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**lakeheadca****Member**- Registered: 2010-01-31
- Posts: 37

ABCD is a square with area 9. efgc is a square in which EC=1/3(BC). HIJC is s square in which HC=1/3(EC). If this pattern is continued so that there are a total of ten squares with common vertex at C. and in which each square lies inside its predecessor, what is the area of the tenth square?

Answer: 9^-8

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Hi lakeheadca;

That answer is correct.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**lakeheadca****Member**- Registered: 2010-01-31
- Posts: 37

how to solve this problem. please explain!

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,393

ABCD has area 3 x 3 = 9 = 9 ^ 1

EFCG has area 1 x 1 = 1 = 9 ^ 0

HICJ has area 1/3 x 1/3 = 1/9 = 9 ^ -1

KLCM has area 1/9 x 1/9 = 1/81 = 9 ^ -2

..... .... .....

10th square has area 3^ -8 x 3 ^ -8 = 9 ^ -8

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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