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Date November 2020 Marks available 4 Reference code 20N.2.SL.TZ0.T_3
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Show that Question number T_3 Adapted from N/A

Question

Using geometry software, Pedro draws a quadrilateral ABCDABCD. AB=8cmAB=8cm and CD=9cmCD=9cm. Angle BAD=51.5°BAD=51.5° and angle ADB=52.5°ADB=52.5°. This information is shown in the diagram.

CE=7cmCE=7cm, where point EE is the midpoint of BDBD.

Calculate the length of BDBD.

[3]
a.

Show that angle EDC=48.0°EDC=48.0°, correct to three significant figures.

[4]
b.

Calculate the area of triangle BDCBDC.

[3]
c.

Pedro draws a circle, with centre at point EE, passing through point CC. Part of the circle is shown in the diagram.

Show that point AA lies outside this circle. Justify your reasoning.

[5]
d.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

BDsin51.5°=8sin52.5°BDsin51.5°=8sin52.5°      (M1)(A1)


Note: Award (M1) for substituted sine rule, (A1) for correct substitution.


(BD=)  7.89 (cm)  (7.89164)(BD=)  7.89 (cm)  (7.89164)      (A1)(G2)


Note: If radians are used the answer is 9.587239.58723 award at most (M1)(A1)(A0).


[3 marks]

a.

cosEDC=92+3.945822-722×9×3.94582cosEDC=92+3.945822722×9×3.94582      (A1)(ft)(M1)(A1)(ft)


Note: Award (A1) for 3.945823.94582 or 7.8916427.891642 seen, (M1) for substituted cosine rule, (A1)(ft) for correct substitutions.


(EDC=)  47.9515°(EDC=)  47.9515°      (A1)

48.0°48.0° (33 sig figures)      (AG)


Note: Both an unrounded answer that rounds to the given answer and the rounded value must be seen for the final (M1) to be awarded.
Award at most (A1)(ft)(M1)(A1)(ft)(A0) if the known angle 48.0°48.0° is used to validate the result. Follow through from their BD BD in part (a).


[4 marks]

b.

Units are required in this question.


(area=) 12×7.89164×9×sin48.0°(area=) 12×7.89164×9×sin48.0° 
     (M1)(A1)(ft)


Note: Award (M1) for substituted area formula. Award (A1) for correct substitution.


(area=) 26.4cm2  (26.3908)(area=) 26.4cm2  (26.3908)      (A1)(ft)(G3)


Note: Follow through from part (a).


[3 marks]

c.

AE2=82+(3.94582)2-2×8×3.94582cos(76°)AE2=82+(3.94582)22×8×3.94582cos(76°)        (A1)(M1)(A1)(ft)


Note: Award (A1) for 76°76° seen. Award (M1) for substituted cosine rule to find AEAE, (A1)(ft) for correct substitutions.


(AE=)  8.02(cm)  (8.01849)(AE=)  8.02(cm)  (8.01849)      (A1)(ft)(G3)


Note: Follow through from part (a).


OR


AE2=9.784242+(3.94582)2-2×9.78424×3.94582cos(52.5°)AE2=9.784242+(3.94582)22×9.78424×3.94582cos(52.5°)        (A1)(M1)(A1)(ft)


Note: Award (A1) for AD (9.78424)AD (9.78424) or 76°76° seen. Award (M1) for substituted cosine rule to find AEAE (do not award (M1) for cosine or sine rule to find ADAD), (A1)(ft) for correct substitutions.

 

(AE=)  8.02(cm)  (8.01849)(AE=)  8.02(cm)  (8.01849)      (A1)(ft)(G3)


Note: Follow through from part (a).


8.02>78.02>7.      (A1)(ft)

point AA is outside the circle.      (AG)


Note: Award (A1) for a numerical comparison of AEAE and CECE. Follow through for the final (A1)(ft) within the part for their 8.028.02. The final (A1)(ft) is contingent on a valid method to find the value of AEAE.
Do not award the final (A1)(ft) if the (AG) line is not stated.
Do not award the final (A1)(ft) if their point AA is inside the circle.


[5 marks]

d.

Examiners report

[N/A]
a.
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b.
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c.
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d.

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.2—2d and 3d trig
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Topic 3—Geometry and trigonometry

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