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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, 05:58
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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.

Attachment:

Region D.png [ 14.72 KiB | Viewed 14426 times ]
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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, 06:03
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Attachment:

Region D2.png [ 14.54 KiB | Viewed 14273 times ]
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, 22:40
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, 08:46
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, 08:52
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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11 Dec 2017, 14:23
Hi All,

Since this is a DS question, we CANNOT trust the picture. We can trust any descriptions and numbers that we are given though - so we know that the triangle is a RIGHT triangle, it's area IS 4 and the shape A IS a square. We're asked for a side length of square A.

1) The area of square region B is 9.

Since shape B is a square - and its area is 9 - we know that its sides are each 3. Knowing the height of the triangle is 3, we can figure out its base:

Area = (1/2)(Base)(Height)
4 = (1/2)(B)(3)
4 = 1.5B
4/1.5 = B

With the base and the height, we CAN figure out the length of the diagonal (by using the Pythagorean Theorem), so we would have the answer to the question.
Fact 1 is SUFFICIENT

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

While the information in Fact 2 might look a big 'weird', we know from the work that we did in Fact 1 that if we have either the base or the height of the triangle, then we can figure out the other side and then solve the diagonal. Thus, Fact 2 is also enough information to answer the question.
Fact 2 is SUFFICIENT

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

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26 Apr 2018, 08:47
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.

why are dividing 8 by 3 can you explain the logic ? $$y=\frac{8}{x}=\frac{8}{3}$$.
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Re: In the figure above, if the area of triangular region D is 4  [#permalink]

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26 Apr 2018, 09:43
1
dave13 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.

why are dividing 8 by 3 can you explain the logic ? $$y=\frac{8}{x}=\frac{8}{3}$$.

Hello

Bunuel explained in his solution that the area of the right triangle will be 1/2 * x * y or xy/2, this area is give as 4, so xy/2 = 4 or xy = 8 which in turn tells us that y = 8/x. He has explained in the highlighted part.
Re: In the figure above, if the area of triangular region D is 4 &nbs [#permalink] 26 Apr 2018, 09:43
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