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.

$\displaystyle \text{diagonal 2}=2\sqrt{15^2-(\frac{16}{2})^2}=2\sqrt{161}$

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.

$\displaystyle \text{Area of Rhombus}=\frac{(16)(2\sqrt{161})}{2}=203.02$

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

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.

$\displaystyle \text{diagonal 2}=2\sqrt{7^2-(\frac{4}{2})^2}=2\sqrt{45}$

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.

$\displaystyle \text{Area of Rhombus}=\frac{(4)(2\sqrt{45})}{2}=26.83$

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

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.

$\displaystyle \text{diagonal 2}=2\sqrt{6^2-(\frac{3}{2})^2}=2\sqrt{\frac{135}{4}}$

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.

$\displaystyle \text{Area of Rhombus}=\frac{(3)(2\sqrt{\frac{135}{4}})}{2}=17.43$

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

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.

$\displaystyle \text{diagonal 2}=2\sqrt{5^2-(\frac{1}{2})^2}=2\sqrt{\frac{99}{4}}$

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.

$\displaystyle \text{Area of Rhombus}=\frac{(1)(2\sqrt{\frac{99}{4}})}{2}=4.97$

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

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.

$\displaystyle \text{diagonal 2}=2\sqrt{12^2-(\frac{2}{2})^2}=2\sqrt{143}$

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.

$\displaystyle \text{Area of Rhombus}=\frac{(2)(2\sqrt{143})}{2}=23.92$

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

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.

$\displaystyle \text{diagonal 2}=2\sqrt{10^2-(\frac{4}{2})^2}=2\sqrt{96}$

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.

$\displaystyle \text{Area of Rhombus}=\frac{(4)(2\sqrt{96})}{2}=39.19$

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

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.

$\displaystyle \text{diagonal 2}=2\sqrt{11^2-(\frac{3}{2})^2}=2\sqrt{\frac{475}{4}}$

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.

$\displaystyle \text{Area of Rhombus}=\frac{(3)(2\sqrt{\frac{475}{4}})}{2}=32.69$

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

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.

$\displaystyle \text{diagonal 2}=2\sqrt{14^2-(\frac{6}{2})^2}=2\sqrt{187}$

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.

$\displaystyle \text{Area of Rhombus}=\frac{(6)(2\sqrt{187})}{2}=82.05$

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

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.

$\displaystyle \text{diagonal 2}=2\sqrt{15^2-(\frac{14}{2})^2}=2\sqrt{176}$

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.

$\displaystyle \text{Area of Rhombus}=\frac{(14)(2\sqrt{176})}{2}=185.73$

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

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.

$\displaystyle \text{diagonal 2}=2\sqrt{13^2-(\frac{9}{2})^2}=2\sqrt{\frac{595}{4}}$

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.

$\displaystyle \text{Area of Rhombus}=\frac{(9)(2\sqrt{\frac{595}{4}})}{2}=109.77$

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