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

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!

### 4 Responses

1. Graeme says:

Hi there –
nice question.

In the interest of speed, could you also do this by just working out the area of the big square (ABCD) and subtracting the four small triangles in the corner cut out by the octagon?

ie. given that we know AB=BC=CD=DA= 2 + sqrt(2), the area of ABCD = 6 + 4*sqrt(2)

The area of the four small triangles is quite easy, just A = 1/2*1*1*4 = 2

So area of the octagon is: 6+4*sqrt(2) – 2 = 4 ( 1 + sqrt(2)) … Ans D

I tend to think that way anyway, only cause I don’t always remember the formula for area of a trapezoid :-)

Cheers,
GB

• Karishma says:

Yes, absolutely. By habit, I like to add areas. You can just as well get your answer by the subtraction you used.

2. Jason says:

This way would be easier,

Subtracting the area of four triangles from the square ABCD.
Then,
(2+root2)^2 – (1x1x1/2)x4 = 4+4root2

3. Gourav says:

I think the easiest way to solve this problem would be to figure out that the octagon in question is regular and that the area of each triangle inside the octagon is 1/2*root2*(1+1/root2) and then multiply this value by 8 as there are 8 triangles.