Draw a kite that has a triangle with sides 1, 1, 2x next to a triangle with sides 2x, 2, 2. Find the area A and the x that maximizes it.

The area of the kite is

a = x(√(1-x^2) + √(4-x^2))

da/dx =
[√(1-x^2) + √(4-x^2)][√(1-x^2)√(4-x^2)-x^2]
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√(1-x^2)√(4-x^2

da/dx = 0 when x = 2/√5

thanks for your help

To solve this problem, we start by drawing the given kite.

First, draw a triangle with sides 1, 1, 2x. Let's call this triangle ABC, with side AB = 2x, side AC = 1, and side BC = 1.

Next, draw another triangle with sides 2x, 2, 2. Let's call this triangle BCD, with side BC = 2x, side BD = 2, and side CD = 2.

Connect vertices A and D to form a kite shape, where A and D are adjacent vertices and B and C are opposite vertices of the kite.

Now let's find the area of the kite and determine the value of x that maximizes it.

The area of a kite can be found by multiplying the lengths of its diagonals and dividing by 2. In this case, the diagonals are AD and BC.

To find the length of AD, we can use the Pythagorean theorem on triangle ABC. The Pythagorean theorem states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In triangle ABC, the hypotenuse is side AB = 2x, and the other two sides are AC = 1 and BC = 1. Therefore, we have:

(2x)^2 = 1^2 + 1^2
4x^2 = 2
x^2 = 2/4
x^2 = 1/2
x = √(1/2)
x = 1/√2
x = √2/2
x = 0.707 (rounded to three decimal places)

Now that we have the value of x, we can find the length of AD:

AD = 2x = 2 * 0.707 = 1.414 (rounded to three decimal places)

To find the length of BC, we can observe that triangle BCD is an isosceles triangle with side lengths 2, 2, and 2x. By symmetry, we can see that BC is the base of this triangle, and its length is equal to 2. Therefore:

BC = 2

Now we can find the area of the kite. Using the formula for the area of a kite, we have:

A = (AD * BC) / 2
A = (1.414 * 2) / 2
A = 2.828 / 2
A = 1.414 (rounded to three decimal places)

Therefore, the area of the kite is approximately 1.414 square units.

To maximize the area, we need to find the maximum value of A with respect to x. Since we already found x to be approximately 0.707, we have already found the value of x that maximizes the area.

So, the area A is approximately 1.414 square units, and the value of x that maximizes it is approximately 0.707.