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Difficulty:
55%
(hard)
Question Stats:
64% (02:23) correct
36%
(02:10)
wrong
based on 120
sessions
History
Date
Time
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Not Attempted Yet
The square ACEG shown below (in the image attached) has an area of 36 units squared. What is the value of x that maximizes the area of the polygon ABDFG?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Source: https://www.GMATH.net
Attachment:
GMATH_figure0271.gif [ 4.58 KiB | Viewed 4352 times ]
fskilnik
Look at the attached figure..
so we have two triangles whose area when added should be minimum....
\(\frac{1}{2}*x*x+\frac{1}{2}(6-x)(6-x)=\frac{1}{2}*(x^2+x^2-12x+36)=x^2-6x+18\)
now when \(x^2-6x+18\) is minimum, will the required area be MAX..
ways to find the minimum of the equation..
A) solve the equation
\(x^2-6x+18=x^2-6x+9+9=(x-3)^2+9\)
(x-3)^2 is least when it is 0, so x=3
B) derivative..
derivative of x^2-6x+18 is 2x-6=0...x=3
C) formula
this is equation of parabola of form ax^2+bx+c and is minimum at \(\frac{-b}{2a}=\frac{-(-6)}{2*1}=\frac{6}{2}=3\)
ways 1 and 3 are the best
ans C
Attachments
GMATfigure0271.gif [ 4.05 KiB | Viewed 3807 times ]
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General Discussion
Posts: 1,348
fskilnik
Almost no math is needed here.
Answer choice A implies the following:
BCD is a 45-45-90 triangle in which each leg has a length of 1.
DEF is a 45-45-90 triangle in which each leg has a length of 5.
Answer choice E implies the following:
BCD is a 45-45-90 triangle in which each leg has a length of 5.
DEF is a 45-45-90 triangle in which each leg has a length of 1.
In each case, one of the 45-45-90 triangles has a leg of 1, while the other 45-45-90 triangle has a leg of 5.
Implication:
The two answer choices will each yield the same total area for the two triangles and thus the same area for polygon ABDFG.
Since A and E cannot both be correct, eliminate A and E.
Answer choice B implies the following:
BCD is a 45-45-90 triangle in which each leg has a length of 2.
DEF is a 45-45-90 triangle in which each leg has a length of 4.
Answer choice D implies the following:
BCD is a 45-45-90 triangle in which each leg has a length of 4.
DEF is a 45-45-90 triangle in which each leg has a length of 2.
In each case, one of the 45-45-90 triangles has a leg of 2, while the other 45-45-90 triangle has a leg of 4.
Implication:
The two answer choices will each yield the same total area for the two triangles and thus the same area for polygon ABDFG.
Since B and D cannot both be correct, eliminate B and D.
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fskilnik
\(?\,\,:\,\,x\,\,{\text{to}}\,\,\max \,\,S\left( {ABDFG} \right)\,\,\,\, \Leftrightarrow \,\,\,x\,\,{\text{to}}\,\,\min \,\,\,\left[ {S\left( {\Delta {\text{BCD}}} \right) + S\left( {\Delta DEF} \right)} \right]\)
\(S\left( {\Delta {\text{BCD}}} \right) + S\left( {\Delta DEF} \right) = \frac{{{x^2}}}{2} + \frac{{{{\left( {6 - x} \right)}^2}}}{2} = \frac{{2{x^2} - 12x + 36}}{2} = {x^2} - 6x + 18\)
\(?\,\,\,:\,\,\,x\,\,{\text{to}}\,\,\min \,\,{x^2} - 6x + 18\,\,\,\,\, \Leftrightarrow \,\,\,\,\,? = x = {x_{vert}} = - \frac{b}{{2a}} = - \frac{{ - 6}}{2} = 3\)
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.
fskilnik
Great question. +1
Attachment:
GMATH_figure0271edit.jpg [ 32.36 KiB | Viewed 3827 times ]
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
\(x = 6-x\)
\(2x=6\)
\(x=3\)
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
\(x=3\)
Answer
Attachments
GMATH_figure0271edit.jpg [ 32.36 KiB | Viewed 3653 times ]
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generis
Hi, generis!
Thank you for the kudos and for your nice contributions!
Regards,
Fabio.
fskilnik
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
fskilnik
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
dave13
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.
fskilnik
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.
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