A rectangular yard (white area) is surrounded by a hedge (shaded regio

Question

Tough and Tricky questions: Geometry. https://gmatclub.com/forum/download/file.php?id=24730 Note: Figure not drawn to scale. A rectangular yard (white area) is surrounded by a hedge (shaded region) with a length of 15 meters and a width of 12 meters, as shown in the figure above. The area of the yard is equal to the area of the hedge. If the ratio of the length of the hedge to the width of the hedge is the same as the ratio of the length of the yard to the width of the yard, what is the width, in meters, of the yard? A. 6\sqrt{2} B. \frac{5}{4} C. \frac{6}{\sqrt{2}} D. 7\frac{\sqrt{2}}{2} E. \frac{6}{2} Kudos for a correct solution . s95-22.png

PareshGmat · Accepted Answer

Answer = A  = 6\sqrt{2}

Total Area = 15*12 = 180

Given that Area of rectangular yard = Area of Hedge (shaded region)  = \frac{180}{2} = 90

Lets say width of the yard = a

then length of the yard  = \frac{90}{a}

Equating the ratios:

\frac{15}{12} = \frac{\frac{90}{a}}{a}

a^2 = 72

a = 6\sqrt{2}

DangerPenguin · Answer

Given a ratio of 5:4 I used length of 10 and width of 8 for the yard as a starting point. The area of the yard would then be 10x8=80m and the area of the hedge would be (15x12)-80 = 180-80=100. This tells us that with length 10 and width 8, the yard is not as big as the hedge. Therefore, the width must be larger than 8.

A. knowing that the square root of 2 is roughly 1.4, this answer is approx. 8.4. this is what we are looking for but we need to check to see if any other choices are larger than 8
B. too small
C. too small
D. too small
E. too small

Answer A!

BassanioGratiano · Answer

This was majorly brilliant. I was going through a ridiculous quadratic equation only to think, "There has to be an easier way to solve this," and here it is.

kinghyts · Answer

The ratio of the length to width of the yard = 5: 4

Let length = 5x and width = 4x

therefore we know , 5x * 4x = 15*12 - 5x*4x 
=> x^2 = 4.5 
Therefore, x = 3 * sqrt (1/2)

Therefore, width of the the yard = 4x = 4 * 3 * sqrt (1/2) = 6 * sqrt(2)

Hence A) is the answer

Bunuel · Answer

Official Solution:

https://gmatclub.com/forum/download/file.php?id=24730 
 Note:  Figure not drawn to scale.

A rectangular yard (white area) is surrounded by a hedge (shaded region) with a length of 15 meters and a width of 12 meters, as shown in the figure above. The area of the yard is equal to the area of the hedge. If the ratio of the length of the hedge to the width of the hedge is the same as the ratio of the length of the yard to the width of the yard, what is the width, in meters, of the yard?

A.  6\sqrt{2} 
B.  \frac{5}{4} 
C.  \frac{6}{\sqrt{2}} 
D.  7\frac{\sqrt{2}}{2} 
E.  \frac{6}{2}

Let the length of the yard be  x , and let the width of the yard be  y .

We are told that the length of the hedge is 15 meters and that the width of the hedge is 12 meters, so we can find the area of the yard and hedge combined using the formula for area of a rectangle:  A_{	ext{yard and hedge}} = 	ext{length} 	imes 	ext{width} = 15 	imes 12 = 180 . Similarly, we can find the area of the yard:  A_{yard} = xy .

We are told that the area of the yard and the hedge are equal, so the area of the hedge must also be  A_{hedge} = A_{yard} = xy . We can use this information to write  xy + xy = 2xy = 180 , or  xy = 90 .

We know that the length and width of the yard are in the same ratio as the length and width of the hedge. We express this with a proportion:  \frac{15}{12} = \frac{x}{y} . Cross-multiply to get  15y = 12x . Simplify to get  5y = 4x .

Now we have two equations for  x  and  y . We will solve for these unknowns using substitution. Since we want to find the width,  y , we first solve for  x  in terms of  y . From the equation  xy = 90 , we get  x = \frac{90}{y} . We plug this into  5y = 4x  and get  5y = 4 	imes \frac{90}{y} = \frac{360}{y} . Dividing both sides by 5 and multiplying by  y , we find  y^2 = 72 . We now take the square root of both sides to get  y = \sqrt{72} .

To simplify this square root, we look for a perfect square that is a factor of 72. We note that  72 = 36 	imes 2 , and so  y = \sqrt{72} = \sqrt{36 	imes 2} = \sqrt{36} 	imes \sqrt{2} = 6\sqrt{2} .
 
