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Date	May 2017	Marks available	2	Reference code	17M.2.SL.TZ1.T_2
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Show that	Question number	T_2	Adapted from	N/A

Question

The base of an electric iron can be modelled as a pentagon ABCDE, where:

$\begin{array}{*{20}{l}} {{\text{BCDE is a rectangle with sides of length }}(x + 3){\text{ cm and }}(x + 5){\text{ cm;}}} \\ {{\text{ABE is an isosceles triangle, with AB}} = {\text{AE and a height of }}x{\text{ cm;}}} \\ {{\text{the area of ABCDE is 222 c}}{{\text{m}}^{\text{2}}}{\text{.}}} \end{array}$

M17/5/MATSD/SP2/ENG/TZ1/02

Insulation tape is wrapped around the perimeter of the base of the iron, ABCDE.

F is the point on AB such that ${\text{BF}} = {\text{8 cm}}$ . A heating element in the iron runs in a straight line, from C to F.

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

[2]

a.i.

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

[2]

a.ii.

Find the length of CD.

[2]

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

[3]

Find the length of the perimeter of ABCDE.

[3]

Calculate the length of CF.

[4]

Markscheme

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

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

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

a.i.

$222 = \frac{1}{2}{x^2} + \frac{3}{2}x + {x^2} + 3x + 5x + 15$ (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]

a.ii.

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

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

$180 - \frac{{67.4}}{2}$ (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)

${\rm{G\hat BF}} = \frac{{67.4}}{2} = 33.7^\circ$ (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.