Question Bank › IGCSE Shape & Space › 2D Area & Perimeter
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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 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 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
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