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Find the sum of the areas of the squares ...................

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by pzazz12 » Wed Sep 22, 2010 3:59 am
A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in the same way and this process continues indefinitely. If the side of the first square is 4 cm .determine the sum of the areas of the squares.

A. 32 sq.units
B. 48 sq.units
C. 64 sq.units
D. 12 sq.units
E. 120 sq.units


Help me to solve it .............
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by msbinu » Wed Sep 22, 2010 5:31 am
pzazz12 wrote:A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in the same way and this process continues indefinitely. If the side of the first square is 4 cm .determine the sum of the areas of the squares.

A. 32 sq.units
B. 48 sq.units
C. 64 sq.units
D. 12 sq.units
E. 120 sq.units


Help me to solve it .............
Are the options given correct ?

I got the answer as 28 :(
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by beatthegmatinsept » Wed Sep 22, 2010 5:42 am
msbinu wrote:
pzazz12 wrote:A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in the same way and this process continues indefinitely. If the side of the first square is 4 cm .determine the sum of the areas of the squares.

A. 32 sq.units
B. 48 sq.units
C. 64 sq.units
D. 12 sq.units
E. 120 sq.units


Help me to solve it .............
Are the options given correct ?

I got the answer as 28 :(
How do you get to 28? How do you know when to stop making squares within squares, since the problem saying its indefinite?
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by lkm » Wed Sep 22, 2010 6:12 am
IMO the answer should be (A).

I don't remember the exact mathematical rule but the area cannot be go beyond the double of the initial square.

Some observations:-

Side of 1st square = 4
Area of 1st square = 16

Side of 2nd square = 2 root 2
Area of 2nd square = 8

Side of 3rd square = 2
Area of 3rd square = 4

So if you see, each time area of next square is reduced by half. So you can easily make a Geometrical progression.

Sum = 16 + 8 + 4 + 2 + 1 + 1/2 + 1/4 + ....

You can rewrite the above equation as:

Sum = 16 [1 + (1/2)^1 + (1/2)^2 + (1/2)^3 + (1/2)^4 + .... infinity]

So formula for sum of infinite series is a/(1-r)

Here a = 16
and r = 1/2

So sum = 16 (1-1/2) = 16 x 2 = 32

Hence answer is (A)
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by kaushiksin » Wed Sep 22, 2010 6:16 am
Area of first square = 4*4 = 16
Area of second square = 16 / 2 = 8

So the area gets decreased by the factor of 1/2.

This is taking a form of infinite geometric progression i.e. a, ar, ar*r,.....
where r is the common ratio.

Sum of areas = a / (1-r)
= 16 / (1-1/2)
= 16 * 2
= 32

Answer is choice (A).
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by msbinu » Wed Sep 22, 2010 8:42 am
beatthegmatinsept wrote:
msbinu wrote:
pzazz12 wrote:A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in the same way and this process continues indefinitely. If the side of the first square is 4 cm .determine the sum of the areas of the squares.

A. 32 sq.units
B. 48 sq.units
C. 64 sq.units
D. 12 sq.units
E. 120 sq.units


Help me to solve it .............
Are the options given correct ?

I got the answer as 28 :(
How do you get to 28? How do you know when to stop making squares within squares, since the problem saying its indefinite?
Again my mistake. I missed the indefinite part in the question :(
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