Bunuel

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
DImage attached with labeling that has been used through the question.
Statement 1: SufficientSince 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: SufficientConsider 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 \(X\).
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