Can you please help me understand how to solve a trig ratio like this?
1 answer:
You might be interested in
Here we need to use what we know about rectangles to make a system of equations.
By solving that system we found that the tile has a length of 36 inches and a width of 21 inches.
Remember that for a rectangle of length L and width W, the perimeter is:
P = 2*(L + W)
And the area is:
A = W*L
Here we know that the perimeter is 114 inches, then we can write:
114in = 2*(L + W)
We also know that the area is 756 in^2, then we can write:
756 in^2 = L*W
So we found two equations, which means that we have a system of two equations with two variables:
114in = 2*(L + W)
756 in^2 = L*W
To solve this, the first step is to isolate one of the variables in one of the equations, we can isolate L in the first equation:
114in = 2*(L + W)
114in/2 = (L + W)
57in = L + W
57in - W = L
Now that we have an expression equivalent to L, we can replace it in the other equation to get:
756 in^2 = L*W
756 in^2 = (57in - W)*W
Now we can solve this for W.
756 in^2 = W*57in - W^2
W^2 - W*57in + 756 in^2 = 0
The solutions are given by the Bhaskara's formula:
Then the two possible values of the width will be:
W = (57in + 15in)/2 = 36 in
W = (57in - 15in)/2 = 21 in
Suppose that we choose the second solution, W = 21in
Now using the equation 57in - W = L we can find the value of L
L = 57in - W = 57in - 21in = 36in
L = 36in
Then we found that the tile has a length of 36 inches and a width of 21 inches.
If you want to learn more, you can read: