Puzzle 149
Which square, which circle, and which triangle has the closest area to the doughnut shape on the left?
The drawings are to scale, so you might be able to judge it, as well as working the actual areas out.
Puzzle Copyright © Kevin Stone
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Hint
The area of a circle is π x Radius2.
Answer
Circle: 3.5.
Square: 3.1.
Triangle: 4.7.
Doughnut
The area of a circle is π x Radius2.
The larger circle has diameter = 4, therefore the radius is 2, and the area is π x 22 = 4π.
The smaller circle has diameter = 2, therefore the radius is 1, and the area is π x 12 = π.
Therefore the shaded area is 4π - π = 3π ≈ 9.42.
Circle
The area of a circle is π x Radius2.
The circle with diameter 3.1 has a radius of 1.55 and an area of π x 1.552 = 7.55.
The circle with diameter 3.5 has a radius of 1.75 and an area of π x 1.752 = 9.62 (closest match).
The circle with diameter 3.9 has a radius of 1.95 and an area of π x 1.952 = 11.95.
Square
The area of a square is Side x Side.
The square with side 3.1 has an area of 3.1 x 3.1 = 9.61 (closest match).
The square with side 3.5 has an area of 3.5 x 3.5 = 12.25.
The square with side 3.9 has an area of 3.9 x 3.9 = 15.21.
Triangle
The area of a triangle is ½ x Base x Height. Using Pythagoras' theorem it can be shown that the area of an equilateral triangle is Sqrt(3) x Base2 ÷ 4.
The triangle with side 3.9 has an area of Sqrt(3) x 3.9 x 3.9 ÷ 4 = 6.59.
The triangle with side 4.3 has an area of Sqrt(3) x 4.3 x 4.3 ÷ 4 = 8.01.
The triangle with side 4.7 has an area of Sqrt(3) x 4.7 x 4.7 ÷ 4 = 9.57 (closest match).
Puzzle 150
Can you draw a line through all of the edges in this picture?
Each side is broken into 2 or 3 edges, and there are also 7 edges inside that you have to cross. The line must be continuous, and cross each edge exactly once.
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Hint
Try this with a piece of paper.
Answer
There is no possible way to complete the line, there will always be one edge left — or you have to cross an edge twice.
Puzzle 151
The other day I was sitting in my local tavern, The Spyglass, which overlooks the sea, when in sailed my old friend the pirate Captain Conan Drum. "Well, shiver me barnacles!" he roared on seeing me. He too is a bit of a puzzle addict and so, after joining me for a glass of milk and telling me about his latest exploits on the high seas, he couldn't resist showing me his latest conundrum.
He reached into one of his jacket pockets and produced seven gleaming £5 coins, which he then proceeded to arrange on the table in front of me exactly as shown below. "Now, me lad." he said, with a mischievous look in his eyes. "I'll wager you'll not be able to solve this one. Take away two coins from this here arrangement to leave five coins across and three coins going down."
It was clear the wily old sea dog still had one or two tricks up his sleeve, as I couldn't for the life of me see how it could be done. Can you see through his skulduggery and solve it?
Puzzle Copyright © Lloyd King
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Hint
Try saying the puzzle out loud.
Answer
Take away the two coins on the right end of the row of five coins to leave 'five coins, a cross and three coins going down'.
I fell into his trap and misinterpreted what he was actually asking me to do!
Puzzle 152
For being good 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?
Puzzle Copyright © Kevin Stone
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Hint
Start by looking at Jesse's sweets.
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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