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Date	November 2019	Marks available	2	Reference code	19N.2.SL.TZ0.S_4
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	S_4	Adapted from	N/A

Question

The following diagram shows a right-angled triangle, ${\text{ABC}}$ , with ${\text{AC}} = 10\,{\text{cm}}$ , ${\text{AB}} = 6\,{\text{cm}}$ and ${\text{BC}} = 8\,{\text{cm}}$ .

The points ${\text{D}}$ and ${\text{F}}$ lie on $\left[ {{\text{AC}}} \right]$ .
$\left[ {{\text{BD}}} \right]$ is perpendicular to $\left[ {{\text{AC}}} \right]$ .
${\text{BEF}}$ is the arc of a circle, centred at ${\text{A}}$ .
The region $R$ is bounded by $\left[ {{\text{BD}}} \right]$ , $\left[ {{\text{DF}}} \right]$ and arc ${\text{BEF}}$ .

Find ${\text{B}}\widehat {\text{A}}{\text{C}}$ .

[2]

a.

Find the area of $R$ .

[5]

b.

Markscheme

correct working (A1)

eg ${\text{sin}}\,\alpha = \frac{8}{{10}}$ , ${\text{cos}}\,\theta = \frac{6}{{10}}$ , ${\text{cos}}\,{\text{B}}\widehat {\text{A}}{\text{C}} = \frac{{{6^2} + {{10}^2} - {8^2}}}{{2 \times 6 \times 10}}$

$0.927295$

${\text{B}}\widehat {\text{A}}{\text{C}} = 0.927$ $\left( { = 53.1^\circ } \right)$ (A1) N2

[2 marks]

a.

Note: There may be slight differences in the final answer, depending on the approach the candidate uses in part (b). Accept a final answer that is consistent with their working.

correct area of sector ${\text{ABF}}$ (seen anywhere) (A1)

eg $\frac{1}{2} \times {6^2} \times 0.927$ , $\frac{{53.1301^\circ }}{{360^\circ }} \times \pi \times {6^2}$ , ${\text{16}}{\text{.6913}}$

correct expression (or value) for either $\left[ {{\text{AD}}} \right]$ or $\left[ {{\text{BD}}} \right]$ (seen anywhere) (A1)

eg ${\text{AD}} = 6\,{\text{cos}}\left( {{\text{B}}\widehat {\text{A}}{\text{C}}} \right)\,\,\,\left( { = 3.6} \right)$

${\text{BD}} = 6\,{\text{sin}}\left( {53.1^\circ } \right)\,\,\,\left( { = 4.8} \right)$

correct area of triangle ${\text{ABD}}$ (seen anywhere) (A1)

eg $\frac{1}{2} \times 6\,{\text{cos}}\,{\text{B}}\widehat {\text{A}}{\text{D}} \times 6\,{\text{sin}}\,{\text{B}}\widehat {\text{A}}{\text{D}}$ , $9\,{\text{sin}}\left( {{\text{2}}\,{\text{B}}\widehat {\text{A}}{\text{C}}} \right)$ , $8.64$ (exact)

appropriate approach (seen anywhere) (M1)

eg ${{\text{A}}_{{\text{triangle ABD}}}} - {{\text{A}}_{{\text{sector}}}}$ , their sector − their triangle ${\text{ABD}}$

$8.05131$

area of shaded region $= 8.05$ (cm²) A1 N2

[5 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.4—Circle: radians, arcs, sectors

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Topic 3— Geometry and trigonometry