The relationship between the areas of the floors is [tex]A_s = \frac{3\sqrt 3}{4} A_t[/tex]
Let the side length of the base of the square building be l, and the side length of the base of the equilateral triangle building be s
So, the perimeter of both buildings is
Both perimeters are equal.
So, we have:
4l = 3s
Divide both sides by 4
[tex]l = \frac 34s[/tex]
Square both sides
[tex]l^2 = \frac{9}{16}s\²[/tex]
The area of an equilateral triangle is:
[tex]A_t = \frac{\sqrt 3}{4}s^2[/tex]
And the area of a square is
[tex]A_s = l^2[/tex]
Make s^2 the subject in [tex]A_t = \frac{\sqrt 3}{4}s^2[/tex]
[tex]s^2 = \frac{4A_t}{\sqrt 3}[/tex]
Substitute [tex]s^2 = \frac{4A_t}{\sqrt 3}[/tex] and [tex]A_s = l^2[/tex] in [tex]l^2 = \frac{9}{16}s\²[/tex]
[tex]A_s = \frac{9}{16} * \frac{4A_t}{\sqrt 3}[/tex]
Evaluate
[tex]A_s = \frac{9}{4} * \frac{A_t}{\sqrt 3}[/tex]
Rationalize
[tex]A_s = \frac{3\sqrt 3}{4} A_t[/tex]
Hence, the relationship between the areas of the floors is [tex]A_s = \frac{3\sqrt 3}{4} A_t[/tex]
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