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Higher June 2025 Paper 1R Q20
20 The diagram shows cuboid \(ABCDEFGH\)
Diagram NOT accurately drawn
\(AB = 9\) cm \(AF = 7\) cm \(FC = 18\) cm
Calculate the length of \(BC\)
(3)
Mark scheme
Mark scheme Scheme Marks eg
\((AC^2 =)\;18^2 - 7^2 (= 275)\) or
\((AC =)\sqrt{18^2 - 7^2}\left(= \sqrt{275} \text{ or } 5\sqrt{11} \text{ or } 16.5(831\ldots)\right)\)
or \((FB^2 =)\;9^2 + 7^2 (= 130)\) or
\((FB =)\sqrt{9^2 + 7^2}\left(= \sqrt{130} \text{ or } 11.4(017\ldots)\right)\) or
\((GC^2 =)\;18^2 - 9^2 (= 243)\) or
\((GC =)\sqrt{18^2 - 9^2}\left(= \sqrt{243} \text{ or } 9\sqrt{3} \text{ or } 15.5(884\ldots)\right)\)
or
\(18^2 = (BC)^2 + 7^2 + 9^2\) oe
M1 eg
\(\text{``}{275}\text{''} - 9^2 (= 194)\) or \(\text{``}{16.5\ldots}\text{''}^2 - 9^2 (= 194)\) or
\(18^2 - \text{``}{130}\text{''} (= 194)\) or \(18^2 - \text{``}{11.4\ldots}\text{''}^2 (= 194)\)
\(\text{``}{243}\text{''} - 7^2 (= 194)\) or \(\text{``}{15.5\ldots}\text{''}^2 - 7^2 (= 194)\) or
\(18^2 - 7^2 - 9^2 (= 194)\)
or \(\angle FCB = \sin^{-1}\left(\dfrac{\text{``}{11.4}\text{''}}{18}\right)(= 39.3(036\ldots))\) and
\(\cos\text{``}{39.3}\text{''} = \dfrac{(BC)}{18}\) or \(\tan\text{``}{39.3}\text{''} = \dfrac{\text{``}{11.4}\text{''}}{(BC)}\) oe
M1 Correct answer scores full marks (unless from obvious incorrect working) A1 (3) (3 marks)
Notes M1: for method to find \(AC^2\) or \(AC\) or \(FB^2\) or \(FB\) or \(GC^2\) or \(GC\) or
for a correct equation using \(BC^2\) and 18 and 7 and 9
other longer ways to find \(AC\), \(FB\), \(GC\) may be used but must be a complete method eg
\(\angle FCA = \sin^{-1}\left(\dfrac{7}{18}\right)(= 22.88\ldots)\) and
\(AC = \dfrac{7}{\tan\text{``}{22.88\ldots}\text{''}}\)
M1: for complete method to find \(BC^2\)
A1: accept 13.8 to 14
Higher June 2025 Paper 2R Q11
11 The diagram shows two triangles, \(ADE\) and \(CDB\)
Diagram NOT accurately drawn
\(ABD\) is a straight line.
\(AE = 28\) cm \(ED = 45\) cm \(AB = 21\) cm \(CD = 35\) cm
angle \(AED\) = angle \(CBD = 90^\circ\)
Work out the area of triangle \(CDB\)
(5)
Mark scheme
Mark scheme Scheme Marks \((AD^2 =)\;28^2 + 45^2\;(= 784 + 2025 = 2809)\)
or \((EDA =)\;\tan^{-1}\left(\dfrac{28}{45}\right)\;(= 31.8(9\ldots))\)
or \((EAD =)\;\tan^{-1}\left(\dfrac{45}{28}\right)\;(= 58.1(09\ldots))\)
M1 eg
\((AD =)\sqrt{28^2 + 45^2}\;\left(= \sqrt{784 + 2025} = \sqrt{2809} = 53\right)\) oe
or \((AD =)\;\dfrac{28}{\sin 31.8(9\ldots)}\;(= 53\ldots)\) or \((AD =)\;\dfrac{45}{\cos 31.8(9\ldots)}\;(= 53\ldots)\) oe
or \((AD =)\;\dfrac{45}{\sin 58.1\ldots}\;(= 53\ldots)\) or \((AD =)\;\dfrac{28}{\cos 58.1\ldots}\;(= 53\ldots)\) oe
M1 eg
\((BC =)\sqrt{35^2 - (\text{``}{53}\text{''} - 21)^2}\;\left(= \sqrt{1225 - 1024} = \sqrt{201} = 14.1(7\ldots)\right)\)
or \((BDC =)\;\cos^{-1}\left(\dfrac{\text{``}{53}\text{''} - 21}{35}\right)\;(= 23.8(9\ldots))\)
or \((BCD =)\;\sin^{-1}\left(\dfrac{\text{``}{53}\text{''} - 21}{35}\right)\;(= 66.1\ldots)\)
M1 eg
\(\dfrac{1}{2} \times \text{``}{14.1(7\ldots)}\text{''} \times (\text{``}{53}\text{''} - 21)\;(= 16\sqrt{201})\)
or \(\dfrac{1}{2} \times 35 \times (\text{``}{53}\text{''} - 21) \times \sin(\text{``}{23.8(9\ldots)}\text{''})\)
or \(\dfrac{1}{2} \times 35 \times \text{``}{14.1(7\ldots)}\text{''} \times \sin(\text{``}{66.1\ldots}\text{''})\)
M1 Correct answer scores full marks (unless from obvious incorrect working) A1 (5) (5 marks)
Notes M1: for a correct method to find \(AD^2\) or angle \(EDA\) or angle \(EAD\)
M1: for a correct method to find \(AD\)
M1: for correct method to find \(BC\) or angle \(BDC\) or angle \(BCD\)
M1: for a correct method to find the area of triangle \(CDB\)
A1: awrt 227
accept \(16\sqrt{201}\)
