Problem:

ABCDE$\ is an equilateral pentagon with side length 2 in which angle B= angle E=120 degrees. The area of ABCDE can be expressed in the form square root of x + square root of y for some integers x and y. Find x+y.

I don't understand. I really need help.

Draw diagonals AD and AC

Now you have two isosceles triangles AED and ABC with vertex angle 120° and equal sides of 2. Those can easily be solved.

Now triangle CAD is isosceles, with base CD=2 and the equal sides the base of the other two triangles.

Now just add up the areas of the three triangles.

To solve this problem, we can use the formula for the area of an equilateral triangle.

First, let's divide the pentagon ABCDE into two congruent triangles by drawing a diagonal from A to C. Let's call the midpoint of AC as point M.

Since ABCDE is an equilateral pentagon, angle B = 120 degrees, and angle E = 120 degrees. Therefore, angles A and C must also be equal to 120 degrees each.

Now, we have an equilateral triangle AMC with side length 2 and angles A and C = 120 degrees each. The formula for the area of an equilateral triangle is given by:

Area = (side length^2 * √3) / 4

Substituting the given values, we get:

Area = (2^2 * √3) / 4
= (√3 * 4) / 4
= √3

So, the area of triangle AMC is √3.

Since triangle AMC is congruent to triangle EMC, the total area of the pentagon ABCDE is twice the area of triangle AMC:

Area of ABCDE = 2 * √3 = 2√3

Thus, x = 3 and y = 0, so x + y = 3 + 0 = 3.