Answer:
210 m
Step-by-step explanation:
Perimeter is the distance around the outside of a two-dimensional shape.
Area is the space inside of a two-dimensional shape. We can also think of area as the amount of space a shape covers.
Christen ran around the soccer field, so she ran around the perimeter.
We can find the perimeter by adding all of the side lengths.
Perimeter=60 m+45 m+60 m+45 m
Christen ran 210m.
It is all of them but except (0,4)
<span>Maximum area = sqrt(3)/8
Let's first express the width of the triangle as a function of it's height.
If you draw an equilateral triangle, then a rectangle using one of the triangles edges as the base, you'll see that there's 4 regions created. They are the rectangle, a smaller equilateral triangle above the rectangle, and 2 right triangles with one leg being the height of the rectangle and the other 2 angles being 30 and 60 degrees. Let's call the short leg of that triangle b. And that makes the width of the rectangle equal to 1 minus twice b. So we have
w = 1 - 2b
b = h/sqrt(3)
So
w = 1 - 2*h/sqrt(3)
The area of the rectangle is
A = hw
A = h(1 - 2*h/sqrt(3))
A = h*1 - h*2*h/sqrt(3)
A = h - 2h^2/sqrt(3)
We now have a quadratic equation where A = -2/sqrt(3), b = 1, and c=0.
We can solve the problem by using a bit of calculus and calculating the first derivative, then solving for 0. But since this is a simple quadratic, we could also take advantage that a parabola is symmetrical and that the maximum value will be the midpoint between it's roots. So let's use the quadratic formula and solve it that way. The 2 roots are 0, and 1.5/sqrt(3).
The midpoint is
(0 + 1.5/sqrt(3))/2 = 1.5/sqrt(3) / 2 = 0.75/sqrt(3)
So the desired height is 0.75/sqrt(3).
Now let's calculate the width:
w = 1 - 2*h/sqrt(3)
w = 1 - 2* 0.75/sqrt(3) /sqrt(3)
w = 1 - 2* 0.75/3
w = 1 - 1.5/3
w = 1 - 0.5
w = 0.5
The area is
A = hw
A = 0.75/sqrt(3) * 0.5
A = 0.375/sqrt(3)
Now as I said earlier, we could use the first derivative. Let's do that as well and see what happens.
A = h - 2h^2/sqrt(3)
A' = 1h^0 - 4h/sqrt(3)
A' = 1 - 4h/sqrt(3)
Now solve for 0.
A' = 1 - 4h/sqrt(3)
0 = 1 - 4h/sqrt(3)
4h/sqrt(3) = 1
4h = sqrt(3)
h = sqrt(3)/4
w = 1 - 2*(sqrt(3)/4)/sqrt(3)
w = 1 - 2/4
w = 1 -1/2
w = 1/2
A = wh
A = 1/2 * sqrt(3)/4
A = sqrt(3)/8
And the other method got us 0.375/sqrt(3). Are they the same? Let's see.
0.375/sqrt(3)
Multiply top and bottom by sqrt(3)
0.375*sqrt(3)/3
Multiply top and bottom by 8
3*sqrt(3)/24
Divide top and bottom by 3
sqrt(3)/8
Yep, they're the same.
And since sqrt(3)/8 looks so much nicer than 0.375/sqrt(3), let's use that as the answer.</span>
L= length= 1 1/2 yards
A= area= 3 1/2 yards^2
w= width
Area= length * width
plug in numbers you know
3 1/2= (1 1/2)(w)
convert to improper fractions
(3*2+1)/2= ((1*2+1)/2)(w)
7/2= 3/2w
divide both sides by 3/2
7/2 ÷ 3/2= w
to divide fractions, multiply by the reciprocal/inverse of 3/2
7/2 * 2/3= w
(7*2)/(2*3)= w
14/6= w
reduce by 2
7/3= width as improper fraction
OR
2 1/3= width as mixed fraction
ANSWER: The width should be 2 1/3 yards wide (or 7/3 yards wide).
Hope this helps! :)
Make into improper fraction
5 and 1/6=5+1/6=30/6+1/6=31/6
3 and 2/4=3+2/4=3+1/2=6/2+1/2=7/2
5 and 1/6 times 2 and 2/4=
31/6 times 7/2=
217/12=
216/12+1/12=
18+1/12=
18 and 1/12
