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In the figure above, the length of AC is 12 and the length of AB is 15

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goodyear2013
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In the figure above, the length of AC is 12 and the length of AB is 15 [#permalink] New post   21 Jan 2014, 11:10
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In the figure above, the length of AC is 12 and the length of AB is 15. The area of right triangle DEF is equal to the area of the shaded region. If the ratio of the length of DF to the length of EF is equal to the ratio of the length of AC to the length of BC, what is the length of EF?

A. (9√2)/4
B. 9/2
C. (9√2)/2
D. 6√2
E. 12

OE
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3:4:5 right triangle ∆ABC → AB = 15 = (3 × 5) & AC = 12 = (4 × 3) → BC = (3 × 3) = 9

Area of ∆DEF = Area of Shaded region & (∆DEF + Shaded region) = ∆ABC → Area of ∆DEF = (1/2) × Area of ∆ABC
Area of ∆ABC = (½ x 9 x 12) = 54
Area of ∆DEF = (1/2)(54) = 27

DF : EF = AC : BC → AC : BC = 12 : 9 = 4 : 3
DF : EF = 4 : 3 = 4x : 3x
∆DEF 3:4:5 Triangle as ∆ABC
Area of ∆DEF = 27 = (1/2)(EF)(DF) = (1/2)(3x)(4x) = 6x2 → 6x2 = 27 → x2 = (9/2) → x = 3/√2
EF = 3x = 3×(3/√2) = (9√2)/2
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PareshGmat
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Re: In the figure above, the length of AC is 12 and the length of AB is 15 [#permalink] New post   05 Jun 2014, 01:43
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Refer diagram below:

\(BC^2 = 15^2 - 9^2 = 81\)

BC = 9

Area of big triangle ABC\(= \frac{1}{2} * 9 * 12 = 54\)

Area of small triangle DEF (shaded region) = Area of unshaded region\(= \frac{54}{2} = 27\)

For Triangle DEF with area 27, if one side is x, then the other side would be 54/x

This can be proved:
\(Area = 27 = \frac{1}{2} * x * \frac{54}{x} = 27\)
So our assumptions are correct (shown in the diagram)

Now, placing the variables in ratios as stated in the problem

\(\frac{x}{\frac{54}{x}} = \frac{12}{9}\)

\(x^2 = 54 * \frac{12}{9}\)

\(x = 6 \sqrt{2}\)

\(EF = \frac{54}{x} = \frac{54}{6 \sqrt{2}} = \frac{9}{\sqrt{2}}\)

Answer \(= C = \frac{9\sqrt{2}}{2}\)
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mikemcgarry
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Re: In the figure above, the length of AC is 12 and the length of AB is 15 [#permalink] New post   21 Jan 2014, 11:28
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goodyear2013 wrote:
Attachment:
Triangle.png

In the figure above, the length of AC is 12 and the length of AB is 15. The area of right triangle DEF is equal to the area of the shaded region. If the ratio of the length of DF to the length of EF is equal to the ratio of the length of AC to the length of BC, what is the length of EF?
(9√2)/4
9/2
(9√2)/2
6√2
12

OE
Show Spoiler
1) In right triangle ABC, AB = 15 & AC = 12. Right triangle ABC is a multiple of 3:4:5 right triangle with each side of right triangle ABC being 3 times the corresponding member of the 3:4:5 ratio.
AB = 15 = 3 × 5 & AC = 12 = 4 × 3 → BC = 3 × 3 = 9.

2) Since area of triangle DEF is equal to area of shaded region, and triangle DEF and shaded region together make up triangle ABC, area of triangle DEF is 1/2 of area of triangle ABC.
Area of right triangle ABC = ½ x 9 x 12 = 54
area of triangle DEF = (1/2)(54) = 27

3) ratios DF:EF and AC:BC are equal. Ratio of length of AC to length of BC is 12/9 = 4/3
ratio of length of DF to length of EF is also 4/3.
Let length of DF = 4x, length of EF = 3x.
(Notice that DEF is also a 3:4:5 triangle.)
Area of triangle DEF = 27. Since DEF is a right triangle just like ABC:
Area of triangle DEF = (1/2)(EF)(DF) = (1/2)(3x)(4x) = 6x^2 → 6x^2 = 27 → x^2 = 9/2 → x = 3/√2
EF = 3x = 9/√2 = (9√2)/2


Hi, I want to know if we have more simple way to solve this question, please.

