Foundation January 2021 Paper 1R Q25
25 The diagram shows one face of a wall.
This face is in the shape of a pentagon with exactly one line of symmetry.
Diagram NOT accurately drawn
Omondi is going to paint this face of the wall once.
He has to buy all the paint that he needs to use.
The paint in each tin of paint Omondi is going to buy will cover 16 m2 of the face of the wall.
Work out the least number of tins of paint Omondi will need to buy.
Show your working clearly.
(5)
| Scheme | Marks |
|---|---|
\(7^2 - (10 \div 2)^2\) (= 24) or \(\dfrac{\sin\left(\frac{1}{2}x\right)}{5} = \dfrac{\sin 90}{7}\) oe or \(\cos x = \dfrac{7^2 + 7^2 - 10^2}{2 \times 7 \times 7}\) oe or \(\sin\left(\dfrac{1}{2}x\right) = \dfrac{5}{7}\) oe or \(\cos y = \dfrac{5}{7}\) oe | M1 |
\(\sqrt{7^2 - (10 \div 2)^2}\) (\(= \sqrt{24} = 2\sqrt{6}\) = 4.898...) or (\(x\) =) \(2 \times \sin^{-1}\left(\dfrac{5 \times \sin 90}{7}\right)\) (= 91.169…) oe or (\(x\) =) \(2 \times \sin^{-1}\left(\dfrac{5}{7}\right)\) (= 91.169…) oe or (\(x\) =) \(\cos^{-1}\left(\dfrac{7^2 + 7^2 - 10^2}{2 \times 7 \times 7}\right)\) (= 91.169…) oe or (\(x\) =) \(2\left(90 - \cos^{-1}\left(\dfrac{5}{7}\right)\right)\) (= 2(90 – 44.415)... = 91.169...) Allow 5 from correct working | M1 |
E.g. \(6 \times 10 + \dfrac{(10 \div 2) \times \text{‘}\sqrt{24}\text{’}}{2} \times 2\) (\(= 60 + 10\sqrt{6}\) = 84.494...) or \(5 \times (6 + 6 + \text{‘}\sqrt{24}\text{’})\) (\(= 60 + 10\sqrt{6}\) = 84.494...) or \(\left(\dfrac{1}{2} \times 7 \times 7 \times \sin \text{‘91.169...’} + 10 \times 6\right)\) (\(= 60 + 10\sqrt{6}\) = 84.494...) | M1 |
E.g. ‘84.494’ ÷ 16 (= 5.28...) or \((60 + 10\sqrt{6}) \div 16\) (= 5.28...) | M1 |
| Working required Answer: 6 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: or use of sine rule or cosine rule to find angle (\(x\)) of the apex or angle \(y\) \(\left(= 90 - \dfrac{1}{2}x\right)\)
M1: for complete method to find height of triangle or the angle (\(x\)) of the apex
\(\cos^{-1}\left(\dfrac{5}{7}\right)\) (= 44.415...) and
\(5 \times \tan\) ‘44.415...’ (= 4.898...) or \(7 \times \sin\) ‘44.415...’ (= 4.898...)
or
\(\sin^{-1}\left(\dfrac{5}{7}\right)\) (= 45.584...) and
\(\dfrac{5}{\tan \text{‘45.584...’}}\) (= 4.898...) or \(7 \times \cos\) ‘45.584...’ (= 4.898...)
M1: for method to find the total area of the pentagon allow answers in the range 84.49 – 85
M1: for method to find the number of tins required using their area
A1: dep on at least M2