Question 1082138: From the point inside the square, the distance in three corners are 4,5, and 6, respectively. Find the length of each side of the square.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Drawing
Image generated by GeoGebra (free graphing software).
Let x be the side length of the square ABCD with point P inside the square. Negative lengths do not make sense so we'll make x > 0.
Let's make
PA = 4
PB = 5
PC = 6
Which are the given distances from P to each of the three corners.
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Focus on triangle APB only for now
Let t = angle ABP which is some angle in degrees.
Note: we don't need to know or solve for t itself, but we will use it indirectly through cos(t).
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Using the law of cosines, we can say
c^2 = a^2+b^2-2*a*b*cos(C)
4^2 = x^2+5^2-2*x*5*cos(angle ABP)
16 = x^2+25-10x*cos(t)
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Isolate the expression cos(t)
16 = x^2+25-10x*cos(t)
16-25 = x^2+25-10x*cos(t)-25
-9 = x^2-10x*cos(t)
x^2-10x*cos(t) = -9
x^2-10x*cos(t)-x^2 = -9-x^2
-10x*cos(t) = -9-x^2
10x*cos(t) = x^2+9
cos(t) = (x^2+9)/(10x) ... call this equation (1)
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Move onto triangle BPC. Focus only on this triangle.
Since angle ABC is a right angle, this makes
angle ABP+angle PBC = 90
t+angle PBC = 90
angle PBC = 90-t
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By the law of cosines,
c^2 = a^2+b^2-2*a*b*cos(C)
6^2 = x^2+5^2-2*x*5*cos(angle PBC)
6^2 = x^2+5^2-2*x*5*cos(90-t)
36 = x^2+25-10x*cos(90-t)
36 = x^2+25-10x * [ cos(90-t) ]
36 = x^2+25-10x * [ cos(90)*cos(t)+sin(90)*sin(t) ]
36 = x^2+25-10x * [ 0*cos(t)+(1)*sin(t) ]
36 = x^2+25-10x*sin(t)
36-25 = x^2+25-10x*sin(t)-25
11 = x^2-10x*sin(t)
x^2-10x*sin(t) = 11
-10x*sin(t) = 11-x^2
sin(t) = (11-x^2)/(-10x)
sin(t) = (x^2-11)/(10x)
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Now substitute in the pythagorean trig identity
sin(t) = (x^2-11)/(10x)
sqrt(1-cos^2(t)) = (x^2-11)/(10x)
1-cos^2(t) = [(x^2-11)/(10x)]^2
Then plug in cos(t) = (x^2+9)/(10x) which is equation (1) shown above and get everything to one side.
1-cos^2(t) = [(x^2-11)/(10x)]^2
1-(cos(t))^2 = [(x^2-11)/(10x)]^2
1-((x^2+9)/(10x))^2 = ((x^2-11)/(10x))^2
1-((x^2+9)/(10x))^2 - ((x^2-11)/(10x))^2 = 0
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Let
g(x) = 1-((x^2+9)/(10x))^2 - ((x^2-11)/(10x))^2
solving g(x) = 0, aka finding the roots of g(x), will yield us the potential solutions for x.
This function is a bit messy fourth degree polynomial so let's turn to a graphing calculator.
I'm using GeoGebra to find the roots, but any graphing calculator or software will do.
I get the following approximate solutions:
{x = -7.069593879182, x = -1.421563358924, x = 1.421563358924, x = 7.069593879182}
which is shown visually here
We can toss out x = -7.069593879182 and x = -1.421563358924 since x > 0.
We can also eliminate x = 1.421563358924 because x must be larger than 6 so P is contained inside the square.
The only possible answer is approximately 7.069593879182
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Answer: The side length of the square is approximately 7.069593879182 units.
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