### Area and Perimeter of Rhombuses and Kites

Recall that a **rhombus** is a quadrilateral with four congruent sides and a **kite** is a quadrilateral with distinct adjacent congruent sides. Both of these quadrilaterals have perpendicular diagonals, which is how we are going to find their areas.

Notice that the diagonals divide each quadrilateral into 4 triangles. If we move the two triangles on the bottom of each quadrilateral so that they match up with the triangles above the horizontal diagonal, we would have two rectangles.

So, the height of these rectangles is half of one of the diagonals and the base is the length of the other diagonal.

The **area of a rhombus or a kite** is \begin{align*}A=\frac{1}{2} d_1 d_2\end{align*}

What if you were given a kite or a rhombus and the size of its two diagonals? How could you find the total distance around the kite or rhombus and the amount of space it takes up?

### Examples

#### Example 1

Find the perimeter and area of the kite below.

In a kite, there are two pairs of congruent triangles. Use the Pythagorean Theorem to find the lengths of sides or diagonals.

\begin{align*}&\text{Smaller diagonal portion} && \text{Larger diagonal portion}\\
20^2+d_s^2&=25^2 && 20^2+d_l^2=35^2\\
d_s^2&=225 && \qquad \ \ d_l^2=825\\
d_s&=15 \ units && \qquad \quad d_l=5\sqrt{33} \ units\end{align*}

\begin{align*}A=\frac{1}{2} \left(15+5 \sqrt{33} \right)(40) \approx 874.5 \ units^2 && P=2(25)+2(35)=120 \ units\end{align*}

#### Example 2

Find the area of a rhombus with diagonals of 6 in and 8 in.

The area is \begin{align*}\frac{1}{2}(8)(6)=24 \ in^2\end{align*}

#### Example 3

Find the perimeter and area of the rhombus below.

In a rhombus, all four triangles created by the diagonals are congruent.

To find the perimeter, you must find the length of each side, which would be the hypotenuse of one of the four triangles. Use the Pythagorean Theorem.

\begin{align*}12^2+8^2 &=side^2 && A=\frac{1}{2} \cdot 16 \cdot 24\\
144+64 &= side^2 && A=192 \ units^2\\
side &= \sqrt{208}=4 \sqrt{13}\\
P &= 4 \left( 4\sqrt{13} \right)=16 \sqrt{13} \ units\end{align*}

#### Example 4

Find the perimeter and area of the rhombus below.

In a rhombus, all four triangles created by the diagonals are congruent.

Here, each triangle is a 30-60-90 triangle with a hypotenuse of 14. From the special right triangle ratios the short leg is 7 and the long leg is \begin{align*}7 \sqrt{3}\end{align*}

\begin{align*}P &= 4 \cdot 14=56 \ units && A=\frac{1}{2} \cdot 14 \cdot 14\sqrt{3}=98\sqrt{3} \ units^2\end{align*}

#### Example 5

The vertices of a quadrilateral are \begin{align*}A(2, 8), B(7, 9), C(11, 2)\end{align*}, and \begin{align*}D(3, 3)\end{align*}. Show \begin{align*}ABCD\end{align*} is a kite and find its area.

After plotting the points, it looks like a kite. \begin{align*}AB = AD\end{align*} and \begin{align*}BC = DC\end{align*}. The diagonals are perpendicular if the slopes are negative reciprocals of each other.

\begin{align*}m_{AC} &= \frac{2-8}{11-2}=-\frac{6}{9}=-\frac{2}{3}\\ m_{BD} &= \frac{9-3}{7-3}=\frac{6}{4}=\frac{3}{2}\end{align*}

The diagonals are perpendicular, so \begin{align*}ABCD\end{align*} is a kite. To find the area, we need to find the length of the diagonals, AC and BD.

\begin{align*}d_1&= \sqrt{(2-11)^2+(8-2)^2} && d_2=\sqrt{(7-3)^2+(9-3)^2}\\ &= \sqrt{(-9)^2+6^2} && \quad =\sqrt{4^2+6^2}\\ &= \sqrt{81+36}=\sqrt{117}=3\sqrt{13} && \quad =\sqrt{16+36}=\sqrt{52}=2\sqrt{13}\end{align*}

Plug these lengths into the area formula for a kite. \begin{align*}A=\frac{1}{2} \left(3 \sqrt{13} \right)\left( 2\sqrt{13} \right)=39 \ units^2\end{align*}

### Review

- Do you think all rhombi and kites with the same diagonal lengths have the same area?
*Explain*your answer.

Find the area of the following shapes. *Round your answers to the nearest hundredth.*

Find the area and perimeter of the following shapes. *Round your answers to the nearest hundredth.*

For Questions 12 and 13, the area of a rhombus is \begin{align*}32 \ units^2\end{align*}.

- What would the product of the diagonals have to be for the area to be \begin{align*}32 \ units^2\end{align*}?
- List two possibilities for the length of the diagonals, based on your answer from #12.

For Questions 14 and 15, the area of a kite is \begin{align*}54 \ units^2\end{align*}.

- What would the product of the diagonals have to be for the area to be \begin{align*}54 \ units^2\end{align*}?
- List two possibilities for the length of the diagonals, based on your answer from #14.

Sherry designed the logo for a new company, made up of 3 congruent kites.

- What are the lengths of the diagonals for one kite?
- Find the area of one kite.
- Find the area of the entire logo.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 10.6.