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by artstudent » Wed Jul 13, 2011 12:29 am
Can someone explain this:
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by MBA.Aspirant » Wed Jul 13, 2011 1:00 am
artstudent wrote: Image
The side of the square and half of it form a triangle where AC is the hypotenuse.

so say square side = x, and half = 1/2x

x^2 + (1/2x)^2 = 20^4

x^2 + x^2/4 = 400

5/4 x^2 = 400

x = √320 <= square side

area of square = side^2 = (√320)^2 = 320
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by artstudent » Wed Jul 13, 2011 1:03 am
How do you know that AC is half?
MBA.Aspirant wrote:
artstudent wrote: Image
The side of the square and half of it form a triangle where AC is the hypotenuse.

so say square side = x, and half = 1/2x

x^2 + (1/2x)^2 = 20^4

x^2 + x^2/4 = 400

5/4 x^2 = 400

x = √320 <= square side

area of square = side^2 = (√320)^2 = 320
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by Frankenstein » Wed Jul 13, 2011 1:21 am
artstudent wrote:How do you know that AC is half?
MBA.Aspirant wrote:
artstudent wrote: Image
The side of the square and half of it form a triangle where AC is the hypotenuse.

so say square side = x, and half = 1/2x

x^2 + (1/2x)^2 = 20^4

x^2 + x^2/4 = 400

5/4 x^2 = 400

x = √320 <= square side

area of square = side^2 = (√320)^2 = 320
Hi,
Consider, P and Q as the other 2 vertices of square.
PC^2 = AC^2 - AP^2
QC^2 = BC^2 - BQ^2. Giebn that AC = BC. AP and BQ are sides of the square. So, they are equal. This is enough to prove that PC and QC are equal.
Cheers!

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by amit2k9 » Wed Jul 13, 2011 11:32 pm
from C drop a perpendicular on AB.

thus 400 = (0.5x)^2 + x^2

x= 320^(1/2). AB=x and altitude = AB = x respectively.
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by MBA.Aspirant » Thu Jul 14, 2011 3:23 am
Image

Another way to prove that PC = CQ is through the angles

given AC =BC so ACB is an isosceles and angle CAB = CBA. call both angles x

since this a square so we know its angles = 90, and angle CAP = CBQ = 90 -x, making the sides opposite them equal as well. so PC = CQ
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