Mark’s vegetable garden is in the shape of a rectangle, and he can enclose it with 96 feet of fencing. if the length of his garden is 16 feet more than 3 times the width, what are the dimensions of the garden?

length = 40 ft; width = 8 ft

Step-by-step explanation:

Let the width = x.

The fence has to go around the entire perimeter of the garden.

A rectangle has 4 sides: 2 lengths and 2 widths.

The perimeter is the sum of the lengths of the 4 sides.

perimeter = length + length + width + width

We are told the perimeter is 96 ft, so we have:

length + length + width + width = 96

Above, we let the width = x.

The length of the rectangle is 16 ft more than 3 times the width.

Since the width is x, the length is 3 times the width, 3x, plus 16, so the width is 3x + 16. We substitute x for each width and 3x + 16 for each length.

length + length + width + width = 96

3x + 16 + 3x + 16 + x + x = 96

Add like terms on the left side.

3x + 3x + x + x + 16 + 16 = 96

8x + 32 = 96

Subtract 32 from both sides.

8x = 64

Divide both sides by 8.

x = 8

Since we let the width = x, now we know the width is 8 ft.

The length is 16 ft more than 3 times the width.

length = 3x + 16 = 3(8) + 16 = 24 + 16 = 40

length = 40 ft; width = 8 ft

Check: 40 ft is indeed 16 more than 3 times 8 since 3(8) + 16 = 40.

Also, 2 lengths of 40 ft and 2 widths of 8 ft do add to 96 ft.

Our answer is correct.

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.