Sometime what you need is some logically deduction base on, of course, some mathematical principles.
Below is one good example of "deduction" type of math solving.
Let start the challenge, and have some fun!
Above you will find 3 squares.
Do note that the 2 yellows are of the same area and 1 blue of area bigger than the yellow ones.
If the total area of the 3 squares are 57 sq cm, determine the area of the bigger blue square.
I believe you will enjoy this math question.
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15 comments:
Any hint? I can't imagine this is possible...
Don't give up. It is easy. Just a bit of your thinking juice will do.
Hint: What is the formula for the area of a square?
Have fun.
As I see it you can't solve this as long as you can't deduce the ratio between the size of these two squares.
The area of the blue one is 25sq cm
Yes, finally the challenge is correctly answered.
The yellow areas are (4x4)x 2 pieces = 32 sq cm.
If the total is 57 sq cm, the blue area will be 57 - 32 = 25 sq cm.
Simply isn't it.
With a few intelligent guesses / deductions (or even with one guess only), everyone should be able to come to this conclusion.
:-)
any no graeter than 29
any natural no greater than 28
Hello maths,
any number greater than 29?
Hmmm....
Interesting. Maybe there is a lead somewhere for that answer.
What is it? How did you arrive at that answer?
49 works as the area of the blue square as well.
7x7=49
2x2 =4
2x2=4
49+4+4 =57 !!!
Well done! Yes, fully agree that 49 is also an answer.
Now hope that everyone understands what really is "deduction" maths. Interesting?
area of yellow squares is x
area of blue square is y
2x + y =57sq cm
y has to be larger than x, y>x
if x=18 then y=21 satisfies y>x
if x=19 then y=19 does not satisfy y>x
therefore the blue square is 21 centimetres squared or a little bigger because of the way the proportions look in the diagram. Does anyone else agree?
area of yellow squares is x
area of blue square is y
2x + y =57sq cm
y has to be larger than x, y>x
if x=18 then y=21 satisfies y>x
if x=19 then y=19 does not satisfy y>x
therefore the blue square is 21 centimetres squared or a little bigger because of the way the proportions look in the diagram. Does anyone else agree?
The logic in solving the question is correct. There are many answers to it. Your 2x + y tells the real thinking behind this question. Great effort. Kudos to you, slominski.
Maths is interesting, right?
wow i really enjoyed this one, it has multiple answers, but the simple logic is take, the area of the yellow boxes as 'Y'=y*y (where y are the sides of the yellow box), and similarly take the area of the blue box as 'B'=b*b (where b would be the area of Blue box), now lets see it in an equation,
Y+Y+B=57;
2Y+B=57;
now look at our 2Y, anything multiplied with 2 will yield an even number, what are even numbers? 2,4,6,8,10,12,14,16,18,20, or, simply:-
x2,x4,x6,x8,x0 where x can be any natural number. therefore to get a x7(57 actually) in our total sum of areas the area of the blue square should be something which adds to a x'even' to produce an x7. giving us results x7-x2=x5, x7-x4=x3, x7-x6=x1, x7-x8=x9(borrowing a 1 from x(in x7)). x5,x3,x1,x9. Remember these areas should be less than 57 and also give enough space for 2 more equal squares. So areas less than 57 with x5,x3,x1,x9 are 5*5=25,7*7=49,3*3=9. now we cant take 3*3=9 because this is the area of the blue box, and we know the area of the yellow box should be less than the area of the blue box that leaves us to 1 case, (2*2)*2 + (3*3) = 17(not equal to 57). whereas taking 5*5=25(blue box area), we get 4*4=16(yellow box area), and if we take 7*7=49 as blue box area, we get 2*2=4(area of yellow box).. I hope that was not so confusing..
Hi anildhan,
My respect to you.
You are able to read my intention of posting this maths challenge well.
Yes, it is the "even" number for the 2Y that starts the game. I had this in mind to make the solution simpler, and mentally done with.
Good explanation, though a bit long. I do not mind. Any serious reader will want to get to the truth and "fun" of solving any maths question.
Thanks for sharing your answers in the comment. :-)
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