This is one of the ways of solving this problem. Since, the octagon is mentioned to be a regular octagon, it can be inscribed inside a square. This can be imagined by knowing that the sides of the octagon are equal and when extended meet each other at same distance at right angles. See figure for example.


Now, consider that the side of the extended triangle is x and since both sides of the triangle are equal, as shown, the triangle is a 45-45-90 triangle. Hence, the hypotenuse is x\sqrt{2}.
Now one side of the octagon will consist of two such sides of triangles and one side of octagon which has length equal to the length of the hypotenuse. Hence,

side of the square= 2x+x\sqrt{2}
= x\sqrt{2}(\sqrt{2} +1)

Now, area of the square=side^2
= (x\sqrt{2}(\sqrt{2} +1))%^2
= 6*x^2+4\sqrt{2}*x^2

Now, the area of the small triangles at the edges of the square when subtracted will give us the area of the octagon.
Area of the triangle= 0.5*x^2
Area of 4 triangles = 2x^2

Hence, the area of the octagon = 4*x^2 + 4\sqrt{2}*x^2 = approx 9.5*x^2 //by estimating \sqrt{2}=1.4

Now, there are 8 equilateral sides with their length equal to the sides of the octagon.
Are of an equilateral triangle=0.5*base(x\sqrt{2})*height (0.5*(\sqrt{3})*x^2)
Hence, the are of the 8 triangles can be solved to be 2\sqrt{3}*x^2
Now, this is approx= 3.5*x^2 //by estimating \sqrt{3}=1.7

Hence, the area left uncovered= 6x^2

finding the percentage reduction

percentage reduction = area left uncovered/total area of the octagon
=6*100/9.5 >60%
Hence, A must be the answer.

Let me know if you need some clarification


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Kris
Instructor at Aspire4GMAT

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HI

Can you please explain the highlighted statements of your explanation? How did you derive these values?
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