how do i solve this?help! 2 problems.?

Question

1.) A rectangular garden has a length which is 4m more than its width. A 2m wide path surrounds this garden. The area of the garden plus the path is 77m sq.Find the dimensions of the garden. Show working.



2.) A garden plot 5m by 15m has one of its longer sides next to a wall. The area of the plot is to be doubled by digging up a stip of uniform width along the other three sides. How wide should this strip be? Show working.

...Please reply within the next two hours. Thanks in advance!

Answer 1

Problem 1)

Let x represent the width of the rectangular garden. If the length is 4m more than the width, the length is x+4.

Now, you have a 2m wide path around the garden. This would make the width 4m wider, and it would make the length 4m wider:

width + 4 = x + 4
length + 4 = x + 4 + 4 = x + 8

We know the area of the garden including the path is 77m sq. The formula for this is the length * width:

length * width = 77
(x+8) * (x+4) = 77
x^2 + 4x + 8x + 32 = 77
x^2 + 12x + 32 = 77
x^2 + 12x - 45 = 0
(x - 3)(x + 15) = 0
x = 3 or x = -15

Now, we can't have a negative length or width, so we throw -15 out and just use x=3.

Our original width was x, so the width of the garden is 3m.
Our original length was x+4, so the length is 3+4 = 7m.



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Problem 2)

The garden is 5m by 15m, so the area is 5*15 = 75m sq. We need to double this area, which would make the area 75*2 = 150m sq.

We have to add uniform width to all three sides. Let this uniform width be represented by x. If we add x to the sides of the garden that are next to the wall, this would make the length of the garden 15 + x + x = 15 + 2x. Since we will only be adding the width to the front of the garden and not the back (since the back is against a wall), we will be adding x to the width. The width will be 5 + x.

The area of this larger garden will be 150m sq. The length will be 15 + 2x and the width will be 5 + x. We can re-write these so that the length is 2x + 15 and the width is x + 5. The area is:

length * width = area
(x+5)(2x+15) = 150
2x^2 + 15x + 10x + 75 = 150
2x^2 + 25x + 75 = 150
2x^2 + 25x - 75 = 0
Use quadratic equation to solve for x
x = 2.5 or x = -15

Again, length can't be negative, so we will throw out -15 and just use 2.5

So, the width of the strip to be added around the three sides in order to double the garden's area is 2.5m

Answer 2

1. Length = Width + 4 or L = W + 4
Path around garden = 2 m wide

Area of Garden = G = L*W = (W + 4)W

The area of the path is the total area using the outside of the path minus the area of the garden
Area of Path = P = (W + 4)(L + 4) - G

Total Area = G + P = G + (W + 4)(L + 4) - G = 77
(W + 4)(L + 4) = (W + 4)(W + 4 + 4) = 77
(W + 4)(W + 8) = 77
W^2 + 12W + 32 = 77
W^2 + 12W - 45 = 0
(W + 15)(W - 3) = 0

W = 3m (the negative answer does not apply)
and L = 7m

2. Area = 5*15 = 75

The longer side is against the wall so I will refer to that as the length.
If the strip is W wide then the length of the new garden = 15 + 2W
The width will then be 5 + W

New area = (15 + 2W)(5 + W) = 2*75 = 150

75 + 25W + 2W^2 = 150
2W^2 + 25W - 75 = 0
(W + 15)(2W - 5) = 0
So W = 5/2 meters or 2 and 1/2 meters

Answer 3

1). let the width of the garden be"w", so its length would be "w+4"
the width of the garden and the path is (w + 2+2) = (w + 4) as the path is on both sides and length would be (w+ 4+ 4) = (w+8).
So area of garden and path = 77
(w + 4)(w+8) = 77
w^2 + 12w + 32 = 77
w^2 + 12w - 45 = 0
w^2 -3w + 15w -45 =0
w(w -3) + 15(w-3) = 0
(w +15)(w-3) = 0
w = -15 or w =3
since width can't be negative,
w = 3
so the width of the garden = 3m
and length = 3 +4 = 7m

2. This is similar to the above one
area = 15 x 5
= 75 m^2
to double the are = 75 x 2
= 150 m^2
let the width of strip be "s"
so the new length = (15 + 2s)
and the new width = (5 + s).. as one of the sides is next to a wall
Area = 150 = (5+s)(15+2s)
2s^2 + 25s + 75 = 150
2s^2 + 25s - 75 = 0
2s^2 + 30s - 5s - 75 = 0
2s(s + 15) - 5(s + 5) = 0
(2s -5)(s +5) = 0
s = 5/2 or s = -5
again since width can't be negative
s = 5/2 or 2.5 m is the width of the strip

Answer 4

hi
1)lets assume that the width is x
so length is x+4
and the width of the garden path the is x+2
and its length is x+4+2=x+6
so area is basically width times length
so in this case area=(x+2)*(x+6)=X^2+8x+12
by using subistitution
x^2+8x+12=77
so x=5 or -13(rejected due to negative sign)
so since x=5 then width is 5 and length is 5+4=9

Answer 5

1)
x, x+4 (x+2)(x+6) = 77 =X^2+8X +12 so (x+4)^2 = 77+ 4
note (a+b)^2= a^2+2ab+b^2
x= -13 or 5 mesure is + so Garden is 5m * 9m
2)
5*15 = 75 , 2y^2+ 25y = 75 , y= 2.5m garden =20m * 7.5m

Answer 6

l=w+4
a=77=(l+2)(w+2)
now substitute,
a=77=(w+6)(w+2)
you should be able to take it from there.

original area=225=Lw=15x5
new area=450=2Lw
2Lw=(L+2x)(w+x)
where x is the width of the strip
you should be able to take it from there.