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 06:19:19 · 3 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 06:25:16 · answer #1 · answered by raj 7 · 0⤊ 0⤋
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 14:24:29 · answer #2 · answered by Amy F 5 · 0⤊ 0⤋
x= original lot length
x^2=lot size
(x-3)(x-5)=x^2-185...solve for x
x=25
new dimensions are 25-3 and 25-5 or 22 and 20
Answer is D
2006-11-15 14:26:10 · answer #3 · answered by runningman022003 7 · 0⤊ 0⤋