The question is from regular polygon. It tests our understanding of regular octagon. CAT Geometry questions are heavily tested in CAT exam. Make sure you master Geometry problems. Triangles are heavily tested, the wonderful infinite-sided polygon that is the circle is also heavily tested. In between these two lies this great mass of regular polygons.
Question 11: What is the ratio of longest diagonal to the shortest diagonal in a regular octagon?
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
a2 = BP2 + AP2 => a2 = 2 AP2 {since BP=AP}
a = √2AP => AP = a/(√2)
AD =AP + PQ + QD = a/(√2) + a + a/(√2)
=> a + a√2
AE2 = AD2 + DE2
AE2 = (a + a√2) 2 + a2
AE2 = (a2 + 2 * a * √2 + 2a2) + a2
AE2 = a2 (1 + 2√2 + 2) + a2
=> a2 (4 + 2√2)
Shortest diagonal = AC or CE
AC2 = AB2 + BC2 – 2AB × BC cos135 degree
(Alternatively, we can deduce this using AC2 = AQ2 + QC2. We use cosine rule just to get some practice on a different method.)
= a2 + a2 – 2a2 * ((−1)/√2)
= 2a2 + √2a2
= a2 (2 + √2)
AE2 = a2 (4 + 2√2)
AE2/AC2 = a2 (4 + 2√2) / a2 (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.
The question is "What is the ratio of longest diagonal to the shortest diagonal in a regular octagon?"
Choice D is the correct answer.
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