Lesson Find the side of a square if distances are given from an interior point to 3 its vertices

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<H2>Find the side of a square if distances are given from an interior point to 3 its vertices</H2>

<H3>Problem 1</H3>From the point inside a square, &nbsp;the distance to three corners are &nbsp;4, &nbsp;5 &nbsp;and &nbsp;6 m, &nbsp;respectively. 
Find the length of the side of a square.

<B>Solution</B>

<pre>
Let ABCD be the square with the side of the length "a" in a coordinate plane,

A = (0,0),  B = (a,0),  C = (a,a)  and  D = (0,a).

Let (x,y) be the point inside the square ABCD with the distance 4 from A, 5 from D  and  6 from B.

Thus we have these three equations ("distances")

x^2     + y^2     = 4^2,      (1)

(a-x)^2 + y^2     = 6^2,      (2)

x^2     + (y-a)^2 = 5^2.      (3)

Making FOIL in equations (2) and (3), I can re-write them in this form

x^2             + y^2 = 16,   (4)    (= same as (1) )

a^2 - 2ax + x^2 + y^2 = 36,   (5)

x^2 + y^2 - 2ay + a^2 = 25.   (6)

Replacing  x^2 + y^2 by 16  in equations (5) and (6), I obtain new equations instead of them

a^2 - 2ax = 20                (7)

a^2 - 2ay =  9                (8)

From equations (7) and (8),  x = {{{(a^2-20)/2a}}},  y = {{{(a^2 - 9)/2a}}}.

Substituting these expressions for x and y into equation (4), you get

{{{(a-20)^2}}} + {{{(a^2 -9)^2}}} = {{{(4a^2)*16}}},

or, simplifying

{{{a^4 - 40a^2 + 400}}} + {{{a^4 - 18a^2 + 81}}} = {{{64a^2}}},

{{{2a^4 - 122a^2 + 481}}} = 0.

From this bi-quadratic equation, you get for {{{a^2}}}, by applying the quadratic formula

{{{a^2}}} = {{{(122 +- sqrt(122^2 - 4*2*481))/(2*2)}}} = {{{(122 +- sqrt(11036))/4}}}.

The smaller value does not work for "a" (it is easy to check), leaving the larger value

{{{a^2}}} = {{{(122 + sqrt(11036))/4}}}

as the only meaningful solution.

Thus  a = {{{sqrt(122 + sqrt(11036))/2}}} = 7.534 (approximately).
</pre>

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