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 55
For being well-behaved at the garden fayre, four children were each given two sweets.
Jesse had an orange sweet. One child who had a red sweet also had a blue one. No child had two sweets of the same colour. A child who had a green sweet also had a red one. Jamie didn't have a red sweet, and Jo had a green one. Jordan didn't have an orange one, and Jesse had no blue sweets.
Knowing that there were two sweets of each colour, can you tell the colours of the sweets each child had?
Answer
Jamie Orange Blue
Jesse Orange Green
Jo Green Red
Jordan Red Blue
Reasoning
The children were: Jamie, Jesse, Jo, Jordan.
The colours were: Blue, Green, Orange, Red.
There were two sweets of each colour, and by (3) no child had two sweets of the same colour.
By (1), Jesse had an Orange sweet …
… by (6), didn't have Blue …
… by (2) and (4), Red was paired with Blue and Green …
… therefore, Jesse's other sweet can't have been Red, so was Green.
Jamie
Jesse Orange Green
Jo
Jordan
By (4) the other Green sweet was paired with Red, by (5) this must have been Jo.
Jamie
Jesse Orange Green
Jo Green Red
Jordan
By (2) the other Red sweet was paired with Blue, but by (5) this wasn't Jamie, so must have been Jordan.
Jamie
Jesse Orange Green
Jo Green Red
Jordan Red Blue
Leaving Jamie with Orange and Blue.
Jamie Orange Blue
Jesse Orange Green
Jo Green Red
Jordan Red Blue
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Puzzle 56
Can you scramble each of these 6-letter words, and then split them into two common 3-letter words?
For example: notify = fin toy. heliumlatestalmondeldestplacedradiumdesertwealthtrickyfunnel
Answers helium = hum + lielatest = let + satalmond = man + oldeldest = led + setplaced = cap + ledradium = air + muddesert = red + setwealth = law + thetricky = cry + kitfunnel = elf + nun Alternative Answers
The following answers use less common words.
helium = him + leu helium = hue + mil helium = hum + lei
latest = els + tat
almond = don + lam almond = lam + nod
eldest = eld + set eldest = els + ted
placed = pec + lad
radium = mud + ria radium = amu + rid radium = aid + rum
wealth = awl + the wealth = haw + let
There are more answers with even rarer words, but I've chosen not to include those.
