4MA0 Higher June 2017 Paper 3R Q14
14 \(ABCDE\) is a regular pentagon with sides of length 10 cm.
Diagram NOT accurately drawn
Calculate the area of triangle \(ACD\).
Give your answer correct to 3 significant figures.
(6)
Scheme 1 (interior angle and angle ADC and an angle in triangle ADC)
Let \(X\) be midpoint of \(DC\)
| Scheme | Marks |
|---|---|
| (Angle \(DEA =\)) \(\dfrac{(5 - 2) \times 180}{5} = 108\) | M1 |
| Angle \(EDA\) or \(EAD = (180 - 108) \div 2 = 36\) | M1 |
Angle \(ADC\) or \(ACD = 108 - 36 = 72\) or Angle \(DAC = 108 - 2 \times 36\) (=36) or Angle \(DAC = \cos^{-1}\left(\dfrac{\text{“}16.18\text{”}^2 + \text{“}16.18\text{”}^2 - 10^2}{2 \times \text{“}16.18\text{”} \times \text{“}16.18\text{”}}\right)\) Angle \(DAX\) or \(CAX = (108 - 2 \times 36) \div 2\) (=18) | M1 |
E.g. \(AX = 5 \times \tan 72\) (=15.38…) or \(AX = 5 \div \tan 18\) (=15.38…) or \(AX = 5 \times \tan ADC\) or \(AX = 5 \div \tan DAX\) or \(AD^2 = 10^2 + 10^2 - 2 \times 10 \times 10 \cos 108\) (=261.8) or \(AD = \sqrt{10^2 + 10^2 - 2 \times 10 \times 10 \times \cos 108}\) (=16.18) or \(AD = \dfrac{10 \times \sin 72}{\sin 36}\) (=16.18) | M1 |
E.g. Area = \(\dfrac{1}{2} \times 10 \times\) “15.38…” oe or 0.5 × 10 × “16.18” × sin72 or 0.5 × 10 × “16.18” × “16.18” × sin36 or 0.5 × 10 × “16.18” × sin\(ADC\) or 0.5 × “16.18” × “16.18” × sin\(DAX\) | M1 |
| 76.9 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: or for 108 seen as an interior angle
M1: Angles may be seen on diagram
M1: or for \(AD = \dfrac{10}{\sin 36} \times \sin 108\) (=16.18…) or
\(AD = \dfrac{10}{\sin 36} \times \sin 72\)
Allow 16 or 16.2 for \(AD\) throughout
NB: Allow the value on the diagram for angle \(ADC\) or \(DAX\) if used in an otherwise correct trig statement
M1: dep on previous M1
NB: Allow the value on the diagram for angle \(ADC\) or \(DAX\) if used in an otherwise correct area statement
A1: for answer in the range 76.5 – 77
SC: B4 for an answer in the range 53 – 53.5
Scheme 2 (only interior angle needed)
Let \(X\) be midpoint of \(DC\)
| Scheme | Marks |
|---|---|
| (Angle \(DEA =\)) \(\dfrac{(5 - 2) \times 180}{5} = 108\) | M1 |
\(AD^2 = 10^2 + 10^2 - 2 \times 10 \times 10 \cos 108\) (=261.8) or \(AD = \sqrt{10^2 + 10^2 - 2 \times 10 \times 10 \times \cos 108}\) (=16.18) | M1 |
| \(AX^2 = \text{“}261.8\text{”} - 5^2\) (= 236.8) or \(AX^2 = \text{“}16.18\text{”}^2 - 5^2\) (= 236.8) | M1 |
| \(\sqrt{\text{“}236.8\text{”}}\) (=15.38..) | M1 |
| 0.5 × 10 × “15.38” | M1 |
| 76.9 | A1 |
Notes
M1: or for 108 seen as an interior angle
M1: Allow 16 or 16.2 for \(AD\) throughout
A1: for answer in the range 76.5 – 77
SC: B4 for an answer in the range 53 – 53.5