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 November 2024 Paper 1 Q10
10 Here are two similar triangles.
Diagrams NOT accurately drawn
Work out the value of \(x\)
(5)
Mark scheme
Mark scheme Scheme Marks eg \(51^2 = (DE)^2 + 24^2\) oe or \(2601 = (DE)^2 + 576\) oe
or \((DE^2 =)\;51^2 - 24^2\;(= 2025)\) oe or \((DE^2 =)\;2601 - 576\;(= 2025)\) oe or
\(\cos(DFE) = \dfrac{24}{51}\) or \(\sin(DEF) = \dfrac{24}{51}\)
M1 \((DE =)\sqrt{51^2 - 24^2}\;\left(= \sqrt{2025} = 45\right)\) or \((DE =)\sqrt{2601 - 576}\;\left(= \sqrt{2025} = 45\right)\)
or \((DFE =)\cos^{-1}\left(\dfrac{24}{51}\right)(= 61.9\ldots)\) or \((DEF =)\sin^{-1}\left(\dfrac{24}{51}\right)(= 28.0\ldots)\)
M1 \(\dfrac{\textit{their } DE}{7.5}(= 6)\) oe or \(\dfrac{7.5}{\textit{their } DE}\left(= \dfrac{1}{6}\right)\) or \(\dfrac{x}{24} = \dfrac{7.5}{\textit{their } DE}\) oe
or \(\tan\text{“}\textit{their } 61.9\ldots\text{”} = \dfrac{7.5}{(x)}\) or \(\tan\text{“}\textit{their } 28.0\ldots\text{”} = \dfrac{(x)}{7.5}\) or
\(\dfrac{(x)}{\sin(\textit{their } 28.0)} = \dfrac{7.5}{\sin(\textit{their } 61.9)}\)
NB Their \(ED\) or their 61.9 or their 28.0 must be clearly identified
Their 61.9.. or their 28.0.. cannot be used as lengths of the triangle
Their 45 cannot be used as an angle of the triangle
M1 \(24 \div\) “6” oe or \(24 \times \text{“}\dfrac{1}{6}\text{”}\) or 24 × “1.67” or \((x =)\;\dfrac{7.5}{\textit{their } ED} \times 24\)
or \((x =)\;\dfrac{7.5}{\tan\text{“}\textit{their } 61.9\ldots\text{”}}\) oe or \((x =)\;7.5 \times \tan\text{“}\textit{their } 28.0\ldots\text{”}\) oe
or \((x =)\;\dfrac{7.5}{\sin(\textit{their } 61.9)} \times \sin(\textit{their } 28.0)\) oe
or \(51 \times \text{“}\dfrac{1}{6}\text{”}\;(= 8.5)\) and \((x =)\sqrt{8.5^2 - 7.5^2}\;\left(= \sqrt{16}\right)\)
M1 Working required A1 (5) (5 marks)
Notes M1: for applying Pythagoras theorem correctly
M1: for square rooting
M1: for a correct method to find the scale factor
Allow correct use of sine rule/cosine rule/Pythagoras theorem
Allow 0.17 or better for \(\dfrac{1}{6}\)
Special case Allow \((DE =)\sqrt{51^2 + 24^2}\;\left(= \sqrt{3177} = 3\sqrt{353} = 56.3\ldots\right)\) for “45” for this mark
M1: dep on previous M1 for a correct method to find \(x\) or for finding \(BC\) and using Pythagoras theorem to find \(x\)
Allow \(24 \times \text{“}\dfrac{7.5}{56(.3\ldots)}\text{”}\) or \(24 \div \text{“}\dfrac{56(.3\ldots)}{7.5}\text{”}\) for scale factor for this mark
A1: dep on M2
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