IMC 2008 Q20
20. What, in cm2, is the area of this quadrilateral?
- A 48
- B 50
- C 52
- D 54
- E 56
Intermediate Mathematical Challenge 2008, Question 20. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: A 48
Let the vertices of the quadrilateral be \(A\), \(B\), \(C\), \(D\) as shown. Then, by Pythagoras' Theorem:
\(AC^2 = AD^2 + DC^2 = (7^2 + 9^2)\) cm2 = 130 cm2.
Similarly, \(AB^2 + BC^2 = AC^2\), so
\(AB^2 = (130 - 9)\) cm2 = 121 cm2. Therefore \(AB\) has length 11 cm and the area, in cm2, of quadrilateral \(ABCD\) is \(\tfrac{1}{2} \times 9 \times 7 + \tfrac{1}{2} \times 3 \times 11 = 48\).
Solution © UK Mathematics Trust, from the IMC 2008 solutions.