A rhombus is a four-sided figure where all sides are straight and equal in length. All opposite sides are parallel. A square is considered to be a rhombus.

Solving for the area of rhombus $\displaystyle MNOP$ requires knowledge of the equation for finding the area of a rhombus. The equation is $\displaystyle \small A=\frac{pq}{2}$, where $\displaystyle p$ and $\displaystyle \small q$ are the two diagonals of the rhombus. Since both of these values are given to us in the original problem, we merely need to substitute these values into the equation to obtain:

$\displaystyle \small A=\frac{(29)(14)}{2}=\frac{406}{2}=203$

The area of rhombus $\displaystyle \small MNOP$ is therefore $\displaystyle \small 203$ square units. 

The formula for the area of a rhombus from the diagonals is half the product of the diagonals, or in mathematical terms:

$\displaystyle A=\frac{p\cdot q}{2}$ where $\displaystyle p$ and $\displaystyle q$ are the lengths of the diagonals.

Substituting our values yields, 

$\displaystyle A=\frac{8*12}{2}=\frac{96}{2}=48$

One of the formulas for a rhombus is base times height,

$\displaystyle A=b \cdot h$

Since a rhombas has equal sides, the base is 5 and the height of the rhombus is the same as the height of the rectangle, 4.

Substituting in these values we get the following:

$\displaystyle A=5 \cdot 4 =20$

To solve for the area of the rhombus $\displaystyle \small ABCD$, we must use the equation $\displaystyle \small A=\frac{pq}{2}$, where $\displaystyle \small p$ and $\displaystyle \small q$ are the diagonals of the rhombus. Since the perimeter of the rhombus is $\displaystyle \small 40$, and by definition all 4 sides of a rhombus have the same length, we know that the length of each side is $\displaystyle \small 10$. We can find the length of the other diagonal if we recognize that the two diagonals combined with a side edge form a right triangle. The length of the hypotenuse is $\displaystyle \small 10$, and each leg of the triangle is equal to one-half of each diagonal. We can therefore set up an equation involving Pythagorean's Theorem as follows:

$\displaystyle \small 10^2=x^2+6^2$, where $\displaystyle \small x$ is equal to one-half the length of the unknown diagonal.

We can therefore solve for $\displaystyle \small x$ as follows:

$\displaystyle \small x^2=10^2-6^2=100-36=64$

$\displaystyle \small x$ is therefore equal to 8, and our other diagonal is 16. We can now use both diagonals to solve for the area of the rhombus:

$\displaystyle \small A=\frac{(16)(12)}{2}=\frac{192}{2}=96$

The area of rhombus $\displaystyle \small ABCD$ is therefore equal to $\displaystyle \small 96$

The area of a rhombus is given below. Plug in the diagonals and solve for the area.

$\displaystyle A=\frac{d1 \cdot d2}{2} = \frac{\sqrt2 \cdot \sqrt3}{2}=\frac{\sqrt6}{2}$

The area of the rhombus is given below.  Substitute the diagonals into the formula.

$\displaystyle A=\frac{d1\cdot d2}{2} = \frac{(1)(2)}{2}=1$

Write the equation for the area of a rhombus.

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

Substitute the diagonals and evaluate the area.

$\displaystyle A= \frac{\sqrt{2x}\:cm\sqrt{4x}\:cm}{2} = \frac{\sqrt{8x^2}}{2}\:cm^2 = \frac{2x\sqrt2}{2}\:cm^2=x\sqrt2\:cm^2$

Write the formula for the area of a rhombus.

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

Since both diagonals are equal, $\displaystyle d_1=d_2$.  Plug in the diagonals and reduce.

$\displaystyle A=\frac{(\sqrt x+2)(\sqrt x+2)}{2}=\frac{\sqrt{x+2}^2}{2}= \frac{x+2}{2}$

Write the formula for an area of a rhombus.

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

Substitute the diagonal lengths provided into the formula.

$\displaystyle A=\frac{d1\cdot d2}{2} = \frac{\frac{6}{x}\cdot\frac{6}{x^2}}{2}$

Multiply the two terms in the numerator.

$\displaystyle A= \frac{\frac{36}{x^3}}{2}$

You can consider the outermost division by two as multiplying everything in the numerator by $\displaystyle \frac{1}{2}$.

$\displaystyle A= \frac{1}{2}(\frac{36}{x^3})$

Multiply across and reduce to arrive at the correct answer.

$\displaystyle A= \frac{1}{2}(\frac{36}{x^3})=\frac{36}{2x^3}=\frac{18}{x^3}$