Write the formula for the area of a rhombus.

\(\displaystyle A=\frac{d_1\cdot d_2}{2}\)

Substitute the given diagonal lengths:

\(\displaystyle A=\frac{(2x+2)(4x+4)}{2}\)

Use FOIL to multiply the two parentheticals in the numerator:

First: \(\displaystyle 2x\cdot4x=8x^2\)

Outer: \(\displaystyle 2x\cdot4=8x\)

Inner: \(\displaystyle 2\cdot4x=8x\)

Last: \(\displaystyle 2\cdot 4=8\)

Add your results together:

\(\displaystyle A=\frac{8x^2+16x+8}{2}\)

Divide all elements in the numerator by two to arrive at the correct answer:

\(\displaystyle A=(4x^2+8x+4) \:in^2\)

Construct the other diagonal of the rhombus, which, along with the first one, form a pair of mutual perpendicular bisectors.

By the Pythagorean Theorem, 

\(\displaystyle n = \sqrt{12^{2}- 10^{2}} =\sqrt{144- 100} = \sqrt{44}= \sqrt{4}\cdot \sqrt{11} = 2 \sqrt{11}\)

The rhombus can be seen as the composite of four congruent right triangles, each with legs 10 and \(\displaystyle 2 \sqrt{11}\), so the area of the rhombus is 

\(\displaystyle A = 4 \cdot \frac{1}{2}\cdot 10 \cdot 2\sqrt{11} =40 \sqrt{11}\).

Each side of a rhombus is congruent, so if it has perimeter 48, it has sidelength 12. Also, the diagonals of a rhombus are each other's perpendicular bisectors, so if they are both constructed, and their point of intersection is called \(\displaystyle M\), then \(\displaystyle AM = 6\). The following figure is formed by the rhombus and its diagonals.

\(\displaystyle \bigtriangleup AMB\) is a right triangle with its short leg half the length of its hypotenuse, so it is a 30-60-90 triangle, and its long leg measures \(\displaystyle 6\sqrt{3}\) by the 30-60-90 Theorem. Therefore, \(\displaystyle BD = 12\sqrt{3}\). The area of a rhombus is half the product of the lengths of its diagonals:

\(\displaystyle \frac{1}{2} \times 12 \times 12 \sqrt{3} = 72 \sqrt{3}\)

The equation for the area of a rhombus is given by:

\(\displaystyle A= \frac{p \cdot q}{2}\)

where \(\displaystyle p\) and \(\displaystyle q\) are the two diagonal lengths. 

This problem very quickly becomes one of the "plug and chug" type, where the given values just need to be substituted into the equation and the equation then solved. By plugging in the values given, we get:

\(\displaystyle A = \frac{12\:cm \cdot 6\:cm}{2}\)

\(\displaystyle A = \frac{72\:cm^2}2{}\)

\(\displaystyle A= 36\:cm^2\)

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 rhombus will vary as the angles made by its sides change. The "flatter" the rhombus is (with two very small angles and two very large angles, say 2, 178, 2, and 178 degrees), the smaller the area is. There is, of course, a lower bound of zero for the area, but the area can get arbitrarily small. This implies that the correct answer would be the largest choice. In fact, the largest area of a rhombus occurs when all four angles are equal, i.e. when the rhombus is a square. The area of a square of side length 5 is 25, so any value bigger than 25 is impossible to acheive.

Finding the area of a rhombus follows the formula: 

\(\displaystyle \frac{p\times q}{2}\)

In this rhombus, you will find that \(\displaystyle p=4-(-2)\), which are the two x coordinates.

The length of q is a more involved process. You can find q by using the Pythagorean Theorem. 

\(\displaystyle 1^2+8^2=65\)

\(\displaystyle \sqrt{65}=8.06\).

Therefore, the area is 

\(\displaystyle \frac{6\times 8.06}{2}=24.18\).

1) The given rhombus is divded into two congruent isosceles triangles.

2) Each isosceles triangle has a height \(\displaystyle h = \frac{1}{2}d_{1}\) and a base \(\displaystyle b = d_{2}\).

3) The area \(\displaystyle A\) of each isosceles triangle is \(\displaystyle A = \frac{1}{2}bh\).

4) The areas of the two isosceles triangles are added together,

\(\displaystyle A + A\)

\(\displaystyle = \frac{1}{2}(\frac{1}{2}d_{1})(d_{2}) + \frac{1}{2}(\frac{1}{2}d_{1})(d_{2})\)

\(\displaystyle = \frac{1}{4}d_{1}d_{2} +\frac{1}{4}d_{1}d_{2}\)

\(\displaystyle =\frac{1}{2}d_{1}d_{2}\)

Recall that the diagonals of the rhombus are perpendicular bisectors. From the given side and the given diagonal, we can find the length of the second diagonal by using the Pythagorean Theorem.

\(\displaystyle \text{side}^2=(\frac{\text{diagonal 1}}{2})^2-(\frac{\text{diagonal 2}}{2})^2\)

Let the given diagonal be diagonal 1, and rearrange the equation to solve for diagonal 2.

\(\displaystyle (\frac{\text{diagonal 2}}{2})^2=\text{side}^2-(\frac{\text{diagonal 1}}{2})^2\)

\(\displaystyle (\frac{\text{diagonal 2}}{2})=\sqrt{\text{side}^2-(\frac{\text{diagonal 1}}{2})^2}\)

\(\displaystyle \text{diagonal 2}=2\sqrt{\text{side}^2-(\frac{\text{diagonal 1}}{2})^2}\)

Plug in the given side and diagonal to find the length of diagonal 2.

\(\displaystyle \text{diagonal 2}=2\sqrt{13^2-(\frac{10}{2})^2}=2\sqrt{144}=24\)

Now, recall how to find the area of a rhombus:

\(\displaystyle \text{Area of Rhombus}=\frac{(\text{diagonal 1})(\text{diagonal 2})}{2}\)

Plug in the two diagonals to find the area.

\(\displaystyle \text{Area of Rhombus}=\frac{(10)(24)}{2}=120\)