Geometry - How to find the area of a rhombus

Rhombus_1

The above figure shows a rhombus . Give its area.

$40 \sqrt{11}$

$40\sqrt{61}$

Explanation

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

Rhombus_1

By the Pythagorean Theorem,

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

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

$A = 4 \cdot \frac{1}{2}\cdot 10 \cdot 2\sqrt{11} =40 \sqrt{11}$ .

Find the area of the rhombus.

Explanation

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

$\text{Area}=\text{base}\times\text{height}$

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.

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

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

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

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

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

$\text{Area}=(5)^2\sin 33=13.62$

Make sure to round to places after the decimal.

Find the area of a rhombus if the both diagonals have a length of $\sqrt{x+2}$ .

$\frac{x+2}{2}$

$\frac{\sqrt{2x+4}}{2}$

$\sqrt{2x+4}$

Explanation

Write the formula for the area of a rhombus.

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

Since both diagonals are equal, . Plug in the diagonals and reduce.

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

Find the area of the rhombus below.

Explanation

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.

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

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

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

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

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

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

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

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

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

Plug in the two diagonals to find the area.

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

Make sure to round to places after the decimal.

Which of the following shapes is a rhombus?

Shapes

Explanation

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.

Find the area of the rhombus.

Explanation

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

$\text{Area}=\text{base}\times\text{height}$

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.

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

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

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

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

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

$\text{Area}=(7)^2\sin 42=32.79$

Make sure to round to places after the decimal.

Find the area of the rhombus below.

Explanation

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.

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

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

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

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

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

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

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

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

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

Plug in the two diagonals to find the area.

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

Find the area of the rhombus.

Explanation

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

$\text{Area}=\text{base}\times\text{height}$

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.

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

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

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

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

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

$\text{Area}=(2)^2\sin 104=3.88$

Make sure to round to places after the decimal.

Find the area of a rhombus if the diagonals lengths are and .

Explanation

Write the formula for the area of a rhombus:

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

Substitute the given lengths of the diagonals and solve:

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

Show algebraically how the formula for the area of a rhombus is developed.

Varsity4

$\frac{1}{2}(\frac{1}{2}d_{1})(d_{2}) + \frac{1}{2}(\frac{1}{2}d_{1})(d_{2})$

$= \frac{1}{4}d_{1}d_{2} +\frac{1}{4}d_{1}d_{2}$

$=\frac{1}{2}d_{1}d_{2}$

$\frac{1}{2}\cdot (d_{1}\cdot d_{2})$

$= \frac{1}{2}d_{1}d_{2}$

$\frac{1}{4}d_{1}+\frac{1}{4}d_{2}$

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

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

$(\frac{1}{2}d_{1}+\frac{1}{2}d_{1})(\frac{1}{2}d_{2}+\frac{1}{2}d_{2})$

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

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

$\frac{1}{4}d_{1}\cdot \frac{1}{4}d_{2}$

$=\frac{1}{2}d_{1}d_{2}$

Explanation

The given rhombus is divded into two congruent isosceles triangles.
Each isosceles triangle has a height $h = \frac{1}{2}d_{1}$ and a base $b = d_{2}$ .
The area of each isosceles triangle is $A = \frac{1}{2}bh$ .
The areas of the two isosceles triangles are added together,