The square ACEG shown below has an area of 36 units squared. What is t

\(?\,\,:\,\,x\,\,{\text{to}}\,\,\max \,\,S\left( {ABDFG} \right)\,\,\,\, \Leftrightarrow \,\,\,x\,\,{\text{to}}\,\,\min \,\,\,\left[ {S\left( {\Delta {\text{BCD}}} \right) + S\left( {\Delta DEF} \right)} \right]\)

\(S\left( {\Delta {\text{BCD}}} \right) + S\left( {\Delta DEF} \right) = \frac{{{x^2}}}{2} + \frac{{{{\left( {6 - x} \right)}^2}}}{2} = \frac{{2{x^2} - 12x + 36}}{2} = {x^2} - 6x + 18\)

\(?\,\,\,:\,\,\,x\,\,{\text{to}}\,\,\min \,\,{x^2} - 6x + 18\,\,\,\,\, \Leftrightarrow \,\,\,\,\,? = x = {x_{vert}} = - \frac{b}{{2a}} = - \frac{{ - 6}}{2} = 3\)


Almost no lines, almost no arguments, almost no effort were needed here. Just the old and powerful good math.


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Great question. +1

I used symmetry first.
I doodled for a few seconds and changed what would be ∆ BDF into an isosceles right triangle.

Testing numbers works, too.

Symmetry
• If a perimeter is fixed, the area of a polygon is maximized when it is most symmetrical.

The most symmetrical rectangle also has the greatest area: a square
The most symmetrical triangle also has the greatest area: an equilateral triangle
A convex regular polygon has both the most symmetry and the greatest area

• Make the figure a pentagon that has one line of symmetry
Its "roof" consists of two congruent isosceles right triangles
Its rectangle consists of two squares with side length \(s=3\)

• \(x = (6 - x)\) (from the diagram)
The "height" of the outer square has length of 6.
The symmetric right isosceles triangles must split that length equally.
So
\(x = 6-x\)
\(2x=6\)


Test numbers

•If \(x=4\), then ∆ BDH with legs of length 4 has area,
\(A=\frac{s^2}{2}=\frac{4^2}{2}=8\)
∆ DFH with legs of length 2 has area, \(A=\frac{2^2}{2}=2\)
Combined area of right triangles: \((8+2)=10\)
Area of polygon: (area of square) - (area of triangles)
Area of polygon: \((36-10)=26\)

•Try \(x=3\)
Combined area of the two right triangles BDH and DFH,
\(A=(2*\frac{3^2}{2})=9\)
(Area of square) - (area of triangles) =
Area of polygon: \((36-9)=27\)

That is the maximum area.

•If we use \(x=5\), ∆ BDH alone will have area \(\frac{25}{2}=12.5\), and
Polygon area will = \((36-12.5)=23.5\)

The farther apart that \(x\) gets from \((6-x)\),
the more that the area of the polygon ABDFG decreases.

The area of the polygon is maximized when

Answer

Hi, generis!
Thank you for the kudos and for your nice contributions!
Regards,
Fabio.

here is my train of thoughts

if area is 36 then side is 6

looking at the figure, x definately cant be four or five. it looks like the length of x is 2. (i thought to myself) but since it is GMAT question i clicked on C

that is my approach

i liked this line:

"Almost no lines, almost no arguments, almost no effort were needed here" Figuratively sounds like a nice slogan in the marketing campaign for selling everything in the 21st century

Hi, Dave13!

I totally agree... to be honest (as usual), I know (and I am happy to say) it takes time, effort, dedication and good guidance to learn anything deeply and, as a sub-product, to find quick and powerful solutions "easily" and "instantaneously".
The sentence: "Seriousness and discipline for hard work ARE prerequisites, tough." is in the FAQ present in the homepage of my website, by the way.
Thank you for your contributions!

Regards,
Fabio.

­It is question where one needs to change the value of x within some limits(here 0 < x < 6) such that some condition is fulfilled(max area ABDFG). Thus, DF too changes. If we apply the logic wherein in a circle the the max area of a rectangle is that of a square then here too the value of x should be equal to DF. So, we can simply make a hypothetical rectangle(square eventually) BDFH where H is at AG such that BD = HF and DF = HA. 

This rectangle BDFH would have max area only if its vertices are at the mid point of the sides of square ACEG. 
Hence, 
x = 3.

Answer C.