Numeracy Higher November 2020 Unit 2 Q12
12. A water trough has a uniform semi-circular cross-section with a radius of 30·9 cm.
It has a length of 600 cm.
Water has been poured into the trough.
The diagram below shows the cross-section of the trough.
The water level is shown by the line \(AB\).
\(O\) is the centre of the semi-circle and \(A\hat{O}B\) = 150º.
Diagram not drawn to scale
(a) Calculate the area of the triangle \(AOB\). [2]
(b) Calculate how many more litres of water can be poured into the trough before the trough is full. [7]
Volume of water that can be added = ………………………… litres
| Scheme | Marks | Comments |
|---|---|---|
| (Area of triangle AOB =) \(\dfrac{1}{2}\) × 30.92 × sin150(º) or equivalent full method | M1 | |
= 238.7(025) or 239 (cm2) | A1 |
| Scheme | Marks | Comments |
|---|---|---|
| Adding two equal sectors to the area of the triangle | S1 | |
| (Area of 2 sectors =) (2 ×) \(\dfrac{(180 - 150)/2}{360}\) × π × 30.92 | M1 | May be embedded in a volume calculation i.e.× 600 |
= 249.8 to 250(.00…) (cm2) | A1 | Or 31827π/400 (cm2) May be implied by 488.5 to 489 (cm2) (total area) May be implied by volume of 149800 to 150001 (cm3) |
| (Number of litres that can be added =) (249.9(686…) + 238.7(025)) × 600 (÷1000) or (488.5 to 489) × 600 (÷1000) | M2 | FT ‘their 238.7(025)’ and ‘their 249.9(686…)’ provided M1 previously awarded M1 (may be embedded within incorrect work) for
M1 (may be embedded within incorrect work) for
|
= 293 to 293.4 (litres) | A2 | FT from M2 only A1 for 293 000 to 293 400 (cm3) |
Alternative
Alternative method:
| Scheme | Marks | Comments |
|---|---|---|
| Area of semicircle subtract (area of the large sector – area of triangle) | S1 | |
| (Area of large sector =) \(\dfrac{150}{360}\) × π × 30.92 or equivalent | M1 | May be embedded in a volume calculation i.e.× 600 |
= 1249 to 1250 (cm2) | A1 | Or 31827π/80 (cm2) 749 400 to 750 000 (cm3) if volumes considered |
| (Number of litres that can be added =) (π × 30.92 ÷ 2 – (1249.8(…) – 238.7(025))) × 600 (÷1000) or ((1499 to 1500.01) – (1010…to 1011.3)) × 600 (÷1000) | M2 | FT ‘their 238.7(025)’ and ‘their 1249.8(…)’ provided M1 previously awarded M1 (may be embedded within incorrect work) for
M1 (may be embedded within incorrect work) for
|
= 293 to 293.4 (litres) | A2 | FT from M2 only A1 for 293 000 to 293 400 (cm3) |