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Sagot :
Answer: [tex]\frac{1}{\sqrt{2}}[/tex] which is the same as [tex]\frac{\sqrt{2}}{2}[/tex]
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Explanation:
Let's make point J the center of both the octagon and the square.
Connect points E and F to this center to form triangle EFJ.
Angle FJE is 360/n = 360/8 = 45 degrees. We simply split the central angle 360 into 8 equal parts because we have a regular octagon.
Without loss of generality, let's make segment JE to be 1 unit long. This means JF is also 1 unit long. The distance from the center J to any corner point on the octagon is the same. We could scale this radial distance up or down, and still arrive at the same final answer. So it really doesn't matter what JE is.
Use the SAS triangular area formula to get the area of triangle EFG.
[tex]\text{area} = 0.5*\text{side1}*\text{side2}*\sin(\text{angle})\\\\\text{area} = 0.5*JF*JE*\sin(\text{angle FJE})\\\\\text{area} = 0.5*1*1*\sin(45^\circ})\\\\\text{area} = 0.5*\frac{\sqrt{2}}{2}\\\\\text{area} = \frac{\sqrt{2}}{4}[/tex]
The area of triangle EFJ is exactly [tex]\frac{\sqrt{2}}{4}[/tex] square units when we start off with JE = 1. We have 8 identical such triangles that form the entire octagon.
Multiply the previous area by 8 to find the area of the full octagon.
[tex]8*(\text{area of triangle}) = 8*\frac{\sqrt{2}}{4} = 2\sqrt{2}[/tex]
Let [tex]n = 2\sqrt{2}[/tex] represent the area of the octagon when JE = 1.
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Now to compute the area of the square ACEG.
Like earlier, split the figure into identical four triangles.
Let's find the area of triangle GJE.
[tex]\text{area} = 0.5*\text{side1}*\text{side2}*\sin(\text{angle})\\\\\text{area} = 0.5*JG*JE*\sin(\text{angle GJE})\\\\\text{area} = 0.5*1*1*\sin(90^{\circ})\\\\\text{area} = 0.5[/tex]
Keep in mind that the diagonals of any square are always perpendicular, which is why angle GJE is 90 degrees.
Four such identical triangles compose the square, so 4*0.5 = 2 square units is the area of the square when we let JE = 1. We'll use this square area later, so let m = 2.
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The last step is to divide the values of m and n, which are the square and octagon areas in that order (when JE = 1).
[tex]\frac{m}{n} = \frac{2}{2\sqrt{2}}\\\\\frac{m}{n} = \frac{1}{\sqrt{2}}\\\\\frac{m}{n} = \frac{1*\sqrt{2}}{\sqrt{2}*\sqrt{2}}\\\\\frac{m}{n} = \frac{\sqrt{2}}{2}\\\\[/tex]
Interestingly, this is exactly equal to the value of sin(45). That fraction is roughly equal to 0.707, meaning that the square's area is roughly 70.7% of the octagon's area.
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