Problem of the Week
Problem B and Solution
Icy Star Rings

Problem

Michel makes coloured ice in the shape of a star ring by pouring coloured water into a special mold and freezing it. Some of his ice stars are shown.

Michel’s mold consists of an outer star and an inner star, and water is poured between the inner and outer star walls.

1. The diagram shows the inner and outer stars from Michel’s mold.

   

   Hide/Reveal Description of Stars

   A smaller five-pointed star is placed inside a larger five-pointed star, with their centres aligned. The smaller star has the same shape and orientation as the larger star. Each star has five outer vertices and five inner vertices. The angles at the outer vertices are all equal. In the smaller star, the five inner vertices are used to form a pentagon, and the pentagon is further divided into five congruent triangles, meeting at the centre of the pentagon. Also, the five remaining triangles outside the centre pentagon, forming the five points of the smaller star, are shaded grey. The rest of the figure is white.

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   The inner star is divided into five outer triangles, shown in grey, and five inner triangles, shown in white. The outer triangles each have a base of \(3\) cm and a height of \(1.7\) cm. The inner triangles each have a base of \(3\) cm and a height of \(2\) cm. Determine the area of the inner star.

2. If we ignore the inner star, then we can divide the outer star into five outer triangles and five inner triangles, like we did with the inner star in part (a).

   

   The base and height of each of the triangles in the outer star are twice the base and height of their corresponding triangle in the inner star. Determine the area of the outer star.

3. The shaded region represents the area between the inner and outer star walls.

   

   Use your calculations from parts (a) and (b) to determine the area of this region.

Solution

1. The area of each outer triangle is \(3 \times 1.7 \div 2 = 5.1 \div 2 = 2.55\text{ cm}^2\).
   Since there are \(5\) outer triangles, their total area is \(5 \times 2.55 = 12.75\text{ cm}^2\).
   The area of each inner triangle is \(3 \times 2 \div 2 = 3\text{ cm}^2\).
   Since there are \(5\) inner triangles, their total area is \(5 \times 3 = 15\text{ cm}^2\).
   Therefore, the area of the inner star is \(12.75+15=27.75\text{ cm}^2\).

2. The base of each outer triangle is \(2 \times 3 = 6 \text{ cm}\) and the height is \(2 \times 1.7 = 3.4 \text{ cm}\).
   The area of each outer triangle is then \(6 \times 3.4 \div 2 = 20.4 \div 2 = 10.2\text{ cm}^2\).
   Since there are \(5\) outer triangles, their total area is \(5 \times 10.2 = 51\text{ cm}^2\).
   The base of each inner triangle is \(2 \times 3 = 6 \text{ cm}\) and the height is \(2 \times 2 = 4 \text{ cm}\).
   The area of each inner triangle is then \(6 \times 4 \div 2 = 24 \div 2 = 12\text{ cm}^2\).
   Since there are \(5\) inner triangles, their total area is \(5 \times 12 = 60\text{ cm}^2\).
   Therefore, the area of the outer star is \(51+60=111\text{ cm}^2\).

   Notice that this is \(4\) times the area of the inner star. Does it make sense why the area was multiplied by \(4\) when we doubled the dimensions?

3. The area of the shaded region is the area of the outer star minus the area of the inner star. This is \(111 - 27.75 = 83.25 \text{ cm}^2\).