# The Matter of Squares

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

Solution:

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