Let’s look at a question today which encompasses most of what we have discussed in this topic. This will be the last post on this topic for a while now. We assume that after going through these posts thoroughly, if you come across any question on ‘this inscribed in that’, you should be able to handle it. Just a reminder, keep in mind the symmetry of the figures you are handling.
Question: Two identical squares EFGH and JKLM are inscribed in a square ABCD such that AJ:JE:EB = 1:√2:1. What is the area of the octagon obtained by joining points E, K, F, L, G, M, H and J if AB = (2 + √2) cm?
(A) 8 cm^2
(B) 4 cm^2
(C) 4√2 + 2 cm^2
(D) 4(√2+1) cm^2
(E) 2(√2 + 1) cm^2
We are given the length of side AB = (2 + √2) cm
Also AJ:JE:EB = 1: √2:1
AJ + JE + EB = (2 + √2) = a + √2a + a
a = 1 cm
AJ = 1 cm
JE = √2 cm
EB = 1 cm
Now let’s make the octagon as required.
Since AJ = 1 cm and AH = 1 cm, JH = √(1^2 + 1^2) = √2 cm
Notice that the octagon is a regular octagon: JE = KF = LG = MH = √2 cm. Also, HJ = EK = FL = GM = √2 cm
The area of the octagon = Area of trapezoid MHJG + area of rectangle JELG + Area of trapezoid KFLE
Area of trapezoid MHJG = (1/2) *(Sum of parallel sides)*Altitude = (1/2)*( √2 + 2 + √2)*(1) = (√2+1) cm^2
Area of trapezoid KFLE = (√2 +1) cm^2 (by symmetry)
Area of rectangle JELG = √2*(2 + √2) = 2(√2 + 1) cm^2
Area of the octagon = (√2 + 1) + 2(√2 + 1) + (√2 + 1) = 4(√2 + 1) cm^2
Hope you see that it doesn’t matter how the question setter twists the concepts, they are still easy to apply if you understand them well!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!