Higher June 2024 Paper 6 Q5
5 The diagram shows a regular pentagon made using ten congruent right-angled triangles.
The length of one side of the pentagon is 12 cm.
Not to scale
(a) Show that the area of the pentagon is \(247.75\,\text{cm}^2\), correct to 2 decimal places. [6]
(b) The regular pentagon is the base of a pyramid.
The pyramid has volume \(450\,\text{cm}^3\).
The perpendicular height of the pyramid is \(h\) cm.
Calculate the value of \(h\).
[The volume of a pyramid is \(\frac{1}{3} \times \text{area of base} \times \text{perpendicular height}\).] [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Allocate marks similarly for other methods such as five triangles using an angle of 72. If in doubt, consult TL. There must be evidence of angle or trig work to score any marks e.g. working back from 247.75 to \(h = 8.258\)…is likely to score 0 or B1 | |||
| [angle =] 36 or 54 | B1 | in correct place if only shown on diagram | Do not award 36 or 54 if calculated as an area |
| [\(h =\)] \(\frac{6}{\tan 36}\) or 6 × tan54 or \(\frac{6\sin 54}{\sin 36}\) may be implied by 8.258 to 8.259 or 8.26 following M1 but not from area | M2 | M1 for \(\tan 36 = \frac{6}{h}\) or \(\tan 54 = \frac{h}{6}\) or \(\frac{6}{\sin 36} = \frac{h}{\sin 54}\) or \(\frac{\sin 36}{6} = \frac{\sin 54}{h}\) | Accept other notation used for ‘\(h\)’ |
| \(10 \times \frac{1}{2} \times 6 \times \textit{their } h\) oe | M2dep | M1 for \(\frac{1}{2} \times 6 \times \textit{their } h\) oe may be implied by 24.774 to 24.775 | Their \(h\) dep on previous M2 or M1 Accept correct use of \(\frac{1}{2}ab\sin C\) |
| 247.748 to 247.749 | A1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 5.45 or 5.449 to 5.450 nfww | 3 | M2 for \(h = \frac{450 \times 3}{247.75}\) oe or M1 for \(\frac{1}{3}h \times 247.75 = 450\) oe | 247.75 may be their more accurate 247.748 to 247.749 or 247.7, 247.8 or 247.74 from (a) Use of incorrect formula is not MR |