One base of trapezoid AGIE is four less than three times the other shorter base. The height of the trapezoid is one half of the shorter base. The area of the trapezoid is 210 square inches. Find the lengths of the bases and the height.

average base = (3b-4+b)/2 = 2b-2

(2b-2)(b/2) = 210

b^2-b-210 = 0
(b-15)(b+14) = 0
b = 15
3b-4 =
b/2 =

To solve this problem, we can use the formula for the area of a trapezoid, which is given by:

Area = (1/2) * (base1 + base2) * height

Let's assign variables for the lengths of the bases and the height of the trapezoid. Let x represent the length of the shorter base, and let y represent the height.

We have been given the following information about the trapezoid:

1) One base of the trapezoid (let's call it base1) is four less than three times the length of the other shorter base:
base1 = 3x - 4

2) The height of the trapezoid (y) is one half of the length of the shorter base:
y = (1/2) * x

3) The area of the trapezoid is 210 square inches:
210 = (1/2) * (base1 + base2) * y

Now, let's substitute the values of base1 and y into the formula for the area:

210 = (1/2) * ((3x - 4) + base2) * (1/2) * x

Simplifying, we get:

420 = ((3x - 4) + base2) * x

Dividing both sides of the equation by x, we have:

420/x = 3x - 4 + base2

Now, we can rearrange this equation to solve for base2:

base2 = 420/x - 3x + 4

To find the lengths of the bases and the height, we need to solve for x and y. However, this problem does not provide us with enough information to find the exact values of x and y. We can only write the expressions for each length in terms of x.

Therefore, the lengths of the bases are as follows:

base1 = 3x - 4
base2 = 420/x - 3x + 4

And the height of the trapezoid is:

y = (1/2) * x