Use the cosine rule to show that \({r^2} - 12\sqrt 3 r + 144 - {p^2} = 0\).

Calculate the two corresponding values of PQ.

Hence, find the area of the smaller triangle.

Determine the range of values of \(p\) for which it is possible to form two triangles.

\({p^2} = {12^2} + {r^2} - 2 \times 12 \times r \times \cos (30^\circ )\)     M1A1

\({r^2} - 12\sqrt 3 r + 144 - {p^2} = 0\)     AG

[2 marks]

EITHER

\({r^2} - 12\sqrt 3 r + 80 = 0\)     (M1)

OR

using the sine rule     (M1)

THEN

\({\text{PQ}} = 5.10{\text{ }}({\text{cm}})\) or     A1

\({\text{PQ}} = 15.7{\text{ }}({\text{cm}})\)     A1

[3 marks]

\({\text{area}} = \frac{1}{2} \times 12 \times 5.1008 \ldots \times \sin (30^\circ )\)     M1A1

\( = 15.3{\text{ }}({\text{c}}{{\text{m}}^2})\)     A1

[3 marks]

METHOD 1

EITHER

\({r^2} - 12\sqrt 3 r + 144 - {p^2} = 0\)

discriminant \( = {\left( {12\sqrt 3 } \right)^2} - 4 \times (144 - {p^2})\)     M1

\( = 4({p^2} - 36)\)     A1

\(({p^2} - 36) > 0\)     M1

\(p > 6\)     A1

OR

construction of a right angle triangle     (M1)

\(12\sin 30^\circ = 6\)     M1(A1)

hence for two triangles \(p > 6\)     R1

THEN

\(p < 12\)     A1

\(144 - {p^2} > 0\) to ensure two positive solutions or valid geometric argument     R1

\(\therefore 6 < p < 12\)     A1

METHOD 2

diagram showing two triangles     (M1)

\(12\sin 30^\circ = 6\)     M1A1

one right angled triangle when \(p = 6\)     (A1)

\(\therefore p > 6\) for two triangles     R1

\(p < 12\) for two triangles     A1

\(6 < p < 12\)     A1

[7 marks]