0606 P13 - Jun 2017 - Q4 - 5 marks
8572
In this question, all dimensions are in centimetres.
The diagram shows an isosceles triangle \(ABC\), where \(AB=AC\). The point \(M\) is the mid-point of \(BC\).
Given that \(AM=3+2\sqrt5\) and \(BC=4+6\sqrt5\), find, without using a calculator,
(i) the area of triangle \(ABC\),
(ii) \(\tan ABC\), giving your answer in the form \(\frac{a+b\sqrt5}{c}\), where \(a\), \(b\) and \(c\) are positive integers.
Solution
Answer: (i) \(\text{area}=36+13\sqrt5\text{ cm}^2\). (ii) \(\tan ABC=\frac{24+5\sqrt5}{41}\).
Since \(ABC\) is isosceles and \(M\) is the midpoint of \(BC\), the line \(AM\) is perpendicular to \(BC\). So \(AM\) is the height of the triangle.
(i) Using
\(\text{Area}=\frac12\times\text{base}\times\text{height},\)
we get
\(\text{Area}=\frac12(4+6\sqrt5)(3+2\sqrt5).\)
Expand the brackets:
\((4+6\sqrt5)(3+2\sqrt5)=12+8\sqrt5+18\sqrt5+60=72+26\sqrt5.\)
Therefore
\(\text{Area}=\frac12(72+26\sqrt5)=36+13\sqrt5.\)
(ii) In right-angled triangle \(ABM\),
\(BM=\frac{BC}{2}=\frac{4+6\sqrt5}{2}=2+3\sqrt5.\)
So
\(\tan ABC=\frac{AM}{BM}=\frac{3+2\sqrt5}{2+3\sqrt5}.\)
Rationalise the denominator:
\(\frac{3+2\sqrt5}{2+3\sqrt5}\times\frac{2-3\sqrt5}{2-3\sqrt5} = \frac{(3+2\sqrt5)(2-3\sqrt5)}{4-45}.\)
The numerator is
\(6-9\sqrt5+4\sqrt5-30=-24-5\sqrt5.\)
Hence
\(\tan ABC=\frac{-24-5\sqrt5}{-41}=\frac{24+5\sqrt5}{41}.\)
