Higher June 2024 Paper 2R Q10

\(\sin 30 = \dfrac{(x)}{5}\)

or

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

 

where \(x\) = height of trapezium

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or

\(5\cos 30\left(= \dfrac{5\sqrt{3}}{2} = 4.33...\right)\)

\((x^2 =)\, 5^2 - \text{``}{5\cos 30}\text{''}^2 \;(= 6.25)\)

\((x =)\, 5\sin 30 \;(= 2.5)\) oe

or

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

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or

\((x =)\sqrt{5^2 - \text{``}{5\cos 30}\text{''}^2}\;(= 2.5)\)

\(\dfrac{1}{2} \times (11 + 23) \times \text{``}{2.5}\text{''} (= 42.5)\) oe or \(\left(\dfrac{1}{2} \times \text{``}{2.5}\text{''} \times (23 - 11)\right) + \left(11 \times \text{``}{2.5}\text{''}\right) (= 42.5)\) oe or

\(\left(\dfrac{1}{2} \times \text{``}{2.5}\text{''} \times \left(23 - 11 - \text{``}{4.3}\text{''}\right)\right) + \left(11 \times \text{``}{2.5}\text{''}\right) + \left(\dfrac{1}{2} \times \text{``}{2.5}\text{''} \times \text{``}{4.3}\text{''}\right) (= 42.5)\) oe or

\(\left(11 \times \text{``}{2.5}\text{''}\right) + \left(\dfrac{1}{2} \times 5 \times (23 - 11) \times \sin 30\right) (= 42.5)\) oe or

\(\left(23 \times \text{``}{2.5}\text{''}\right) - \left(\dfrac{1}{2} \times \text{``}{2.5}\text{''} \times \left(23 - 11 - \text{``}{4.3}\text{''}\right)\right) - \left(\dfrac{1}{2} \times \text{``}{2.5}\text{''} \times \text{``}{4.3}\text{''}\right) (= 42.5)\) oe or

\(\left(23 \times \left(\text{``}{2.5}\text{''} + 4.7\right)\right) - \left(\dfrac{1}{2} \times \text{``}{2.5}\text{''} \times \left(23 - 11 - \text{``}{4.3}\text{''}\right)\right) - \left(\dfrac{1}{2} \times \text{``}{2.5}\text{''} \times \text{``}{4.3}\text{''}\right)\) oe