The radius of the circle with centre C is 7 cm and the radius of the circle with centre D is 5 cm. If the length of the chord [AB] is 9 cm, find the area of the shaded region enclosed by the two arcs AB.

\(\alpha = 2\arcsin \left( {\frac{{4.5}}{7}} \right)\) (\( \Rightarrow \alpha = 1.396... = 80.010^\circ ...\))     M1(A1)

\(\beta = 2\arcsin \left( {\frac{{4.5}}{5}} \right)\) (\( \Rightarrow \beta = 2.239... = 128.31^\circ ...\))     (A1)

Note: Allow use of cosine rule.

area \(P = \frac{1}{2} \times {7^2} \times \left( {\alpha - \sin \alpha } \right) = 10.08...\)     M1(A1)

area \(Q = \frac{1}{2} \times {5^2} \times \left( {\beta - \sin \beta } \right) = 18.18...\)     (A1)

Note: The M1 is for an attempt at area of sector minus area of triangle.

Note: The use of degrees correctly converted is acceptable.

area = 28.3 (cm²)     A1

[7 marks]

Whilst most candidates were able to make the correct construction to solve the problem some candidates seemed unable to find the area of a segment. In a number of cases candidates used degrees in a formula that required radians. There were a number of candidates who followed a completely correct method but due to premature approximation were unable to obtain a correct solution.