Dear goodyear2013,
I'm happy to respond. The short answer to your question is: no. There is not a more simple way to approach this problem. This is a difficult problem, and like many difficult GMAT math problems, you are asked to combine a few different pieces of material.

The Pythagorean Triplets should be something you know cold. You may find this blog helpful.
https://magoosh.com/gmat/2012/pythagorea ... -the-gmat/
When you see the 12 & 15, your mind should immediately flash to the 3-4-5 triangle. If that much is not instantaneous, then you don't know your Pythagorean Triplets well enough yet.

Once you have BC = 9, it should be very simple to get the total area, and the area of the smaller triangle.

The final part, setting up the right proportion --- proportional reasoning is one of the huge strategies and time-saving for the GMAT Quant section. Here's a blog you may find helpful:
https://magoosh.com/gmat/2012/pythagorea ... -the-gmat/
Thinking in terms of ratios & proportions is also something that you should practice until you can do it in your sleep.

So, no, there's not a simpler solution to this problem, but there definitely are things you can learn better so that a problem such as this seems simpler.

Does all this make sense?
Mike :-)
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Re: In the figure above, the length of AC is 12 and the length of AB is 15 [#permalink] New post   21 Jan 2014, 22:15
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goodyear2013 wrote:
Attachment:
Triangle.png

In the figure above, the length of AC is 12 and the length of AB is 15. The area of right triangle DEF is equal to the area of the shaded region. If the ratio of the length of DF to the length of EF is equal to the ratio of the length of AC to the length of BC, what is the length of EF?
(9√2)/4
9/2
(9√2)/2
6√2
12

OE
Show Spoiler
3:4:5 right triangle ∆ABC → AB = 15 = (3 × 5) & AC = 12 = (4 × 3) → BC = (3 × 3) = 9

Area of ∆DEF = Area of Shaded region & (∆DEF + Shaded region) = ∆ABC → Area of ∆DEF = (1/2) × Area of ∆ABC
Area of ∆ABC = (½ x 9 x 12) = 54
Area of ∆DEF = (1/2)(54) = 27

DF : EF = AC : BC → AC : BC = 12 : 9 = 4 : 3
DF : EF = 4 : 3 = 4x : 3x
∆DEF 3:4:5 Triangle as ∆ABC
Area of ∆DEF = 27 = (1/2)(EF)(DF) = (1/2)(3x)(4x) = 6x2 → 6x2 = 27 → x2 = (9/2) → x = 3/√2
EF = 3x = 3×(3/√2) = (9√2)/2



You need to use two concepts here to arrive at the answer quickly: Pythagorean triplets, Similar triangles
ABC is right triangle such that AC = 12 and AB = 15. This means BC = 9 (Using the 3-4-5 Pythagorean triplet. 9-12-15 is a multiple of this triplet)

Given: DF/EF = AC/BC
We know that AC/BC = 4/3 = DF/EF. So triangle DEF is also a right triangle with sides in the ratio 3:4:5. So triangles ABC and DEF are similar.
In similar triangles, if the sides are in the ratio 'r', the areas of the triangles are in the ratio '\(r^2\)'.

What does this tell us: "The area of right triangle DEF is equal to the area of the shaded region"?
It tells us that the area of triangle ABC is twice the area of triangle DEF. So the ratio of the sides must be \(\sqrt{2}\) i.e. every side of ABC must be \(\sqrt{2}\) times the side of DEF.
Since BC = 9, EF must be \(9/\sqrt{2} = 9\sqrt{2}/2\)
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Re: In the figure above, the length of AC is 12 and the length of AB is 15 [#permalink] New post   09 Jun 2016, 21:40
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goodyear2013 wrote:
Attachment:
Triangle.png

In the figure above, the length of AC is 12 and the length of AB is 15. The area of right triangle DEF is equal to the area of the shaded region. If the ratio of the length of DF to the length of EF is equal to the ratio of the length of AC to the length of BC, what is the length of EF?
(9√2)/4
9/2
(9√2)/2
6√2
12



Triangle ABC is a right angled triangle. Therefore sides Ab, AC and BC can be one of the pythagorean triplets. we find that hypotenuse AB = 15 and other side AC = 12

They are in a ratio 5:4 (Hint: multiply each by 3, you get actuallengths). This means the third side will be such that BC:AC:AB = 3:4:5.

length of BC is known = 9

Area of triangle ABC is known (1/2) (Base * height) = (1/2) (12*9) = 54

Area of inner right angled triangle and shaded portion is same. Let that area is B

B+B = 54.

