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Date May 2017 Marks available 3 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:

BCDE is a rectangle with sides of length  ( x + 3 )  cm and  ( x + 5 )  cm; ABE is an isosceles triangle, with AB = AE and a height of  x  cm; the area of ABCDE is 222 c m 2 .

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 BF = 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 + 19 x 414 = 0 .

[2]
a.ii.

Find the length of CD.

[2]
b.

Show that angle B A ^ E = 67.4 , correct to one decimal place.

[3]
c.

Find the length of the perimeter of ABCDE.

[3]
d.

Calculate the length of CF.

[4]
e.

Markscheme

222 = 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 ) ( 2 x + 5 ) 2 ( 1 4 ) 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 = 1 2 x 2 + 3 2 x + x 2 + 3 x + 5 x + 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 + 19 x 414 = 0     (AG)

[2 marks]

a.ii.

x = 9   ( and  x = 46 3 )     (A1)

CD = 12  (cm)     (A1)(G2)

[2 marks]

b.

1 2 ( their  x + 3 ) = 6     (A1)(ft)

 

Note:     Follow through from part (b).

 

tan ( B A ^ E 2 ) = 6 9     (M1)

 

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

 

B A ^ E = 67.3801     (A1)

= 67.4     (AG)

 

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

 

OR

1 2 ( their  x + 3 ) = 6     (A1)(ft)

tan ( A B ^ E ) = 9 6     (M1)

 

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

 

B A ^ E = 180 2 ( A B ^ E )

B A ^ E = 67.3801     (A1)

= 67.4     (AG)

 

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

 

[3 marks]

c.

2 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  (cm)  ( 61.6333  (cm) )     (A1)(ft)(G3)

 

Note:     Follow through from part (b).

 

[3 marks]

d.

F B ^ C = 90 + ( 180 67.4 2 )   ( = 146.3 )     (M1)

OR

180 67.4 2     (M1)

C F 2 = 8 2 + 14 2 2 ( 8 ) ( 14 ) cos ( 146.3 )     (M1)(A1)(ft)

 

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

 

CF = 21.1  (cm)  ( 21.1271 )     (A1)(ft)(G3)

OR

G B ^ F = 67.4 2 = 33.7     (A1)

 

Note:     Award (A1) for angle G B ^ F = 33.7 , 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.

M17/5/MATSD/SP2/ENG/TZ1/02.e/M

 

GF = 8 sin 33.7 = 4.4387 AND BG = 8 cos 33.7 = 6.6556     (M1)

 

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

 

C F 2 = ( 14 + 6.6556 ) 2 + ( 4.4387 ) 2     (M1)

 

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

 

CF = 21.1  (cm)  ( 21.1271 )     (A1)(ft)(G3)

[4 marks]

e.

Examiners report

[N/A]
a.i.
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a.ii.
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b.
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c.
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d.
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e.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.2—2d and 3d trig, sine rule, cosine rule, area
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Topic 3— Geometry and trigonometry

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