GK 2015 Q8
8. In the diagram, the centre of the top square is directly above the common edge of the lower two squares. Each square has sides of length 1 cm. What is the area of the shaded region?
- A \(\tfrac{3}{4}\) cm2
- B \(\tfrac{7}{8}\) cm2
- C 1 cm2
- D \(1\tfrac{1}{4}\) cm2
- E \(1\tfrac{1}{2}\) cm2
Grey Kangaroo 2015, Question 8. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 1 cm2
The centre of the top square is directly above the common edge of the lower two squares. Hence a rectangle half the size of the square, and so of area \(\tfrac{1}{2}\) cm2, can be added to the diagram to form a right-angled triangle as shown. The area of the shaded region and the added rectangle is equal to \(\left(\tfrac{1}{2} \times 2 \times 1\tfrac{1}{2}\right)\) cm2 = \(1\tfrac{1}{2}\) cm2. Hence the area of the shaded region, in cm2, is \(1\tfrac{1}{2} - \tfrac{1}{2} = 1\).
Solution © UK Mathematics Trust, from the GK 2015 solutions.