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# The shaded portion of the rectangular lot shown above repres

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The shaded portion of the rectangular lot shown above repres [#permalink]  19 Dec 2012, 05:28
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Question Stats:

80% (02:33) correct 20% (02:22) wrong based on 412 sessions
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Shaded region.png [ 17.6 KiB | Viewed 7310 times ]
The shaded portion of the rectangular lot shown above represents a flower bed. If the area of the bed is 24 square yards and x = y + 2, then z equals

(A) $$\sqrt{13}$$
(B) $$2\sqrt{13}$$
(C) 6
(D) 8
(E) 10
[Reveal] Spoiler: OA
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Re: The shaded portion of the rectangular lot shown above repres [#permalink]  19 Dec 2012, 05:34
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The shaded portion of the rectangular lot shown above represents a flower bed. If the area of the bed is 24 square yards and x = y + 2, then z equals

(A) $$\sqrt{13}$$
(B) $$2\sqrt{13}$$
(C) 6
(D) 8
(E) 10

Given that $$\frac{xy}{2}=24$$ and $$x = y + 2$$ --> $$xy=48$$ --> $$(y+2)y=48$$.

$$z^2=x^2+y^2$$ --> $$z^2=(y + 2)^2+y^2=2y^2+4y+4=2y(y+2)+4$$.

Since $$(y+2)y=48$$, then $$z^2=2y(y+2)+4=2*48+4=100$$ --> $$z=10$$.

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Re: The shaded portion of the rectangular lot shown above repres [#permalink]  17 Nov 2013, 19:25
What is the shaded part, the pentagon or the right angled triangle?
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Re: The shaded portion of the rectangular lot shown above repres [#permalink]  17 Nov 2013, 22:37
Expert's post
What is the shaded part, the pentagon or the right angled triangle?

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Re: The shaded portion of the rectangular lot shown above repres [#permalink]  06 Mar 2014, 01:32
Expert's post
Bumping for review and further discussion.

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Re: The shaded portion of the rectangular lot shown above repres [#permalink]  18 Nov 2014, 19:22
Attachment:
The shaded portion of the rectangular lot shown above represents a flower bed. If the area of the bed is 24 square yards and x = y + 2, then z equals

(A) $$\sqrt{13}$$
(B) $$2\sqrt{13}$$
(C) 6
(D) 8
(E) 10

I actually did this a different way.

For a triangle area=$$\frac{B*H}{2}$$ so solving I get that $$B*H=48$$ now I would solve by algebra, but the problem tells that the difference between x & y is two, so what two numbers multiply to 48 that separated by 2... 6 & 8. Since it doesn't matter which side is which length I saw this was a pythagorean triplet and z or the hypotenuse is 10.

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The shaded portion of the rectangular lot shown above repres [#permalink]  20 Nov 2014, 00:30
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Modified the diagram as shown below:

Attachment:

Shaded%20region.png [ 18.65 KiB | Viewed 1613 times ]

$$Area of shaded region = \frac{1}{2} x (x-2) = 24$$

$$x^2 - 2x - 48 = 0$$

x = 8

Base = 6; Height = 8

$$z = \sqrt{6^2+8^2} = \sqrt{100} = 10$$

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Re: The shaded portion of the rectangular lot shown above repres [#permalink]  11 Jan 2015, 04:03
Hi,

This one was tricky, not because the solution was difficult, but because of my mistakes in the calculations...

I also used Paresh's approach, but instead of solving for y and getting y=x-2, I just used y and y+2. But, I couldn't see the 6*4=48 ... so I couldn't find the possible solutions of the equation easily!

In fact I did this:
Area= (b*2) / 2
24=[y(y+z)] / 2
24= (y^2 + 2y) / 2
48 = y^2 + 2y
y^2 + 2y - 48 = 0.

I didn't see these solutions: -6, 8 and ended up using the quadratic formula, which is not handy when you are not quick with calculations! Haha! So, I passed the time...

Just as a small piece of advice, in similar situations, what makes it easier to find the roots is to make the prime factorization of the constant. Then you end up with this 2*2*2*2*3, so you have fewer values to test and find one that works.
Re: The shaded portion of the rectangular lot shown above repres   [#permalink] 11 Jan 2015, 04:03
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