Given a square ABCD with two consecutive vertices, say A and B on the positive x-axis and positive y-axis respectively. Suppose the other vertex C lying in the first quadrant has coordinates (u , v). Then find the area of the square ABCD in terms of u and v.
Assume:
A is on x-axis
B is on y-axis
C is next in clockwise order in first quadrant.
A: (a,0)
B: (0,b)
C: (u,v) = (b, a+b)
D: (x1,y1) = (a+b, a)
Area = a^2 + b^2, but need this in terms of u and v.
Label the origin 'O'.
Draw a line through C and perpendicular to the y-axis, label that intersection 'E'.
Then triangles AOB and BEC are congruent.
So BE = a and EC = b.
(u,v) = (b, a+b)
a = v-u
b = u
Area = a^2 + b^2 = 2u^2 - 2uv + v^2
https://www.desmos.com/calculator/f1l4ogoqhw
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Posted by Larry
on 2024-08-31 18:59:27 |
solution
