the width of a basketball court is 1 m more than half the length. If the area of the court is 364 m squared, find the length and width. Sounds easy...but I do not know how to start off the problem.

first we represent the unknowns using variables:

let x = length
according to first statement, the width is 1 more than half the length,, thus
let (1/2)x + 1 = width
now we set up the equation,, since we are only given the area, we can use this to solve for length and width. recall that the area of a rectangle (shape of the court) is given by
A = L*W
where L is length and W is width.
substituting,
A = L*W
364 = x*[ (1/2)x + 1 ]
364 = x*(0.5x + 1)
364 = 0.5 x^2 + x
we can transpose 364 to right side:
0 = 0.5x^2 + x - 364
we can multiply 2 to both sides of equation to make the 0.5 to whole number:
0 = x^2 + 2x - 728
we can either factor this or use quadratic formula,, but since it's factorable, we just factor this:
0 = (x-26)(x+28)
thus
x = 26 and x = -28
since measurements are always positive, we take only
x = 26 m (the length) ; and thus
(1/2)x + 1 = 14 m (the width)

hope this helps~ :)

Thanks helped a lot

Well, it seems like basketball courts can do math too! Alright, let's put our thinking caps on and start this problem off.

Let's call the length of the basketball court "L" (in meters) and the width "W" (also in meters).

According to the information given, the width of the court is 1 meter more than half the length. So we can write this as an equation:

W = (L/2) + 1

Now, let's move on to the area of the court. The area of a rectangle is calculated by multiplying its length by its width. So, we can set up another equation:

Area = Length * Width

In this case, we know that the area is 364 square meters, so we can write:

364 = L * W

Here's where the fun begins! We can solve this system of equations using substitution. Let's substitute the value of W from the first equation into the second equation:

364 = L * ((L/2) + 1)

Now, multiply L by (L/2) and L by 1:

364 = (L^2/2) + L

To get rid of that pesky fraction, let's multiply every term in the equation by 2:

728 = L^2 + 2L

Now, rearrange the equation to look nice and square:

L^2 + 2L - 728 = 0

Alright, now it's time to capture this equation's true essence and solve it. I can't wait to see what the length and width of this witty basketball court are!

To solve this problem, let's start by assigning variables to the length and width of the basketball court.

Let's use:
L for the length
W for the width

According to the given information:
The width of the basketball court is 1 meter more than half the length, which can be expressed as W = (1/2)L + 1.

The area of a rectangle is given by the formula A = L × W, where A is the area, L is the length, and W is the width.

We know that the area of the court is 364 m², so we can substitute this into the equation, which gives us the equation:
364 = L × ((1/2)L + 1)

Now we can solve this equation to find the length and width.

To solve this problem, we can start by assigning variables to the length and width of the basketball court.

Let's say:
Length of the court = L
Width of the court = W

The problem tells us that the width of the court is 1 meter more than half the length. So we can write the equation:

W = (1/2)L + 1

The problem also states that the area of the court is 364 square meters. The formula for the area of a rectangle is A = Length * Width. In this case, it is given as:

364 = L * W

Now we have two equations:
W = (1/2)L + 1
364 = L * W

To solve these two equations simultaneously, we can substitute the value of W from the first equation into the second equation:

364 = L * [(1/2)L + 1]

Now, we can simplify and solve for L.