IMC 2008 Q24
24. The diagram has order 4 rotational symmetry about \(D\). If angle \(ABC\) is 15° and the area of \(ABEF\) is 24 cm2, what, in cm, is the length of CD ?
- A 1
- B \(\sqrt{3}\)
- C 2
- D \(\sqrt{5}\)
- E \(2\sqrt{3} - 1\)
Intermediate Mathematical Challenge 2008, Question 24. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 2
As the figure has rotational symmetry of order 4, \(ABEF\) is a square.
Area \(ABEF\) = 4 × area Δ\(BDA\) = 4 × \(\tfrac{1}{2}BD \times DA = 2BD^2\) = 24 cm2 so \(BD = \sqrt{12}\) cm = \(2\sqrt{3}\) cm. As \(ABEF\) is a square, \(\angle ABD\) = 45° so \(\angle CBD\) = 45° − 15° = 30°. Therefore \(\tan 30^\circ = \dfrac{1}{\sqrt{3}} = \dfrac{CD}{BD} = \dfrac{CD}{2\sqrt{3}}\), so \(CD\) = 2 cm.
Solution © UK Mathematics Trust, from the IMC 2008 solutions.