Therefore area of inner Right angled triangle is 27.

Now we are given that

\(\frac{AC}{BC}\) = \(\frac{DF}{FE}\) = \(\frac{12}{9}\) = \(\frac{4}{3}\)

The sides are in the ratio 4:3, say they are 4x and 3x

Then (1/2)*(4x)*(3x) = 27

6 \(x^2\) = 27

x = \(\frac{3}{{\sqrt{2}}}\)

Length EF = 3x = \(\frac{9}{{\sqrt{2}}}\)

C is the answer.
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Re: In the figure above, the length of AC is 12 and the length of AB is 15 [#permalink] New post   15 Sep 2020, 21:56
Concept: Given 2 Similar Triangles:

if the Ratio of AREAS of the 2 Similar Triangles is = a : b

then the Ratio of CORRESPONDING SIDE Lengths of the 2 Similar Triangles is = sqrt (a) : sqrt (b)

1st) Larger Triangle ABC is a Multiple of a 3-4-5 Pythagorean Triplet Right Triangle

Side BC = 9


2nd) Since the Area of Triangle DEF = Area of Shaded Region

and

we are told that the Ratio of Corresponding Sides are Equal, thereby telling us that Triangle DEF ~Similar to~ Triangle ABC


RATIO of:


Area of DEF : Area Shaded Region : TOTAL Area of ABC
_________________________________________________
1 --- 1 --- 2


From the Concept Above:

Corresponding Side of Triangle DEF / Corresponding Side of Triangle ABC = sqrt (1) / sqrt (2)

Side EF Corresponds to Side BC. Setting up the Ratio Proportion:

1 / sqrt(2) = EF / 9

Side EF = 9 * sqrt(2) / 2

-C-
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In the figure above, the length of AC is 12 and the length of AB is 15 [#permalink] New post   Updated on: 16 Sep 2020, 10:39
Area of ACB = 1/2 * 12 *9 = 54.
area of shaded region = area of smaller triangle = 1/2 * area of Bigger triangle ABC = 27 .... ( As both shaded region and smaller triangle have equal area)

We also know that DF/FE = AC/CB = 4x/3x .

Hence area of triangle DFE = 1/2*DF*FE = 27 = 1/2 * 4x * 3x .
6sq(x) = 27 .
x = 3/sqrt(2) .
EF = 3x = 3*3*sqrt(2)/2.

Ans -: C

Originally posted by Harshjha001 on 16 Sep 2020, 06:05.
Last edited by Harshjha001 on 16 Sep 2020, 10:39, edited 1 time in total.
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In the figure above, the length of AC is 12 and the length of AB is 15 [#permalink] New post   16 Sep 2020, 07:28
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goodyear2013 wrote:
Attachment:
Triangle.png

In the figure above, the length of AC is 12 and the length of AB is 15. The area of right triangle DEF is equal to the area of the shaded region. If the ratio of the length of DF to the length of EF is equal to the ratio of the length of AC to the length of BC, what is the length of EF?

A. (9√2)/4
B. 9/2
C. (9√2)/2
D. 6√2
E. 12


ABC is a multiple of a 3-4-5 triangle:
9-12-15
Since AC=12 and AB=15, BC=9.

Since the legs for DEF are in the same ratio as the legs for ABC, the two triangles are SIMILAR.

Rule for SIMILAR TRIANGLES:
If each side of the larger triangle is \(x\) times the corresponding side in the smaller triangle, then the area of the larger triangle is \(x^2\) times the area of the smaller triangle.
Put another way:
If the area of the larger triangle is \(k\) times the area of the smaller triangle, then each side of the larger triangle is \(\sqrt{k}\) times the corresponding side in the smaller triangle.

Since DEF and the shaded region have equal areas, \(DEF = \frac{1}{2}ABC\).
Thus:
\(ABC = 2(DEF)\)
Since BC and EF are corresponding sides, the blue rule above implies the following:
\(BC = \sqrt{2}(EF)\)
Thus:
\(9 = \sqrt{2}(EF)\)
\(EF = \frac{9}{\sqrt{2}} = \frac{9\sqrt{2}}{2}\)

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