fencing three sides, shared fence down middle; find dim.

[max]

Mr. Smith and Mr. Jones each built a similar stone fence to enclose his backyard. They only enclosed three sides, each using his own house as the fourth side and had a community fence between the two yards. If the total cost was $5 per linear foot, the yards were 20 feet wider than they were deep, and the total cost was $950, what were the length and width of each yard?

any help would be appreciated.

 

[arthur ohlsten]

Re: need help understanding how to solve this word problem

I always find it of help to sketch the problem. This isn't neccessary but helps.
sketch a horizontal line about 2 inchs long
sketch a vertical line at the midpoint of the above line
mark one side of the horizontal line with a S and the second 1/2 with a J
sketch vertical lines at each end of the horizontal line 1 inch long. mark these lines with a D for deep, and the central vertical line with a D.
connect the bottom of the 3 lines with a horizontal line, and mark eaxh 1/2 with D+20

the amount of fence used is
length = 3D+2[20+D]
length= 3D+2D+40
length=5d+40
cost = 5[5d+40]
but cost =950
950=25D+200
750=25D
D=30 ft then with 50 ft
Arthur

 

[mmm4444bot]

Ciminy! Who authors these exercises? (This is a rhetorical question.)

We have length, depth, and width. The stone fences form a rectangular prism? (This is a rhetorical question, also.)

Mr. Smith and Mr. Jones each built a similar stone fence to enclose [their own] backyard. They [each] enclosed [only] three sides [of their own backyard,] using [their] own house as the fourth side[,] and [they share] a community fence between [their] two [backyards]. If the total cost [is] $5 per linear foot, the [backyards are each] 20 feet wider than they [are] deep, and the total cost [is] $950, what [are] the [depth] and width of each [backyard]?

'

WHO KNOWS?

Is "width" the measurement of the common side or is "depth" the measurement of the common side?

Do both backyards have the same dimensions?

If I assume that both backyards have the same dimensions, then I see four possibilities.

22 feet by 42 feet

26 feet by 46 feet

30 feet by 50 feet

34 feet by 54 feet

Cheers,

~ Howard I. Noe :?



DOUBLE-CLICK THE IMAGE TO EXPAND

[attachment=0:2hpzfmfv]junk563.JPG[/attachment:2hpzfmfv]

 

[max]

"width" the measurement of the common side or is "depth"

This is where I was stuck. It talks about depth and width in the problem and want you to answer with width and length.
I was wondering if depth = length here.
but then in the problem it would sound like width is greater than length.

My answer was 22 by 42
Can this problem have 4 answers?

 

[mmm4444bot]

... Can this problem have 4 answers?

Hi Max:

I believe that the author of this exercise intended there to be only one answer, but, as you can see from my sloppy sketch, I see four different possibilities.

The exercise could be worded more clearly. Your result matches one of the four possibilies, so I think it is valid.

Cheers,

~ Mark

 

[Bill51]

The words do, in fact, say 'backyard', which would appear to eliminate the side-yard possibilities. There is the one remaining ambiguity about whether the two properties joined side-by-side or back-line-to-back-line. "Depth", perhaps, is in the real-estate sense: distance from street.

 

[mmm4444bot]

The words ... say 'backyard', which would appear to eliminate the side-yard possibilities ... I do not see any "side-yard" possibilities.

... "Depth", perhaps, is in the real-estate sense ... Perhaps. Who knows? The author is very sloppy.

 

[TchrWill]

Mr. Smith and Mr. Jones each built a similar stone fence to enclose his backyard. They only enclosed three sides, each using his own house as the fourth side and had a community fence between the two yards. If the total cost was $5 per linear foot, the yards were 20 feet wider than they were deep, and the total cost was $950, what were the length and width of each yard?

any help would be appreciated.

Mr. Jones and Mr. Smith live on two parallel streets, the backs of their houses facing one another. The community fence between them is the common fence, parallel to the backs of the houses. Two parallel fences run between the backs of the two houses.
The distance from the back of either house to the common fence is X. The length of the common fence is X + 20.

Therefore, 4X + X + 20 = 950/5 = 190

Then, 5X = 170 making X = 34 ft. and the common fence 54 ft.

4(34) + 34 + 20 = 190