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# In the figure above, if the area of triangular region D is 4

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In the figure above, if the area of triangular region D is 4 [#permalink]

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18 Dec 2012, 06:58
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In the figure above, if the area of triangular region D is 4, what is the length of a side of square region A ?

(1) The area of square region B is 9.
(2) The area of square region C is 64/9.
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Re: In the figure above, if the area of triangular region D is 4 [#permalink]

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18 Dec 2012, 07:03
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In the figure above, if the area of triangular region D is 4, what is the length of a side of square region A ?

The area of triangular region D = $$\frac{xy}{2} = 4$$ --> $$xy=8$$.
Question: $$z = \sqrt{x^2+y^2}$$

(1) The area of square region B is 9 --> $$x^2=9$$ --> $$x=3$$ --> $$y=\frac{8}{x}=\frac{8}{3}$$. Sufficient.

(2) The area of square region C is 64/9 --> $$y^2=\frac{64}{9}$$ --> $$y=\frac{8}{3}$$ --> $$x=\frac{8}{y}=3$$. Sufficient.

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Re: In the figure above, if the area of triangular region D is 4 [#permalink]

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03 Aug 2014, 09:35
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Re: In the figure above, if the area of triangular region D is 4 [#permalink]

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03 Aug 2014, 23:40
Attachment:
Region D.png
In the figure above, if the area of triangular region D is 4, what is the length of a side of square region A ?

(1) The area of square region B is 9.
(2) The area of square region C is 64/9.

We have the area of right triangle D is 4, so if we know one of two legs, we could calculate the hypotenuse or the side of square region A ( using Pythagorean theorem c^2 = a^2 + b^2)

(1) From area of B, we could calculate one of the two legs -> sufficient
(2) From area of C, we could calculate one of the two legs -> sufficient

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Re: In the figure above, if the area of triangular region D is 4 [#permalink]

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29 Nov 2014, 09:46
Bunuel wrote:
Attachment:
Region D2.png
In the figure above, if the area of triangular region D is 4, what is the length of a side of square region A ?

The area of triangular region D = $$\frac{xy}{2} = 4$$ --> $$xy=8$$.
Question: $$z = \sqrt{x^2+y^2}$$

(1) The area of square region B is 9 --> $$x^2=9$$ --> $$x=3$$ --> $$y=\frac{8}{x}=\frac{8}{3}$$. Sufficient.

(2) The area of square region C is 64/9 --> $$y^2=\frac{64}{9}$$ --> $$y=\frac{8}{3}$$ --> $$x=\frac{8}{y}=3$$. Sufficient.

Hi Bunnel,
I wonder why aren't we using the hypotenuse as one of the sides when calculating the area??

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Re: In the figure above, if the area of triangular region D is 4 [#permalink]

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29 Nov 2014, 09:52
LaxmiReddy wrote:
Bunuel wrote:
Attachment:
Region D2.png
In the figure above, if the area of triangular region D is 4, what is the length of a side of square region A ?

The area of triangular region D = $$\frac{xy}{2} = 4$$ --> $$xy=8$$.
Question: $$z = \sqrt{x^2+y^2}$$

(1) The area of square region B is 9 --> $$x^2=9$$ --> $$x=3$$ --> $$y=\frac{8}{x}=\frac{8}{3}$$. Sufficient.

(2) The area of square region C is 64/9 --> $$y^2=\frac{64}{9}$$ --> $$y=\frac{8}{3}$$ --> $$x=\frac{8}{y}=3$$. Sufficient.

Hi Bunnel,
I wonder why aren't we using the hypotenuse as one of the sides when calculating the area??

Check here: math-triangles-87197.html
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Re: In the figure above, if the area of triangular region D is 4 [#permalink]

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30 Jun 2016, 06:46
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Re: In the figure above, if the area of triangular region D is 4 [#permalink]

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13 Sep 2017, 06:50
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the figure above, if the area of triangular region D is 4   [#permalink] 13 Sep 2017, 06:50
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