The formula for the area of a kite:

$\displaystyle A = \frac{1}{2} (a)(b)$,

where $\displaystyle a$ represents the length of one diagonal and $\displaystyle b$ represents the length of the other diagonal.

Plugging in our values, we get:

$\displaystyle A = \frac{1}{2} (4\: m)(5\: m)$

$\displaystyle A = \frac{1}{2} (20\: m^2) = 10\: m^2$

The area of the kite is given below.  The FOIL method will need to be used to simplify the binomial.

$\displaystyle A=\frac{d1\cdot d2}{2}=\frac{(x+3)(x-6)}{2}= \frac{x^2-6x+3x-18}{2}$

$\displaystyle A= \frac{x^2-3x-18}{2}$

The formula for the area for a kite is

$\displaystyle A=\frac{d_1\cdot d_2}{2}$, where $\displaystyle d_1$ and $\displaystyle d_2$ are the lengths of the kite's two diagonals. We are given the length of these diagonals in the problem, so we can substitute them into the formula and solve for the area:

$\displaystyle A = \frac{d1\cdot d2}{2} = \frac{\sqrt10 \cdot \sqrt5}{2} = \frac{\sqrt50}{2} = \frac{5\sqrt2}{2}\:cm^2$

Write the formula for the area of a kite.

$\displaystyle A= \frac{d1\cdot d2}{2}$

Plug in the given diagonals.

$\displaystyle A= \frac{(a+b)\cdot (2a-2b)}{2}$

Pull out a common factor of two in $\displaystyle 2a-2b$ and simplify.

$\displaystyle A= \frac{(a+b)\cdot2(a-b)}{2}$

$\displaystyle A=(a+b)(a-b)$

Use the FOIL method to simplify.

$\displaystyle A=(a+b)(a-b)= (a)(a)+(a)(-b)+(b)(a)+(b)(-b)$

$\displaystyle A= a^2-b^2$

Write the formula for the area of a rhombus:

$\displaystyle A=\frac{d_1 \cdot d_2}{2}$

Substitute the given lengths of the diagonals and solve:

$\displaystyle A=\frac{d1 \cdot d2}{2} = \frac{20\:cm \cdot 40\:cm}{2} =\frac{800\:cm^2}{2} = 400\:cm^2$

Write the formula for finding the area of a rhombus. Substitute the diagonals and evaluate.

$\displaystyle A=\frac{d1\cdot d2}{2}= \frac{2a \cdot 5a^2}{2}= 5a^3$

The area of a triangle with two sides of lengths $\displaystyle a$ and $\displaystyle b$ and included angle of measure $\displaystyle \gamma$ can be calculated using the formula

$\displaystyle A = \frac{1}{2} ab \sin \gamma$.

Setting $\displaystyle a = AB = 19$, $\displaystyle b = AC = 25$, and $\displaystyle \gamma = m \angle A = 57^{\circ }$, then evaluating $\displaystyle A$:

$\displaystyle A = \frac{1}{2} (19 )(25)\sin 57^{\circ }$

$\displaystyle A = \frac{1}{2} (19 )(25) (0.8387)$

$\displaystyle A \approx 199$.

Write the formula for the area of a triangle.

$\displaystyle A=\frac{1}{2}bh$

Substitute the base and height into the formula.

$\displaystyle A=\frac{1}{2}(\frac{1}{2})(\frac{1}{4})$

Simplify the fractions.

The answer is:  $\displaystyle \frac{1}{16}$

Write the formula for the circumference of a circle.

$\displaystyle C=2\pi r$

Substitute the circumference.

$\displaystyle 3=2\pi r$

Divide by $\displaystyle 2\pi$ on both sides.

$\displaystyle \frac{3}{2\pi}= r$

We will need the formula for the area of the circle.

$\displaystyle A=\pi r^2$

Substitute the radius to find the area.

$\displaystyle A=\pi (\frac{3}{2\pi})^2 = \pi (\frac{3}{2\pi})(\frac{3}{2\pi}) = \frac{9}{4\pi}$

The answer is:  $\displaystyle \frac{9}{4\pi}$

Write the formula for the area of a square.

$\displaystyle A=s^2$

Substitute the side into the formula.

$\displaystyle A=(\pi\sqrt2)^2 = \pi^2 (2) = 2\pi ^2$

The answer is:  $\displaystyle 2\pi ^2$