C2 January 2013 Q7

EdexcelOld spec12 marksRadians

The triangle \(XYZ\) in Figure 1 has \(XY = 6\) cm, \(YZ = 9\) cm, \(ZX = 4\) cm and angle \(ZXY = \alpha\). The point \(W\) lies on the line \(XY\).

The circular arc \(ZW\), in Figure 1 is a major arc of the circle with centre \(X\) and radius 4 cm.

(a) Show that, to 3 significant figures, \(\alpha = 2.22\) radians. (2)

(b) Find the area, in cm², of the major sector \(XZWX\). (3)

The region enclosed by the major arc \(ZW\) of the circle and the lines \(WY\) and \(YZ\) is shown shaded in Figure 1.

Calculate

(d) the perimeter \(ZWYZ\) of this shaded region. (4)

Mark scheme (a)

Scheme	Marks
\(9^2 = 4^2 + 6^2 - 2 \times 4 \times 6\cos\alpha \Rightarrow \cos\alpha = \ldots..\)	M1
\(\cos\alpha = \dfrac{4^2 + 6^2 - 9^2}{2 \times 4 \times 6}\left(= -\dfrac{29}{48} = -0.604..\right)\)
\(\alpha = 2.22\ \ *\)	A1
(NB \(\alpha = 2.219516005\))
	(2)

M1: Correct use of cosine rule leading to a value for \(\cos\alpha\)

A1: Cso (2.22 must be seen here)

Scheme	Marks
\(XY^2 = 4^2 + 6^2 - 2 \times 4 \times 6\cos 2.22 \Rightarrow XY^2 = \ldots\)	M1
\(XY^2 = 81.01\ldots.\)
\(XY = 9.00\ldots.\)	A1
	(2)

M1: Correct use of cosine rule leading to a value for \(XY^2\)

Mark scheme (b)

B1: \(2\pi - 2.22\) or awrt 4.06

M1: Correct method for major sector area.

A1: Awrt 32.5

B1: Correct expression for circle area

M1: Correct method for circle - minor sector area

A1: Awrt 32.5

Mark scheme (c)

Scheme	Marks
Area of triangle \(= \tfrac{1}{2} \times 4 \times 6 \times \sin 2.22\ (= 9.56)\)	B1
So area required = “9.56” + “32.5”	M1
\(= 42.1\) cm² or \(42.0\) cm²	A1
	(3)

B1: Correct expression for the area of triangle \(XYZ\)

M1: Their Triangle \(XYZ\) + (part (b) answer or correct attempt at major sector)

A1: Awrt 42.1 or 42.0 (Or just 42)

Mark scheme (d)

M1: \(4 \times their\,(2\pi - 2.22)\) Or circumference – minor arc

A1ft: Correct ft expression

M1: \(9 + 2 +\) Any Arc

A1: Awrt 27.2 or awrt 27.3