C2 June 2008 Q7

EdexcelOld spec12 marksRadians

Figure 1: sector ABC with centre A, AB = 7 cm and angle BAC = 0.8 rad; D on AC, and the region R bounded by DB, DC and the arc BC hatched — Figure 1

Figure 1 shows \(ABC\), a sector of a circle with centre \(A\) and radius 7 cm.

Given that the size of \(\angle BAC\) is exactly 0.8 radians, find

(a) the length of the arc \(BC\), (2)

(b) the area of the sector \(ABC\). (2)

The point \(D\) is the mid-point of \(AC\). The region \(R\), shown shaded in Figure 1, is bounded by \(CD\), \(DB\) and the arc \(BC\).

Find

(d) the area of \(R\), giving your answer to 3 significant figures. (4)

Mark scheme (a)

Scheme	Marks
\(r\theta = 7\times 0.8 = 5.6\) (cm)	M1 A1
	(2)

Notes

Units (cm or \(\mathrm{cm}^2\)) are not required in any of the answers.
(a) and (b): Correct answers without working score both marks.

M: Use of \(r\theta\) (with \(\theta\) in radians), or equivalent (could be working in degrees with a correct degrees formula).

Mark scheme (b)

Scheme	Marks
\(\dfrac{1}{2}r^2\theta = \dfrac{1}{2}\times 7^2\times 0.8 = 19.6\ (\mathrm{cm}^2)\)	M1 A1
	(2)

Notes

(a) and (b): Correct answers without working score both marks.

M: Use of \(\dfrac{1}{2}r^2\theta\) (with \(\theta\) in radians), or equivalent (could be working in degrees with a correct degrees formula).

Mark scheme (c)

Scheme	Marks
\(BD^2 = 7^2 + (\text{their } AD)^2 - (2\times 7\times(\text{their } AD)\times\cos 0.8)\)	M1
\(BD^2 = 7^2 + 3.5^2 - (2\times 7\times 3.5\times\cos 0.8)\) (or awrt \(46^\circ\) for the angle) \((BD = 5.21)\)	A1
Perimeter = (their \(DC\)) + “5.6” + “5.21” = 14.3 (cm) (Accept awrt)	M1 A1
	(4)

Notes

1^st M: Use of the (correct) cosine rule formula to find \(BD^2\) or \(BD\).
Any other methods need to be complete methods to find \(BD^2\) or \(BD\).
2^nd M: Adding their \(DC\) to their arc \(BC\) and their \(BD\).

Beware: If 0.8 is used, but calculator is in degree mode, this can still earn M1 A1 (for the required expression), but this gives \(BD = 3.50\ldots\) so the perimeter may appear as 3.5 + 5.6 + 3.5 (earning M1 A0).

Mark scheme (d)

Scheme	Marks
\(\Delta ABD = \dfrac{1}{2}\times 7\times(\text{their } AD)\times\sin 0.8\) (or awrt \(46^\circ\) for the angle) (ft their \(AD\)) \((= 8.78\ldots)\) (If the correct formula \(\dfrac{1}{2}ab\sin C\) is quoted the use of any two of the sides of \(\Delta ABD\) as \(a\) and \(b\) scores the M mark).	M1 A1ft
Area = “19.6” – “8.78…” \(= 10.8\ (\mathrm{cm}^2)\) (Accept awrt)	M1 A1
	(4)
	12

Notes

1^st M: Use of the (correct) area formula to find \(\Delta ABD\).
Any other methods need to be complete methods to find \(\Delta ABD\).
2^nd M: Subtracting their \(\Delta ABD\) from their sector \(ABC\).

Using segment formula \(\dfrac{1}{2}r^2(\theta - \sin\theta)\) scores no marks in part (d).