C2 January 2007 Q9

EdexcelOld spec11 marksRadians

Figure 2: sector PQRS with centre Q, QP = QR = 6 m, arc PSR, and the chord PR of length 6√3 m shown dashed — Figure 2

Figure 2 shows a plan of a patio. The patio \(PQRS\) is in the shape of a sector of a circle with centre \(Q\) and radius 6 m.

Given that the length of the straight line \(PR\) is \(6\sqrt{3}\) m,

(a) find the exact size of angle \(PQR\) in radians. (3)

(b) Show that the area of the patio \(PQRS\) is \(12\pi\ \mathrm{m}^2\). (2)

(d) Find, in \(\mathrm{m}^2\) to 1 decimal place, the area of the segment \(PRS\). (2)

(e) Find, in m to 1 decimal place, the perimeter of the patio \(PQRS\). (2)

Mark scheme (a)

Scheme	Marks
\(\cos PQR = \dfrac{6^2 + 6^2 - (6\sqrt{3})^2}{2\times 6\times 6}\) \(\left\{= -\dfrac{1}{2}\right\}\)	M1, A1
\(PQR = \dfrac{2\pi}{3}\)	A1
	(3)

Notes

N.B. \(a^2 = b^2 + c^2 - 2bc\cos A\) is in the formulae book.

M1: Use of cosine rule for \(\cos PQR\). Allow \(A\), \(\theta\) or other symbol for angle.
(i) \((6\sqrt{3})^2 = 6^2 + 6^2 - 2.6.6\cos PQR\): Apply usual rules for formulae: (a) formula not stated, must be correct, (b) correct formula stated, allow one sign slip when substituting.
or (ii) \(\cos PQR = \dfrac{\pm 6^2 \pm 6^2 \pm (6\sqrt{3})^2}{\pm 2\times 6\times 6}\)
Also allow invisible brackets [so allow \(6\sqrt{3}^2\)] in (i) or (ii)

A1: Correct expression \(\dfrac{6^2 + 6^2 - (6\sqrt{3})^2}{2\times 6\times 6}\) o.e. (e.g. \(-\dfrac{36}{72}\) or \(-\dfrac{1}{2}\))

A1: \(\dfrac{2\pi}{3}\)

9(a) Alternative

\(\sin\theta = \dfrac{a\sqrt{3}}{6}\) where \(\theta\) is any symbol and \(a \lt 6\).	M1
\(\sin\theta = \dfrac{3\sqrt{3}}{6}\) where \(\theta\) is any symbol.	A1
\(\dfrac{2\pi}{3}\)	A1

Mark scheme (b)

Scheme	Marks
Area \(= \dfrac{1}{2}\times 6^2\times\dfrac{2\pi}{3}\ \mathrm{m}^2\)	M1
\(= 12\pi\ \mathrm{m}^2\) (✱)	A1cso
	(2)

Notes

M1: Use of \(\tfrac{1}{2}r^2\theta\) with \(r = 6\) and \(\theta =\) their (a). For M mark \(\theta\) does not have to be exact.
M0 if using degrees.

A1: \(12\pi\) c.s.o. (\(\Rightarrow\) (a) correct exact or decimal value) N.B. Answer given in question

Special case:
Can come from an inexact value in (a)
\(PQR = 2.09 \to\) Area \(= \tfrac{1}{2}\times 6^2\times 2.09 = 37.6\) (or 37.7) \(= 12\pi\) (no errors seen, assume full values used on calculator) gets M1 A1.
\(PQR = 2.09 \to\) Area \(= \tfrac{1}{2}\times 6^2\times 2.09 = 37.6\) (or 37.7) \(= 11.97\pi = 12\pi\) gets M1 A0.

Mark scheme (c)

Scheme	Marks
Area of \(\Delta = \dfrac{1}{2}\times 6\times 6\times\sin\dfrac{2\pi}{3}\ \mathrm{m}^2\)	M1
\(= 9\sqrt{3}\ \mathrm{m}^2\)	A1cso
	(2)

Notes

M1: Use of \(\tfrac{1}{2}r^2\sin\theta\) with \(r = 6\) and their (a).
\(\theta = \cos^{-1}(\text{their } PQR)\) in degrees or radians
Method can be implied by correct decimal provided decimal is correct (corrected or truncated to at least 3 decimal places).
15.58845727

A1cso: \(9\sqrt{3}\) c.s.o. Must be exact, but correct approx. followed by \(9\sqrt{3}\) is okay (e.g. … = 15.58845 = \(9\sqrt{3}\))

9(c) Alternative (using \(\tfrac{1}{2}bh\))

Attempt to find \(h\) using trig. or Pythagoras and use this \(h\) in \(\tfrac{1}{2}bh\) form to find the area of triangle \(PQR\)	M1
\(9\sqrt{3}\) c.s.o. Must be exact, but correct approx. followed by \(9\sqrt{3}\) is okay (e.g. … = 15.58845 = \(9\sqrt{3}\))	A1cso

Mark scheme (d)

Scheme	Marks
Area of segment \(= 12\pi - 9\sqrt{3}\ \mathrm{m}^2\)	M1
\(= 22.1\ \mathrm{m}^2\)	A1
	(2)

Notes

M1: Use of area of sector – area of \(\Delta\) or use of \(\tfrac{1}{2}r^2(\theta - \sin\theta)\).

A1: Any value to 1 decimal place or more which rounds to 22.1

Mark scheme (e)

Scheme	Marks
Perimeter \(= 6 + 6 + \left[6\times\dfrac{2\pi}{3}\right]\) m	M1
\(= 24.6\) m	A1ft
	(2)
	(11)

Notes

M1: \(6 + 6 + [6\times\text{their (a)}]\).

A1 ft: Correct for their (a) to 1 decimal place or more