Answer: A.

Temurkhon · Answer

I missed the fact that one should 180/2=90, so

5x*4x=90
20x^2=90
x^2=90/20 --> x=3*sqrt2/2

finally, 3*sqrt2*4/2=6*sqrt2

A

anulfc · Answer

I don't get it. Why is everyone taking the length and breadth of the hedge as 15 and 12? Wouldn't the area be 180 then? And we know the area is 90. So, shouldn't we equate X/Y = 15-X/12-Y??
Can anyone tell me what silly mistake i am doing?

anulfc · Answer

I don't get it. Why is everyone taking the length and breadth of the hedge as 15 and 12? Wouldn't the area be 180 then? And we know the area is 90. So, shouldn't we equate X/Y = 15-X/12-Y??
Can anyone tell me what silly mistake i am doing?
 Bunuel 
 PareshGmat

Bunuel · Answer

The ratio of the length of the hedge (15) to the width of the hedge (12) is the same as the ratio of the length of the yard (x) to the width of the yard (y): 15/12 = x/y Untitled.png

Bunuel · Answer

Similar questions to practice about rectangles: a-square-wooden-plaque-has-a-square-brass-inlay-in-the-89215.html a-border-of-uniform-width-is-placed-around-a-rectangular-144434.html a-rectangular-yard-is-20-yards-wide-and-40-yards-long-it-is-8453.html the-figure-above-represents-a-square-plot-measuring-x-feet-o-160568.html a-rectangular-photograph-is-surrounded-by-a-border-that-is-72686.html harriet-wants-to-put-up-fencing-around-three-sides-of-her-127082.html Similar questions to practice about circles: marty-s-pizza-shop-guarantees-that-their-pizzas-all-have-a-148428.html a-circular-lawn-with-a-radius-of-5-meters-is-surrounded-by-a-154737.html the-figure-above-shows-a-circular-flower-bed-with-its-cente-144448.html the-figure-shows-the-top-side-of-a-circular-medallion-made-o-101845.html

anulfc · Answer

Ohhh so those aren't the actual dimensions but their ratios. Dimensions will be less than those so as to have an area of 90. My bad. Thank you for swift response, Bunuel

mvictor · Answer

180-lw=lw or 180=2lw or lw=90.

now l/w=5/4

5l/4 * l = 90
l^2 = 90*4/5
l^2 = 72
l= sqrt(72)
sqrt(72) = 6(sqrt2)

A.

gracie · Answer

let L,W=length and width of hedge
let l,w=length and width of yard
LW=180m^2
lw=LW-lw=90m^2
15/12=5/4
(5x)(4x)=90
x^2=9/2
x=3/sqrt2
4x=(6)(sqrt2) meters width of yard

spetznaz · Answer

A it is well explained above !

DarkHorse2019 · Answer

Dear Moderators  chetan2u ,  VeritasKarishma   gmatbusters  
Can you help ? I am unable to understand why did we divide the area of 180 by 2 to get 90

GMATBusters · Answer

Total Area = 15*12 = 180

Given that Area of rectangular yard = Area of Hedge (shaded region) 
So, let Area of rectangular yard = Area of Hedge (shaded region) = x
hence, x+x = total area = 180,
2x = 180, 
or x = 180/2 = 90.

Area of rectangular yard = Area of Hedge (shaded region)  = \frac{180}{2} = 90

KarishmaB · Answer

Area of yard (white rectangle) = Area of hedge (shaded region)

Total area of yard + hedge = 15*12 = 180 (Area of the complete big rectangle)

So area of just the yard = 180/2 = 90

Ratio of length:width = 5:4. Say the length and width are 5x and 4x.

5x*4x = 90 
 x = 3/\sqrt{2}

Width = 4x = 4*3/\sqrt{2} = 6\sqrt{2}

Fdambro294 · Answer

Let L = length of Internal yard

Let W = width of Internal yard

Area of Hedge = (15 * 12) - (L * W)

Area of Yard = L * W

Since Area of Yard = Area of Hedge

(180) - LW = LW

180 = 2 * LW ---- equation 1

(Length of Hedge) / (Width of Hedge) = 15 / 12 = L / W

L = (5/4) * W ----Substituting L in for equation 1 above

2 * (5/4) * w * w = 180

(w)^2 = 72

w = sqrt(72) = 6 * sqrt(2)

Answer -A-

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