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Date	November 2020	Marks available	5	Reference code	20N.2.SL.TZ0.T_3
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Justify and Show that	Question number	T_3	Adapted from	N/A

Question

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

$CE = 7 cm$ , where point $E$ is the midpoint of $BD$ .

Calculate the length of $BD$ .

[3]

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

[4]

Calculate the area of triangle $BDC$ .

[3]

Pedro draws a circle, with centre at point $E$ , passing through point $C$ . Part of the circle is shown in the diagram.

Show that point $A$ lies outside this circle. Justify your reasoning.

[5]

Markscheme

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

$\frac{BD}{\sin 51.5 °} = \frac{8}{\sin 52.5 °}$ (M1)(A1)

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

$(BD =) 7.89 (cm) (7.89164 \dots)$ (A1)(G2)

Note: If radians are used the answer is $9.58723 \dots$ award at most (M1)(A1)(A0).

[3 marks]

$\cos EDC = \frac{9^{2} + 3.94582 \dots^{2} - 7^{2}}{2 \times 9 \times 3.94582 \dots}$ (A1)(ft)(M1)(A1)(ft)

Note: Award (A1) for $3.94582 \dots$ or $\frac{7.89164 \dots}{2}$ seen, (M1) for substituted cosine rule, (A1)(ft) for correct substitutions.

$(EDC =) 47.9515 \dots °$ (A1)

$48.0 °$ ( $3$ 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 $48.0 °$ is used to validate the result. Follow through from their $BD$ in part (a).

[4 marks]

Units are required in this question.

$(area =) \frac{1}{2} \times 7.89164 \dots \times 9 \times \sin 48.0 °$ (M1)(A1)(ft)

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

$(area =) 26.4 {cm}^{2} (26.3908 \dots)$ (A1)(ft)(G3)

Note: Follow through from part (a).

[3 marks]

${AE}^{2} = 8^{2} + {(3.94582 \dots)}^{2} - 2 \times 8 \times 3.94582 \dots \cos (76 °)$ (A1)(M1)(A1)(ft)

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

$(AE =) 8.02 (cm) (8.01849 \dots)$ (A1)(ft)(G3)

Note: Follow through from part (a).

${AE}^{2} = 9.78424 \dots^{2} + {(3.94582 \dots)}^{2} - 2 \times 9.78424 \dots \times 3.94582 \dots \cos (52.5 °)$ (A1)(M1)(A1)(ft)

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

$(AE =) 8.02 (cm) (8.01849 \dots)$ (A1)(ft)(G3)

Note: Follow through from part (a).

$8.02 > 7$ . (A1)(ft)

point $A$ is outside the circle. (AG)

Note: Award (A1) for a numerical comparison of $AE$ and $CE$ . Follow through for the final (A1)(ft) within the part for their $8.02$ . The final (A1)(ft) is contingent on a valid method to find the value of $AE$ .
Do not award the final (A1)(ft) if the (AG) line is not stated.
Do not award the final (A1)(ft) if their point $A$ is inside the circle.

[5 marks]

Examiners report

[N/A]

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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