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Date November 2013 Marks available 2 Reference code 13N.1.sl.TZ0.5
Level SL only Paper 1 Time zone TZ0
Command term Find Question number 5 Adapted from N/A

Question

In triangle ABCABC, AC=20 cmAC=20 cm, BC=12 cmBC=12 cm and AˆBC=90A^BC=90.


     diagram not to scale

Find the length of ABAB.

[2]
a.

DD is the point on ABAB such that tan(DˆCB)=0.6tan(D^CB)=0.6.

Find the length of DBDB.

[2]
b.

DD is the point on ABAB such that tan(DˆCB)=0.6tan(D^CB)=0.6.

Find the area of triangle ADCADC.

[2]
c.

Markscheme

(AB2)=202122(AB2)=202122     (M1)

 

Note: Award (M1) for correctly substituted Pythagoras formula.

 

AB=16 cmAB=16 cm     (A1)     (C2)

[2 marks]

a.

DB12=0.6DB12=0.6     (M1)

 

Note: Award (M1) for correct substitution in tangent ratio or equivalent ie seeing 12×0.612×0.6.

 

DB=7.2 cmDB=7.2 cm     (A1)     (C2)

 

Note: Award (M1)(A0) for using tan31tan31 to get an answer of 7.217.21.

     Award (M1)(A0) for 12sin59=DBsin3112sin59=DBsin31 to get an answer of 7.21037.2103 or any other incorrect answer.

 

[2 marks]

b.

12×12×(167.2)12×12×(167.2)     (M1)

 

Note: Award (M1) for their correct substitution in triangle area formula.

 

OR

12×12×1612×12×7.212×12×1612×12×7.2     (M1)

 

Note: Award (M1) for subtraction of their two correct area formulas.

 

=52.8 cm2     (A1)(ft)     (C2)

 

Notes: Follow through from parts (a) and (b).

     Accept alternative methods.

 

[2 marks]

c.

Examiners report

This question was not well answered. Many candidates could not use Pythagoras’ theorem correctly and many failed to appreciate the significance of tan(<DCB)=0.6 and calculated the size of the angle DCB, rounding it to 31°. Unfortunately, this method led to an inaccurate value for DB. Finding the area of triangle ADC was also difficult for many who did not realize that they needed to do a subtraction of triangle areas. Candidates who tried to find side lengths and angles for triangle ADC were generally unsuccessful in calculating its area.

a.

This question was not well answered. Many candidates could not use Pythagoras’ theorem correctly and many failed to appreciate the significance of tan(<DCB)=0.6 and calculated the size of the angle DCB, rounding it to 31°. Unfortunately, this method led to an inaccurate value for DB. Finding the area of triangle ADC was also difficult for many who did not realize that they needed to do a subtraction of triangle areas. Candidates who tried to find side lengths and angles for triangle ADC were generally unsuccessful in calculating its area.

b.

This question was not well answered. Many candidates could not use Pythagoras’ theorem correctly and many failed to appreciate the significance of tan(<DCB)=0.6 and calculated the size of the angle DCB, rounding it to 31°. Unfortunately, this method led to an inaccurate value for DB. Finding the area of triangle ADC was also difficult for many who did not realize that they needed to do a subtraction of triangle areas. Candidates who tried to find side lengths and angles for triangle ADC were generally unsuccessful in calculating its area.

c.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.0 » Simple two-dimensional shapes and their properties, including perimeters and areas of circles, triangles, quadrilaterals and compound shapes.
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