1) a rectangle has perimeter 42m and area 104m^2. find the dimentions of the rectangle.
2) A square field has 4m added to its length and 2m added to its width, thereby creating a feild with an are of 15m^2. Find the dimensions of the origional field.
2007-04-15 03:19:47 · 11 answers · asked by cast.no.shadow 5 in Science & Mathematics ➔ Mathematics
1. xy=104
2x+2y=42
one side is 13, the other is 8
2. (x+4)(x+2)=15
x=1,
2007-04-15 03:26:50 · answer #1 · answered by leo 6 · 1⤊ 0⤋
1. If the perimeter is 42, the sum of the different sides (say, a and b) is 21. The area, a * b, is 104. So,
a + b = 21
a * b = 104
Substituting the first equation into the second,
a * (21 - a) = 104
-a^2 + 21a - 104 = 0
a^2 - 21a + 104 = 0
This quadratic equation has solutions 8 and 13 (check it).
If a = 8, b = 13, and vice-versa; check it in the original equations.
2. Let x be the square's side. Thus,
(x + 4) * (x + 2) = 15
x^2 + 6x + 8 = 15
x^2 + 6x - 7 = 0
This equation has solutions 1 and -7. Since lengths can't be negative, the only solution is 1: the field was 1m x 1m.
2007-04-15 10:29:39 · answer #2 · answered by jcastro 6 · 0⤊ 0⤋
The first couple sorta left their answer on #2 unfinished. The math on the first is correct. The x=1 solves the equation but doesn't answer the question. The dimensions were 1 m^2.
The last 2 guys way overcomplicated the problem and that's why they couldn't solve it. Use common sense people, you can do that one in your head. A 5 by 3 rectangle has a square meterage of 15 sq m, and fits the equation. Check it by going in reverse:
5-4=1, 3-2=1. x=1
2007-04-15 10:31:06 · answer #3 · answered by ma_tt_00 2 · 0⤊ 0⤋
1) For rectangle, perimeter= 2(length+width)
so 2(l+w) = 42
(l+w) +21
l= 21-w -equation (1)
area= length x width
(lw) =104 m^2 -equation (2)
Substitute (1) into (2):
(21-w)(w) =104
Expand into quadratic equation:
w^2 -21w +104 = 0
Factorise:
(w-13)(w-8) = 0
w= 13 or w=8
Substitute w= 13 or w=8 into equation (1):
(l) = 21- 13 or (l) = 21- 8
so length = 13 or 8 m
dimensions of rectangle: 13m x 8m or 8m x 13m
2) Let L = original lengths of square
Given L1= L+2m and L2= L+4m and A = 15 m^2
(L1)(L2) = A
(L+2)(L+4) = 15
expand into quadratic equation:
L^2 +6L -7 = 0
Factorise:
(L+7)(L-1) =0
so L = -7 or L = 1
since length cannot have a negative value,
original lenght of field is 1 m
therefore, dimensions of field= 1m x1m
Hope its correct
2007-04-15 10:39:14 · answer #4 · answered by siao_commando 1 · 0⤊ 0⤋
1) the perimeter of a rectangle is 2*L+2*l (L,l= the lenght of the sides)
the area is L*l
so you have to solve the sistem:
2L+2l=42
l*L=104
from the first equation L=42/2-l=21-l
in the second:
l(21-l)=104
l^2-21l+104=0
l1=8 and L1= 13
l2=13 and L2=8 (it's the same)
2) (l+4)(l+2)=15
l^2+ 6l-7=0
l1=1
l2=-7 (not good <0)
2007-04-15 10:37:49 · answer #5 · answered by ruxacelul 2 · 0⤊ 0⤋
1) The rectangle is 8m x 13m.
2) The original dimensions of the field were 1m x 1m. If you multiple 5 by 3 you get 15.
2007-04-15 10:34:36 · answer #6 · answered by Glenn N 2 · 0⤊ 0⤋
1) 2(l+b)=42
l*b=21
solving thm bth length=13 m
breadth=8 m
2)let the side be x.
final area: (x+4)*(x+2)=15
on solving x=1.
so dimensions of original field is side=1mtr.
pls give me 10 pts now!!!
2007-04-15 10:45:21 · answer #7 · answered by aniket_bond 2 · 0⤊ 0⤋
1) 13 by 8
2)x=1
2007-04-15 10:27:35 · answer #8 · answered by penguins4869 2 · 0⤊ 0⤋
1) Its dimensions are 8m x 13m.
2) Its dimensions were originally 1m x 1m.
2007-04-15 10:32:36 · answer #9 · answered by Jeremy 2 · 0⤊ 0⤋
1) the dimensions are 13m x 8m
2) 1m x 1m
2007-04-15 10:37:23 · answer #10 · answered by gma 1 · 0⤊ 0⤋