Triangle: 47. Doughnut
The area of a circle is π x Radius2.
The larger circle has diameter = 40, therefore, the radius is 20, and the area is π x 202 = 400π.
The smaller circle has diameter = 20, therefore, the radius is 10, and the area is π x 102 = 100π.
Therefore, the shaded area is 400π – 100π = 300π ≈ 942. Circle
The area of a circle is π x Radius2.
The circle with diameter 31 has a radius of 15.5 and an area of π x 15.52 ≈ 755.
The circle with diameter 35 has a radius of 17.5 and an area of π x 17.52 ≈ 962 (closest match).
The circle with diameter 39 has a radius of 19.5 and an area of π x 19.52 ≈ 1195. Square
The area of a square is Side x Side.
The square with side 31 has an area of 31 x 31 = 961 (closest match).
The square with side 35 has an area of 35 x 35 = 1225.
The square with side 39 has an area of 39 x 39 = 1521. Triangle
The area of a triangle is 1/2 x Base x Height. Using Pythagoras' theorem it can be shown that the area of an equilateral triangle is √3 x Base2 ÷ 4.
The triangle with side 39 has an area of √3 x 39 x 39 ÷ 4 ≈ 659.
The triangle with side 43 has an area of √3 x 43 x 43 ÷ 4 ≈ 801.
The triangle with side 47 has an area of √3 x 47 x 47 ÷ 4 ≈ 957 (closest match).
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Puzzle 27
Start in the bottom left corner and move either up or right, one square at a time, adding up the numbers. What is the largest total you can make?
1
4
2
2
1
2
4
3
1
3
4
3
2
1
4
2
1
2
1
2
2
3
3
1
4
1
3
1
2
1
3
1
4
3
4
2
2
1
1
1
1
3
3
4
2
3
2
2
Note: this puzzle is not interactive, and the numbers cannot be clicked.
At the local sweet shop, three particularly nice sweets are on special offer.
A Nobbler is over three times the price of a Sparkle. Six Sparkles are worth more than a Wibbler. A Nobbler, plus two Sparkles costs less than a Wibbler. A Sparkle, a Wibbler and a Nobbler together cost 40p.
Can you determine the price of each type of sweet?
Reasoning
By (3) a Nobbler, plus two Sparkles costs less than a Wibbler, therefore a Wibbler must be the most expensive sweet.
By (1) a Nobbler is over three times the price of a Sparkle, therefore a Sparkle must be the cheapest sweet.
So the order of sweets, from the least to most expensive, is Sparkle, Nobbler, Wibbler.
If a Sparkle was 1p, by (2) a Wibbler could only be up to 5p, by (4) a Nobbler would cost at least 34p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet.
If a Sparkle was 2p, by (2) a Wibbler could only be up to 11p, by (4) a Nobbler would cost at least 27p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet.
If a Sparkle was 3p, by (2) a Wibbler could only be up to 17p, by (4) a Nobbler would cost at least 20p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet.
So a Sparkle must be at least 4p.
If a Sparkle was 4p, by (1) a Nobbler must be at least 13p, by (4) a Wibbler would cost 23p. This combination matches all of the clues and is a possible solution.
If a Sparkle was 4p and a Nobbler 14p, by (4) a Wibbler would cost 22p. This would not satisfy (3). And if we increase the price of a Nobbler, (3) is never satisfied.
If a Sparkle was 5p, by (1) a Nobbler must be at least 16p, by (4) making a Wibbler at most 19p. This would not satisfy (3).
If we increase the price of a Sparkle or Nobbler further, (3) is will never be satisfied.
Therefore, the only solution we came across must be the correct one.
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