Write down an equation for the area of ABCDE using the above information.

Show that the equation in part (a)(i) simplifies to \(3{x^2} + 19x - 414 = 0\).

Find the length of CD.

Show that angle \({\rm{B\hat AE}} = 67.4^\circ \), correct to one decimal place.

Find the length of the perimeter of ABCDE.

Calculate the length of CF.

\(222 = \frac{1}{2}x(x + 3) + (x + 3)(x + 5)\)     (M1)(M1)(A1)

Note:     Award (M1) for correct area of triangle, (M1) for correct area of rectangle, (A1) for equating the sum to 222.

OR

\(222 = (x + 3)(2x + 5) - 2\left( {\frac{1}{4}} \right)x(x + 3)\)     (M1)(M1)(A1)

Note:     Award (M1) for area of bounding rectangle, (M1) for area of triangle, (A1) for equating the difference to 222.

[2 marks]

\(222 = \frac{1}{2}{x^2} + \frac{3}{2}x + {x^2} + 3x + 5x + 15\)     (M1)

Note:     Award (M1) for complete expansion of the brackets, leading to the final answer, with no incorrect working seen. The final answer must be seen to award (M1).

\(3{x^2} + 19x - 414 = 0\)     (AG)

[2 marks]

\(x = 9{\text{ }}\left( {{\text{and }}x =  - \frac{{46}}{3}} \right)\)     (A1)

\({\text{CD}} = 12{\text{ (cm)}}\)     (A1)(G2)

[2 marks]

\(\frac{1}{2}({\text{their }}x + 3) = 6\)     (A1)(ft)

Note:     Follow through from part (b).

\(\tan \left( {\frac{{{\rm{B\hat AE}}}}{2}} \right) = \frac{6}{9}\)     (M1)

Note:     Award (M1) for their correct substitutions in tangent ratio.

\({\rm{B\hat AE}} = 67.3801 \ldots ^\circ \)     (A1)

\( = 67.4^\circ \)     (AG)

Note:     Do not award the final (A1) unless both the correct unrounded and rounded answers are seen.

OR

\(\frac{1}{2}({\text{their }}x + 3) = 6\)     (A1)(ft)

\(\tan ({\rm{A\hat BE}}) = \frac{9}{6}\)     (M1)

Note:     Award (M1) for their correct substitutions in tangent ratio.

\({\rm{B\hat AE}} = 180^\circ  - 2({\rm{A\hat BE}})\)

\({\rm{B\hat AE}} = 67.3801 \ldots ^\circ \)     (A1)

\( = 67.4^\circ \)     (AG)

Note:     Do not award the final (A1) unless both the correct unrounded and rounded answers are seen.

[3 marks]

\(2\sqrt {{9^2} + {6^2}}  + 12 + 2(14)\)     (M1)(M1)

Note:     Award (M1) for correct substitution into Pythagoras. Award (M1) for the addition of 5 sides of the pentagon, consistent with their \(x\).

\(61.6{\text{ (cm) }}\left( {61.6333 \ldots {\text{ (cm)}}} \right)\)     (A1)(ft)(G3)

Note:     Follow through from part (b).

[3 marks]

\({\rm{F\hat BC}} = 90 + \left( {\frac{{180 - 67.4}}{2}} \right){\text{ }}( = 146.3^\circ )\)     (M1)

OR

\(180 - \frac{{67.4}}{2}\)     (M1)

\({\rm{C}}{{\text{F}}^2} = {8^2} + {14^2} - 2(8)(14)\cos (146.3^\circ )\)     (M1)(A1)(ft)

Note:     Award (M1) for substituted cosine rule formula and (A1) for correct substitutions. Follow through from part (b).

\({\text{CF}} = 21.1{\text{ (cm) }}(21.1271 \ldots )\)     (A1)(ft)(G3)

OR

\({\rm{G\hat BF}} = \frac{{67.4}}{2} = 33.7^\circ \)     (A1)

Note:     Award (A1) for angle \({\rm{G\hat BF}} = 33.7^\circ \), where G is the point such that CG is a projection/extension of CB and triangles BGF and CGF are right-angled triangles. The candidate may use another variable.

\({\text{GF}} = 8\sin 33.7^\circ  = 4.4387 \ldots \)\(\,\,\,\)AND\(\,\,\,\)\({\text{BG}} = 8\cos 33.7^\circ  = 6.6556 \ldots \)     (M1)

Note:     Award (M1) for correct substitution into trig formulas to find both GF and BG.

\({\text{C}}{{\text{F}}^2} = {(14 + 6.6556 \ldots )^2} + {(4.4387 \ldots )^2}\)     (M1)

Note:     Award (M1) for correct substitution into Pythagoras formula to find CF.

\({\text{CF}} = 21.1{\text{ (cm) }}(21.1271 \ldots )\)     (A1)(ft)(G3)

[4 marks]