IMC 2018 Q8
8. The diagram shows a square of perimeter 20 cm inscribed inside a square of perimeter 28 cm.
What is the area of the shaded triangle?
- A 6 cm2
- B 7 cm2
- C 8 cm2
- D 9 cm2
- E 10 cm2
Intermediate Mathematical Challenge 2018, Question 8. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: A 6 cm2
First note that the length of each side of the larger square is 7 cm and that of the smaller square is 5 cm. So their areas are 49 cm2 and 25 cm2 respectively. Let \(\angle BAC = x°\). As the interior angles of a triangle sum to 180° and \(\angle ABC = 90°\), \(\angle ACB = (180 - 90 - x)° = (90 - x)°\).
The angles on a straight line sum to 180°, so \(\angle ECD = [180 - 90 - (90 - x)]° = x°\).
Consider triangles \(ABC\) and \(CDE\). Note that \(\angle BAC = \angle DCE = x°\); \(\angle ABC = \angle CDE = 90°\); \(AC = CE = 5\) cm. Therefore the two triangles are congruent (AAS). By a similar method, it may be shown that all four triangles which lie between the two squares are congruent to each other.
So the area of the shaded triangle is \(\tfrac{1}{4}(49 - 25)\) cm2 = 6 cm2.
Solution © UK Mathematics Trust, from the IMC 2018 solutions. Longer versions with investigations: Solutions and investigations (PDF).