A corner lot that originally was square lost 185 m2 of area when one of the adjacent streets was widened by 3 m and the other was widened by 5 m. Find the new dimensions of the lot. (Hint: Let x equal the length of a side of the original square lot.)
A.
16 m wide, 18 m long
B.
18 m wide, 20 m long
C.
21 m wide, 23 m long
D.
20m wide, 22 m long
2006-11-15 06:24:10 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^2-(x-3)(x-5)=185
x^2-x^2+8x-15=185
8x+-15=185
adding 15
8x=200
x=25m
new dimensions=22m*20m
choice D
2006-11-15 06:33:38 · answer #1 · answered by raj 7 · 0⤊ 0⤋
I already answered this once, but hey, if you're willing to give out points twice...
If it was square, then the original area is x^2. The new area should be (x-5)*(x-3), since one dimension was shortened by 5 m and the other by 3 m. Multiply this out to get x^2 -8x + 15 for the new area. Now we know that the new area is smaller by 185 m^2, so we can write "x^2 (the old area) - (x^2 - 8x + 15) (the new area) = 185." This simplifies to 8x-15 = 185, so 8x = 200, so x = 25 m. The new area is 3 m shorter and 5 m narrower, so it should be 22x20. D.
2006-11-15 14:25:47 · answer #2 · answered by Amy F 5 · 0⤊ 0⤋
Let the original side=xm
original area=x^2 sq.m
New area
=[x-5][x-3]
=x^2-8x+15
Decrease in area
=[x^]-[x^2-8x+15]
=8x-15
=185
8x=200
x=25
The new sides are
22m,20m
D.
2006-11-15 14:34:42 · answer #3 · answered by openpsychy 6 · 0⤊ 0⤋
http://www.freevbcode.com/ShowCode.asp?ID=1106
this might help
2006-11-15 14:26:57 · answer #4 · answered by jdratbull 2 · 0⤊ 0⤋