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2012 Paper 2 Q6
D: 1600.0 B: 1528.8
trigonometry cyclic quadrilateral cosine rule area Brahmagupta's formula Heron's formula algebraic manipulation geometric proof

A cyclic quadrilateral \(ABCD\) has sides \(AB\), \(BC\), \(CD\) and \(DA\) of lengths \(a\), \(b\), \(c\) and \(d\), respectively. The area of the quadrilateral is \(Q\), and angle \(DAB\) is \(\theta\). Find an expression for \(\cos\theta\) in terms of \(a\), \(b\), \(c\) and \(d\), and an expression for \(\sin\theta\) in terms of \(a\), \(b\), \(c\), \(d\) and \(Q\). Hence show that \[ 16Q^2 = 4(ad+bc)^2 - (a^2+d^2-b^2-c^2)^2 \,, \] and deduce that \[ Q^2 = (s-a)(s-b)(s-c)(s-d)\,, \] where \(s= \frac12(a+b+c+d)\). Deduce a formula for the area of a triangle with sides of length \(a\), \(b\) and \(c\).

Solution:

TikZ diagram
\begin{align*} && BD^2 &= a^2+d^2 - 2ad \cos \theta \\ && BD^2 &= b^2+c^2-2bc \cos (\pi - \theta) \\ \Rightarrow && a^2+d^2 - 2ad \cos \theta &= b^2+c^2+2bc \cos \theta \\ \Rightarrow && 2(ad+bc)\cos \theta &= a^2+d^2-b^2-c^2 \\ \Rightarrow && \cos \theta &= \frac{a^2+d^2-b^2-c^2}{2(ad+bc)} \\ \\ && Q &= \frac12 ad \sin \theta + \frac12 bc \sin (\pi - \theta) \\ &&&= \frac12 (ad+bc) \sin \theta \\ \Rightarrow && \sin \theta &= \frac{2Q}{ad+bc} \\ \\ && 1 &= \sin^2 \theta + \cos^2 \theta \\ &&&= \frac{4Q^2}{(ad+bc)^2} + \frac{(a^2+d^2-b^2-c^2)^2}{4(ad+bc)^2} \\ \Rightarrow && 4(ad+bc)^2 &= 16Q^2 + (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= 4(ad+bc)^2- (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= (2ad+2bc - a^2-d^2+b^2+c^2)(2ad+2bc+a^2+d^2-b^2-c^2) \\ &&&= ((b+c)^2-(a-d)^2)((a+d)^2-(b-c)^2) \\ &&&= (b+c-a+d)(b+c+a-d)(a+d+b-c)(a+d-b+c) \\ \Rightarrow && Q^2 &= (s-a)(s-b)(s-c)(s-d) \end{align*} Since all triangles are cyclic, we can place \(D\) at the same point as \(A\) to obtain Heron's formula \(A = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s = \frac12(a+b+c)\)

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