Recall that one of the ways to find the area of a rhombus is with the following formula:

\(\displaystyle \text{Area}=\text{base}\times\text{height}\)

Now, since all four sides in the rhombus are the same, we know from the given side value what the length of the base will be. In order to find the length of the height, we will need to use sine.

\(\displaystyle \sin x=\frac{\text{height}}{\text{side}}\), where \(\displaystyle x\) is the given angle.

\(\displaystyle \text{height}=(\text{side})(\sin x)\)

Now, plug this into the equation for the area to get the following equation:

\(\displaystyle \text{Area}=(side)^2 \sin x\)

Plug in the given side length and angle values to find the area.

\(\displaystyle \text{Area}=(3)^2\sin 112=8.34\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

Recall that one of the ways to find the area of a rhombus is with the following formula:

\(\displaystyle \text{Area}=\text{base}\times\text{height}\)

Now, since all four sides in the rhombus are the same, we know from the given side value what the length of the base will be. In order to find the length of the height, we will need to use sine.

\(\displaystyle \sin x=\frac{\text{height}}{\text{side}}\), where \(\displaystyle x\) is the given angle.

\(\displaystyle \text{height}=(\text{side})(\sin x)\)

Now, plug this into the equation for the area to get the following equation:

\(\displaystyle \text{Area}=(side)^2 \sin x\)

Plug in the given side length and angle values to find the area.

\(\displaystyle \text{Area}=(7)^2\sin 129=38.08\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

Recall that one of the ways to find the area of a rhombus is with the following formula:

\(\displaystyle \text{Area}=\text{base}\times\text{height}\)

Now, since all four sides in the rhombus are the same, we know from the given side value what the length of the base will be. In order to find the length of the height, we will need to use sine.

\(\displaystyle \sin x=\frac{\text{height}}{\text{side}}\), where \(\displaystyle x\) is the given angle.

\(\displaystyle \text{height}=(\text{side})(\sin x)\)

Now, plug this into the equation for the area to get the following equation:

\(\displaystyle \text{Area}=(side)^2 \sin x\)

Plug in the given side length and angle values to find the area.

\(\displaystyle \text{Area}=(9)^2\sin 84=80.56\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

Recall that one of the ways to find the area of a rhombus is with the following formula:

\(\displaystyle \text{Area}=\text{base}\times\text{height}\)

Now, since all four sides in the rhombus are the same, we know from the given side value what the length of the base will be. In order to find the length of the height, we will need to use sine.

\(\displaystyle \sin x=\frac{\text{height}}{\text{side}}\), where \(\displaystyle x\) is the given angle.

\(\displaystyle \text{height}=(\text{side})(\sin x)\)

Now, plug this into the equation for the area to get the following equation:

\(\displaystyle \text{Area}=(side)^2 \sin x\)

Plug in the given side length and angle values to find the area.

\(\displaystyle \text{Area}=(4)^2\sin 82=15.84\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

To find the value of diagonal \(\displaystyle \small \overline{FH}\), we must first recognize some important properties of rhombuses. Since the perimeter is of \(\displaystyle \small EFGH\) is \(\displaystyle \small 24\), and by definition a rhombus has four sides of equal length, each side length of the rhombus is equal to \(\displaystyle \small 6\). The diagonals of rhombuses also form four right triangles, with hypotenuses equal to the side length of the rhombus and legs equal to one-half the lengths of the diagonals. We can therefore use the Pythagorean Theorem to solve for one-half of the unknown diagonal:

\(\displaystyle \small 6^2=5^2+x^2\), where \(\displaystyle \small 6\) is the rhombus side length, \(\displaystyle \small 5\) is one-half of the known diagonal, and \(\displaystyle \small x\) is one-half of the unknown diagonal. We can therefore solve for \(\displaystyle \small x\):

\(\displaystyle \small x^2=6^2-5^2=36-25=11\)

\(\displaystyle \small x\) is therefore equal to \(\displaystyle \small \sqrt{11}\). Since \(\displaystyle \small x\) represents one-half of the unknown diagonal, we need to multiply by \(\displaystyle \small 2\) to find the full length of diagonal \(\displaystyle \small \overline{FH}\).

The length of diagonal \(\displaystyle \small \overline{FH}\) is therefore \(\displaystyle \small 2\sqrt{11}\)

This problem relies on the knowledge of the equation for the area of a rhombus, \(\displaystyle \small \small A=\frac{pq}{2}\), where \(\displaystyle \small A\) is the area, and \(\displaystyle \small p\) and \(\displaystyle \small q\) are the lengths of the individual diagonals. We can substitute the values that we know into the equation to obtain:

\(\displaystyle \small 273=\frac{(13)q}{2}\)

\(\displaystyle \small q=\frac{2 \cdot 273}{13}\)

\(\displaystyle \small q=42\)

Therefore, our final answer is that the diagonal \(\displaystyle \small \overline{KN}=42\)

The area of a rhombus is given below.

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

Substitute the given area and a diagonal.  Solve for the other diagonal.

\(\displaystyle 12=\frac{6\cdot d2}{2}\)

\(\displaystyle 24=6\cdot d2\)

\(\displaystyle 4 =d2\)

The area of a rhombus is given below. Plug in the area and the given diagonal. Solve for the other diagonal.

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

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

\(\displaystyle d2=4\)

Let the shorter diagonal be \(\displaystyle d1\), and the longer diagonal be \(\displaystyle d2\).  The longer dimension is twice as long as the other diagonal.  Write an expression for this.

\(\displaystyle d2=2\cdot d1\)

Write the area of the rhombus.

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

Since we are solving for the shorter diagonal, it's best to setup the equation in terms \(\displaystyle d1\), so that we can solve for the shorter diagonal.  Plug in the area and expression to solve for \(\displaystyle d1\).

\(\displaystyle 5=\frac{d1 \cdot (2\cdot d1)}{2}\)

\(\displaystyle 5=d1^2\)

\(\displaystyle \sqrt5=d1\)

A rhombus is a quadrilateral with four sides of equal length. Rhombuses have diagonals that bisect each other at right angles.

Thus, we can consider the right triangle \(\displaystyle AED\) to find the length of diagonal \(\displaystyle \overline{AC}\). From the problem, we are given that the sides are \(\displaystyle 10\:cm\) and \(\displaystyle \overline{BD} = 12\:cm\). Because the diagonals bisect each other, we know:

\(\displaystyle \overline{AD} = 10\:cm\)

\(\displaystyle \overline{ED} = 6\:cm\)

Using the Pythagorean Theorem,

\(\displaystyle a^2 + b^2 = c^2\)

\(\displaystyle (\overline{AE})^2 + (6\:cm)^2 = (10\:cm)^2\)

\(\displaystyle (\overline{AE})^2 + 36\:cm^2 = 100\:cm^2\)

\(\displaystyle (\overline{AE})^2 = 64\:cm^2\)

\(\displaystyle \overline{AE} = 8\:cm\)

\(\displaystyle \overline{AC} = 16\:cm\)