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Math Expert V
Joined: 02 Sep 2009
Posts: 55274
ABCDEFGH is a regular octagon. What is the area of triangle ABC?  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 67% (01:31) correct 33% (01:46) wrong based on 33 sessions

### HideShow timer Statistics ABCDEFGH is a regular octagon. What is the area of triangle ABC?

(1) AB = 2.
(2) AD = 2(√2 + 1).

Attachment: 2017-07-24_1032.png [ 6.12 KiB | Viewed 6511 times ]

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ABCDEFGH is a regular octagon. What is the area of triangle ABC?  [#permalink]

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Bunuel wrote: ABCDEFGH is a regular octagon. What is the area of triangle ABC?

(1) AB = 2.
(2) AD = 2(√2 + 1).

Attachment:
The attachment 2017-07-24_1032.png is no longer available

It should be D

Image attached with labeling that has been used through the question.

Statement 1: Sufficient

Since it is a regular ocatgon, all sides are equal and equal to $$AB = BC = 2$$. We also know internal angle of regular octagon is $$135$$ and angles marked in Indigo are consequently $$45$$ degrees. $$\triangle ABI$$ and $$\triangle BKC$$ are isosceles and have both the same hypotenuse of 2. We can use these to measure $$BK$$ which is the height of the $$\triangle ABC$$ (Consider $$\triangle ABC$$ as an inverted triangle with $$AB$$ base which makes $$BK$$ height). This can then be used to calculate area.

$$\frac{1}{2}*base * height$$.

Statement 2: Sufficient

Consider isosceles triangles $$\triangle ABI$$ and $$\triangle CDJ$$ and square $$BCIJ$$. Chord $$AD$$ is a sum of $$AI$$, $$IJ$$ (which incidentally is equal to $$AB$$) and $$JD$$. We know the two triangles are 45,45,90 so ratio of Hypotenuse $$(AB = CD = IJ = 2)$$ to arms is $$X$$, $$X$$ and$$X*\sqrt{2}$$

$$X + X + X\sqrt{2} = 2 (\sqrt{2} + 1).$$

$$2X + X\sqrt{2} = 2\sqrt{2} +2$$

$$X (2 + \sqrt{2}) = 2\sqrt{2} +2$$

$$X = (2\sqrt{2} +2)/(2 + \sqrt{2})$$

$$X = \sqrt{2} (2 + \sqrt{2})/(2 + \sqrt{2})$$

$$X = \sqrt{2}$$

$$X$$ is again the height of the inverted $$\triangle ABC$$ and This can be used to calculate $$AB = X*\sqrt{2}$$ and this can be used to calculate $$Area = \frac{1}{2} * base * height$$.

Edit: Original diagram wasn't really representative of a regular Octagon so realized my mistake. Credits to a friend in a WhatsApp group who calculated numeric value of $$X$$.
Attachments octagon.png [ 17.24 KiB | Viewed 4821 times ]

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Originally posted by jedit on 24 Jul 2017, 10:29.
Last edited by jedit on 24 Jul 2017, 12:10, edited 1 time in total.
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Re: ABCDEFGH is a regular octagon. What is the area of triangle ABC?  [#permalink]

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its D
A is easy as we know that regular octagon will have same interior angle, we know the side ab and bc are of same length and we can find the angle B. the other two angles will be of same value as two sides ab and bc are same and hose angles are opposite to these lengths...
Coming back to statement B, it gives the measurement of AD and HE will also be of same length and AD and HE will make a regular rectangle. Similarly BG and CF make another rectangle. working on these, one can find the area of ABC
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Re: ABCDEFGH is a regular octagon. What is the area of triangle ABC?  [#permalink]

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Answer: D
Basically, isosceles trapezoid ABCH (or any other combination) is key to answer the question. Re: ABCDEFGH is a regular octagon. What is the area of triangle ABC?   [#permalink] 24 Jul 2017, 13:08
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# ABCDEFGH is a regular octagon. What is the area of triangle ABC?

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