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.
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.
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.
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).
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.
[Puzzle Code = ZKPQ]
Direct Link: www.brainbashers.com?ZKPQ
Hint: Try this with a piece of paper.
There is no possible way to complete the line, there will always be one edge left - or you have to cross an edge twice.
This puzzle is the same as the famous 'Seven Bridges of Konigsberg' problem first solved by Euler.
In that problem, the task was to find a closed path that crossed each of the seven bridges of Konigsberg (now Kaliningrad, Russia) exactly once.
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?
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!
For being good at the garden fete, four children were each given two sweets.
Jesse had an orange sweet.
One child who had a red one 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.
Jules 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?
Hint: Remember that they were two sweets of each colour.
Jesse had an orange and a green sweet.
Jamie had an orange and a blue one.
Jules had a red and a blue sweet.
Jo had a green and a red one.