Standard +0.3 This is a multi-step problem requiring area formula (½ab sin C), perimeter constraint, and potentially cosine rule, but follows a standard approach for AS-level sine/cosine rule questions. The given information (perimeter, area, one angle's sine) leads to a system of equations that can be solved systematically without requiring novel insight, making it slightly easier than average.
5 The diagram shows the triangle ABC in which \(\mathrm { AC } = 13 \mathrm {~cm}\) and AB is the shortest side. The perimeter of the triangle is 32 cm . The area is \(24 \mathrm {~cm} ^ { 2 }\) and \(\sin \mathrm { B } = \frac { 4 } { 5 }\).
Determine the lengths of AB and BC .
Attempt to solve their non-linear simultaneous equations
Giving \(a = 4\) or \(15\)
A1
Correct roots of quadratic
AB is the shortest side, so \(AB = 4\) cm and \(BC = 15\) cm
A1 [5]
Must be the right way round
# Question 5:
Let $AB = c$ and $BC = a$
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using perimeter $a + c + 13 = 32$ giving $a + c = 19$ | M1 | Forming an equation – need not be simplified |
| Using area $\frac{1}{2}ac\sin B = \frac{1}{2}ac \times \frac{4}{5} = 24$ giving $ac = 60$ | M1 | Forming another equation – need not be simplified |
| Solving simultaneously: $a(19-a) = 60 \Rightarrow a^2 - 19a + 60 = 0$ | M1 | Attempt to solve their non-linear simultaneous equations |
| Giving $a = 4$ or $15$ | A1 | Correct roots of quadratic |
| AB is the shortest side, so $AB = 4$ cm and $BC = 15$ cm | A1 [5] | Must be the right way round |
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5 The diagram shows the triangle ABC in which $\mathrm { AC } = 13 \mathrm {~cm}$ and AB is the shortest side. The perimeter of the triangle is 32 cm . The area is $24 \mathrm {~cm} ^ { 2 }$ and $\sin \mathrm { B } = \frac { 4 } { 5 }$.
Determine the lengths of AB and BC .
\hfill \mbox{\textit{OCR MEI AS Paper 1 2021 Q5 [5]}}