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21 Jan 2014, 11:10
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File comment: Triangle
Triangle.png [ 24.93 KiB | Viewed 20364 times ]
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?Triangle.png [ 24.93 KiB | Viewed 20364 times ]
A. (9√2)/4
B. 9/2
C. (9√2)/2
D. 6√2
E. 12
OE
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
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
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 \(= C = \frac{9\sqrt{2}}{2}\)
\(BC^2 = 15^2 - 9^2 = 81\)
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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tri.jpg [ 21.61 KiB | Viewed 19672 times ]
General Discussion
21 Jan 2014, 11:28
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goodyear2013
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
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
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.
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
