A regular octagon (a polygon with 8 sides of identical lengt

The explanation is available in the attachment. Please check the same

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

Assume length of side = 1
Area of 8 triangles = 8*sqrt(3)/4 = 3.5
Area of a Octagon = 4.8 * side square = 4.8 * 1 = 4.8

Remaining area = 4.8 - 3.5 = Approx(28 %) = Rounding it to 30 %

Answer is D

Please find the solution in attachment

Good question! We can use some estimation techniques to solve this.

Consider the attached image.

Already by looking at the diagram we can estimate that the equilateral triangle covers greater than 2/3rd of the bigger (Octagon's) triangle. That means, lesser than 1/3rd or 33.3% is uncovered. We just need to decide between 20% or 30%. To ascertain which one, we can employ the following technique just using known angles.

We know a regular octagon will have equal central angles, each equal to \(45^{\circ}\). If we create a triangle at one of the octagon's sides, the other two angles will be \(72.5^{\circ}\) each. At the same side, we are folding over an equilateral triangle, so each angle in that as we know is \(60^{\circ}\).

So, the uncovered area is really the difference in the measure of these angles of the two triangles.

We can see that 45/60 = 3/4 or the equilateral triangle almost covers 75% of Octagon's triangle. 25% is uncovered. Plus, \(12.5^{\circ}\) on each side that is left uncovered. To calculate that in percent, we'll calculate the proportion which is covered and subtract from 1. So, \(120^{\circ}\)/\(135^{\circ}\) = 0.88 approx, of covered area or 0.12 of uncovered area.

Thus, total percentage uncovered = 25% + 1.2% = 26.2% which is closer to 30%, hence answer -> D.

HI

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

Hi,

how do we derive.. Side of square = 1/\sqrt{2} + 1 + 1/\sqrt{2} = 1 + \sqrt{2} ?

The best way is to divide the octagon into a rectangle, 2 squares and 4 triangles .

The triangles are right angled , hence if the hypotenuse is 1 , then the side will be 1/root(2) .

The whole area of octagon can be obtained from the same .

I think this Q can be solved by simple guess:

Area uncovered has be less than 50%. More over 20% is too less.
Now the options left: 30% & 40%.
40% in nearer to 50%, possibly will not be the answer.

hence 30 percent.

Solution:

If the side length of the regular octagon is s, then the area of the octagon is given by 2(1 + √2)(s^2). Partitioning the regular octagon into a bunch of isosceles right triangles, rectangles and a square is one way of obtaining this formula.

Since the regular octagon has 8 sides, 8 equilateral triangles with a side length of s will be drawn and folded over the octagon. Since the area of each equilateral triangle is [(s^2)√3]/4 and since there are 8 such equilateral triangles, the total area of the equilateral triangles is 8 * [(s^2)√3]/4 = 2√3(s^2). Hence, the uncovered area equals 2(1 + √2)(s^2) - 2√3(s^2) = 2(s^2)(1 + √2 - √3). It follows that the ratio we need to approximate is [2(s^2)(1 + √2 - √3)]/[2(1 + √2)(s^2)] = (1 + √2 - √3)/(1 + √2). Since √2 is approximately 1.4 and √3 is approximately 1.7, we get:

(1 + √2 - √3)/(1 + √2)

(1 + 1.4 - 1.7)/(1 + 1.4)

0.7/2.4 ≈ 0.29

Expressed as a percentage, this is 29%, and the closest answer choice is D.

Answer: D

because we have a Regular Octagon: Each interior Angle is 135 Degrees and Each Exterior Angle taken at One Vertex = 45 Degrees


(1st) Find the Approximate Area of Regular Octagon

Draw 2 Exterior Angles from Adjacent Vertices and Connect them to Form a Triangle at each of the 4 Corners (of the "Stop sign")

Each of the 4 Triangles will be 45 - 45 - 90 Triangles, and with these 4 Right Triangles Added on to the 4 Corners we have created a Square.

Let the Side of the Regular Octagon = 1

Then the Side of Each Leg of Exterior Isosceles Triangles will measure = (1) / (sqrt(2))

Taking 1 Side of the Square = 1/sqrt(2) + 1 + 1/sqrt(2) = 1 + sqrt(2)

The Area of the Regular Octagon can be Found as follows:

Area of Regular Octagon = (Area of our Created Square of Side = 1 + sqrt(2) ) - (the Areas of the 4 Congruent Isosceles Right Triangles we created on the 4 Corners with Leg = 1/sqrt(2)

Area of Regular Octagon = (1 + sqrt(2))^2 - (4) [ (1/2) * (1/sqrt(2)) * (1/sqrt(2)) ]

Area of Regular Octagon =approx.= (2.4)^2 - (4) * (1) = approximately 4.76


(2nd) Each of the Equilateral Triangles will have the Same Side Length as the Regular Octagon ----> we assigned 1

Area of the 8 Equilateral Triangles that are Folded over into the Regular Octagon:

= (8) * [ (s)^2 * sqrt(3) / 4 ]

= 8 * [ (1) * sqrt(3) / 4]

= 2 * sqrt (3) = approximately = 3.46



(3rd) the Proportion of the Regular Octagon left Uncovered is =

[ (Area of Regular Octagon) - (Area of 8 Triangles) ] / (Area of Regular Octagon)

= [ (4.76 - 3.46) ] / (4.76)

= 1.3 / 4.76 -------> works out to approximately (7/24) which is a little less than 30% (approximately around 28 to 29 percent)

(D) at 30% is the Closest Approximation

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.