GK 2016 Q9
9. Three equilateral triangles are cut from the corners of a large equilateral triangle to form an irregular hexagon, as shown in the diagram.
The perimeter of the large equilateral triangle is 60 cm. The perimeter of the irregular hexagon is 40 cm. What is the sum of the perimeters of the triangles that were cut from the large triangle?
- A 60 cm
- B 66 cm
- C 72 cm
- D 75 cm
- E 81 cm
Grey Kangaroo 2016, Question 9. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: A 60 cm
Let the lengths of the sides of the equilateral triangles that are cut off be \(x\) cm, \(y\) cm and \(z\) cm, as shown in the diagram.
The length of a side of the large equilateral triangle is \(\tfrac{1}{3} \times 60\) cm = 20 cm. The perimeter of the irregular hexagon is 40 cm. Therefore we have
\(40 = x + (20 - x - y) + y + (20 - y - z) + z + (20 - z - x)\). Hence \(40 = 60 - (x + y + z)\) and therefore \(x + y + z = 20\). Therefore the sum of the perimeters of the triangles cut off is \((3x + 3y + 3z)\) cm = 60 cm.
Solution © UK Mathematics Trust, from the GK 2016 solutions.