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

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

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20 Sep 2018, 14:54
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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: http://www.GMATH.net

Attachment:

GMATH_figure0271.gif [ 4.58 KiB | Viewed 727 times ]

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

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21 Sep 2018, 06:25
1
fskilnik wrote:

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: http://www.GMATH.net

Attachment:
The attachment GMATH_figure0271.gif is no longer available

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 460 times ]

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

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20 Sep 2018, 15:48
1
fskilnik wrote:

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: http://www.GMATH.net

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

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20 Sep 2018, 17:21
1
fskilnik wrote:

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: http://www.GMATH.net

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

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

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

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.
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Our high-level "quant" preparation starts here: https://gmath.net

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

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20 Sep 2018, 19:00
2
fskilnik wrote:
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: http://www.GMATH.net

Great question. +1
Attachment:

GMATH_figure0271edit.jpg [ 32.36 KiB | Viewed 566 times ]

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
$$x = 6-x$$
$$2x=6$$
$$x=3$$

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
$$x=3$$

Attachments

GMATH_figure0271edit.jpg [ 32.36 KiB | Viewed 399 times ]

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Re: The square ACEG shown below has an area of 36 units squared. What is t  [#permalink]

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20 Sep 2018, 19:09
1
generis wrote:
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.

Hi, generis!
Thank you for the kudos and for your nice contributions!
Regards,
Fabio.
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Our high-level "quant" preparation starts here: https://gmath.net

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Joined: 09 Mar 2016
Posts: 1287
The square ACEG shown below has an area of 36 units squared. What is t  [#permalink]

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21 Sep 2018, 05:08
fskilnik wrote:

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: http://www.GMATH.net

Attachment:
GMATH_figure0271.gif

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
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Joined: 09 Mar 2016
Posts: 1287
Re: The square ACEG shown below has an area of 36 units squared. What is t  [#permalink]

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21 Sep 2018, 05:45
fskilnik wrote:
fskilnik wrote:

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: http://www.GMATH.net

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

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

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

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.

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
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Re: The square ACEG shown below has an area of 36 units squared. What is t  [#permalink]

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21 Sep 2018, 05:56
dave13 wrote:
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.

Regards,
Fabio.
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Re: The square ACEG shown below has an area of 36 units squared. What is t &nbs [#permalink] 21 Sep 2018, 05:56
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