Foundation June 2024 Paper 2R Q26

\(\sin 30 = \dfrac{(x)}{5}\) or \(\dfrac{(x)}{\sin 30} = \dfrac{5}{\sin 90}\) oe

where \(x\) = height of trapezium

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or

\(5\cos30\) \(\left(= \dfrac{5\sqrt3}{2} = 4.33...\right)\) and (\(x^2 =\)) \(5^2 - \text{``}{5\cos30}\text{''}^2\) (= 6.25)

(\(x =\)) \(5 \sin 30\) (= 2.5) oe or (\(x =\)) \(\dfrac{5}{\sin 90} \times \sin 30\) (= 2.5) oe

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or

(\(x =\)) \(\sqrt{5^2 - \text{``}{5\cos30}\text{''}^2}\) (= 2.5)

\(\tfrac12\times(11+23)\times\text{``}{2.5}\text{''}\) (= 42.5) oe

or \(\left(\tfrac12\times\text{``}{2.5}\text{''}\times(23-11)\right) + (11\times\text{``}{2.5}\text{''})\) (= 42.5) oe

or \(\left(\tfrac12\times\text{``}{2.5}\text{''}\times(23-11-\text{``}{4.3}\text{''})\right) + (11\times\text{``}{2.5}\text{''}) + \left(\tfrac12\times\text{``}{2.5}\text{''}\times\text{``}{4.3}\text{''}\right)\) (= 42.5) oe

or \((11\times\text{``}{2.5}\text{''}) + \left(\tfrac12\times5\times(23-11)\times\sin 30\right)\) (= 42.5) oe

or \((23\times\text{``}{2.5}\text{''}) - \left(\tfrac12\times\text{``}{2.5}\text{''}\times(23-11-\text{``}{4.3}\text{''})\right) - \left(\tfrac12\times\text{``}{2.5}\text{''}\times\text{``}{4.3}\text{''}\right)\) (= 42.5) oe

or \((23\times(\text{``}{2.5}\text{''}+4.7)) - \left(\tfrac12\times\text{``}{2.5}\text{''}\times(23-11-\text{``}{4.3}\text{''})\right) - \left(\tfrac12\times\text{``}{2.5}\text{''}\times\text{``}{4.3}\text{''}\right)\) oe