17N.1.SL.TZ0.S_4

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The following diagram shows triangle ABC, with ${\text{AB}} = 3{\text{ cm}}$ , ${\text{BC}} = 8{\text{ cm}}$ , and ${\rm{A\hat BC = }}\frac{\pi }{3}$ .

N17/5/MATME/SP1/ENG/TZ0/04

Show that ${\text{AC}} = 7{\text{ cm}}$ .

[4]

The shape in the following diagram is formed by adding a semicircle with diameter [AC] to the triangle.

N17/5/MATME/SP1/ENG/TZ0/04.b

Find the exact perimeter of this shape.

[3]

Markscheme / solution

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

evidence of choosing the cosine rule (M1)

eg $\,\,\,\,\,$ ${c^2} = {a^2} + {b^2} - ab\cos C$

correct substitution into RHS of cosine rule (A1)

eg $\,\,\,\,\,$ ${3^2} + {8^2} - 2 \times 3 \times 8 \times \cos \frac{\pi }{3}$

evidence of correct value for $\cos \frac{\pi }{3}$ (may be seen anywhere, including in cosine rule) A1

eg $\,\,\,\,\,$ $\cos \frac{\pi }{3} = \frac{1}{2},{\text{ A}}{{\text{C}}^2} = 9 + 64 - \left( {48 \times \frac{1}{2}} \right),{\text{ }}9 + 64 - 24$

correct working clearly leading to answer A1

eg $\,\,\,\,\,$ ${\text{A}}{{\text{C}}^2} = 49,{\text{ }}b = \sqrt {49}$

${\text{AC}} = 7{\text{ (cm)}}$ AG N0

Note: Award no marks if the only working seen is ${\text{A}}{{\text{C}}^2} = 49$ or ${\text{AC}} = \sqrt {49}$ (or similar).

[4 marks]

correct substitution for semicircle (A1)

eg $\,\,\,\,\,$ ${\text{semicircle}} = \frac{1}{2}(2\pi \times 3.5),{\text{ }}\frac{1}{2} \times \pi \times 7,{\text{ }}3.5\pi$

valid approach (seen anywhere) (M1)

eg $\,\,\,\,\,$ ${\text{perimeter}} = {\text{AB}} + {\text{BC}} + {\text{semicircle, }}3 + 8 + \left( {\frac{1}{2} \times 2 \times \pi \times \frac{7}{2}} \right),{\text{ }}8 + 3 + 3.5\pi$

$11 + \frac{7}{2}\pi {\text{ }}( = 3.5\pi + 11){\text{ (cm)}}$ A1 N2

[3 marks]

Examiners’ report

[N/A]