Consider regular octagon ABCDEFGH
Its longest diagonal would be AE or BF or CG or DH.
Let us try to find out AE.
Join AD and draw BP AD and CQ AD.
PQ = a
AP = QD
a^{2} = BP^{2} + AP^{2} => a^{2} = 2 AP^{2} {since BP=AP}
a = 2AP => AP = a/(2)
AD =AP + PQ + QD = a/(2) + a + a/(2)
=>a + a2
AE^{2} = AD^{2} + DE^{2}
AE^{2} = (a + a2) 2 + a^{2}
AE^{2} = (a^{2} + 2 x a x 2 + 2a^{2}) + a^{2}
AE^{2} = a^{2} (1 + 22 + 2) + a^{2}
=> a^{2} (4 + 22)
Shortest diagonal = AC or CE
AC^{2} = AB^{2} + BC^{2} – 2AB × BC cos135 degree
(Alternatively, we can deduce this using AC^{2} = AQ^{2} + QC^{2}. We use cosine rule just to get some practice on a different method.)
= a^{2} + a^{2} – 2a^{2} × ((−1)/(2))
= 2a^{2} + 2a^{2}
= a^{2} (2 + 2)
AE^{2} = a^{2} (4 + 22)
"AE^{2}" /"AC^{2}" = "a^{2} (4 + 22)" /"a^{2} (2 + 2)" = 2
"AE" /"AC" = 2
Remember, for a regular octagon.
Each internal angle = 135 degrees.
Each external angle = 45 degrees.
So, we get a bunch of squares and isosceles right–angled s if we draw diagonals.
A regular hexagon breaks into equilateral triangles. A regular octagon breaks into isosceles right angled triangles.
Answer choice (d)
Correct Answer: 2 : 1