obvious incorrect working) A1 (3) (3 marks)
Notes M1: a correct trig statement for \(x\) or \(QR\) or a correct Pythagoras statement for \(x^2\)
M1: a fully correct calculation to find \(x\)
A1: awrt 6.3 seen even if then rounded incorrectly
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