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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.

Answer E

Just another way to think about this
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Re: The shaded portion of the rectangular lot shown above repres [#permalink]

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11 Jan 2015, 05:03

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]

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21 Jun 2015, 15:52

Great OG problem. Paresh's solution helped me see the light on this one. The others made sense, but I may not see that relationship on test day. Re-labeling the figure with the formulas will help add clarity. Great explanation!

On certain GMAT questions, you can "brute force" the work and quickly come up with the correct answer. Here's how:

From the prompt and the picture, we know that… 1) The triangle is a right triangle (because it's in the "corner" of a rectangle) 2) The two legs of the triangle differ by 2 3) The area of the triangle is 24 (A = (1/2)(B)(H))

Let's focus on the area = 24 for a moment. We know the two legs differ by 2, so we can probably "brute force" the possibilities and find the match:

If the legs are: 2 and 4, then the area = 4 4 and 6, then the area = 12 6 and 8, then the area = 24 ---> that's THE match

If the legs are 6 and 8, then we can use the Pythagorean Formula to figure out the value of Z. You might also recognize the "Pythagorean Triplet" and solve the problem that way.

As you continue to study for the GMAT, you're going to find that 'shortcuts' vary from question to question, but there are almost always shortcuts to found (and in many cases there are alternative ways to approach the question besides 'just doing the math'). It's to your benefit to learn to spot the 'clues' that hint at these options - and then practice using those approaches. While not every shortcut will be available on every question, the list of possible shortcuts is definable, so you can train to use them again and again on Test Day.

In solving this problem we first must recognize that the flower bed is the right triangle with sides of y yards, x yards, and z yards. We are given that the area of the bed (which is the right triangle) is 24 square yards. Since we know that area of a triangle is ½ Base x Height, we can say:

24 = ½(xy)

48 = xy

We also know that x = y + 2, so substituting in y + 2 for x in the area equation we have:

48 = (y+2)y

48 = y^2 + 2y

y^2 + 2y – 48 = 0

(y + 8)(y – 6) = 0

y = -8 or y = 6

Since we cannot have a negative length, y = 6.

We can use the value for y to calculate the value of x.

x = y + 2

x = 6 + 2

x = 8

We can see that 6 and 8 represent two legs of the right triangle, and now we need to determine the length of z, which is the hypotenuse. Knowing that the length of one leg is 6 and the other leg is 8, we know that we have a 6-8-10 right triangle. Thus, the length of z is 10 yards.

If you didn't recognize that 6, 8, and 10 are the sides and hypotenuse of a right triangle, you would have to use the Pythagorean to find the length of the hypotenuse: 6^2 + 8^2 = c^2 → 36 + 64 = c^2 → 100 = c^2. The positive square root of 100 is 10, and thus the value of z is 10.

Answer is E.
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I thought that the shaded part was isosceles right triangle and came with Z = \(4\sqrt{6}\). Is it because "\(x= y+2\)" that this is not an isos rt. triangle?

I thought that the shaded part was isosceles right triangle and came with Z = \(4\sqrt{6}\). Is it because "\(x= y+2\)" that this is not an isos rt. triangle?

x is 2 units more than y, so x does not equal to y, and the triangle is not isosceles.
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Re: The shaded portion of the rectangular lot shown above repres [#permalink]

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14 Sep 2016, 15:13

Answer E

The easy stuff: (x*y) / 2 = 24 therefore x*y = 48

So since x = y + 2 replace x to product x*y = 48. Hence (y+2) * y = 48 (equation 1) .

IMHO this IS the TRICKY part... and I think this is the moment where you can save so much time the only number that gives you 48 when is multiplied by itself +2 is 6! because 6 (6+2) = 48 HENCE y = 6 and x = 8 ... the rest is history as you apply Pythagoras theorem and you get z^2 = sqrt(100) ..... and so z = 10 BOOM!

The shaded portion of the rectangular lot shown above repres [#permalink]

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08 Apr 2017, 00:41

pacifist85 wrote:

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.

pacifist85 I share the same problem as you do. Finding the roots can be tricky at times and time consuming too. I ended up prime factorizing the constant and saw the 6*8. I started to solve with the quadratic formula but said in my mind "Screw this - I'll be here till tomorrow solving this thing!"
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