C2 June 2013 Q5

EdexcelOld spec9 marksRadians

Figure 2: garden plan, triangle ABE with AB = 23 m and angle ABE = 0.64 rad, joined to a sector BCDE of radius 12 m with centre B, A, B and C on a straight line — Figure 2

Figure 2 shows a plan view of a garden.
The plan of the garden \(ABCDEA\) consists of a triangle \(ABE\) joined to a sector \(BCDE\) of a circle with radius 12 m and centre \(B\).
The points \(A\), \(B\) and \(C\) lie on a straight line with \(AB = 23\) m and \(BC = 12\) m.

Given that the size of angle \(ABE\) is exactly 0.64 radians, find

(a) the area of the garden, giving your answer in m², to 1 decimal place, (4)

(b) the perimeter of the garden, giving your answer in metres, to 1 decimal place. (5)

Mark scheme (a)

Scheme	Marks
Mark (a) and (b) together.
Usually answered in radians: Uses either \(\dfrac{1}{2}ab\sin(\text{angle})\) or \(\dfrac{1}{2}(12)^2(\text{angle})\) or both	M1
Area \(= \dfrac{1}{2}(23)(12)\sin 0.64\) or \(\dfrac{1}{2}(12)^2(\pi - 0.64)\) \(\{= 82.41297091\ldots\) or \(180.1146711\ldots\}\)	A1
Area \(= \dfrac{1}{2}(23)(12)\sin 0.64 + \dfrac{1}{2}(12)^2(\pi - 0.64)\) \(\{= 82.41297091\ldots + 180.1146711\ldots\}\)	A1
\(\{\text{Area} = 262.527642\ldots\} =\) awrt 262.5 (m²) or 262.4(m²) or 262.6 (m²)	A1
	(4)

Notes

M1: uses either area of triangle formula as given in mark scheme, or area of sector or both (may be implied by answer)

A1: one correct area expression (with correct angle used) \(\dfrac{1}{2}(23)(12)\sin 0.64\) or \(\dfrac{1}{2}(12)^2(\pi - 0.64)\) or see awrt 82.4 or awrt 180 (180 may be split as 226.2(semicircle) minus 46.1(small sector))

A1: two correct area expressions (with correct angles) added together (allow 2.5 as implying \(\pi - 0.64\)) or see awrt 82.4 + awrt 180 ( or 226 - 46 )

A1: answers which round to 262.5 or 262.4 or 262.6

Degrees (a)

Uses either \(\dfrac{1}{2}ab\sin(\text{angle})\) or \(\dfrac{\text{angle in degrees}}{360} \times \pi(12)^2\) or both for M1

Area \(= \dfrac{1}{2}(23)(12)\sin 36.7\) or \(\dfrac{(180 - 36.7)}{360} \times \pi(12)^2\ \{= awrt\ 82.4\ldots\) or \(180\}\) A1

Area \(= \dfrac{1}{2}(23)(12)\sin 36.7 + \dfrac{(180 - 36.7)}{360} \times \pi(12)^2\ \ \{= awrt\ 82.4\ldots + 180\}\) A1

Final mark as before

Mark scheme (b)

Scheme	Marks
\(CDE = 12 \times (angle),\ = 12(\pi - 0.64)\ \{\Rightarrow CDE = 30.01911\ldots\}\)	M1, A1
\(AE^2 = 23^2 + 12^2 - 2(23)(12)\cos(0.64) \Rightarrow AE^2 =\) or \(AE =\) \(\{AE = 15.17376\ldots\}\)	M1
Perimeter \(= 23 + 12 + 15.17376\ldots + 30.01911\ldots\)	M1
\(= 80.19287\ldots =\) awrt 80.2 (m)	A1
	(5)
	[9]

Notes

1^st M1 for attempt to use \(s = r\,\theta\) (any angle)

1^st A1 for \(\pi - 0.64\) in the formula (or 2.5)

2^nd M1: Uses correct cosine rule to obtain \(AE\) or \(AE^2\) (this may appear in part (a))

3^rd M1(independent): Perimeter \(= 23 + 12 +\) their \(AE\) + their \(CDE\)

2^nd A1: awrt 80.2 (ignore units – even incorrect units)

Degrees (b)

\(CDE = \dfrac{\text{Angle in degrees}}{360} \times 24\pi,\ = \dfrac{180 - 36.7}{360} \times 24\pi\ \{\Rightarrow CDE = 30.01268\ldots\}\) M1, A1

Final three marks as before