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# Area and Perimeter of Rhombuses and Kites

## Area is half the product of the diagonals while the perimeter is the sum of the sides.

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Area and Perimeter of Rhombuses and Kites

What if you wanted to find the areas of different shapes on the Brazilian flag, pictured below? The flag has dimensions of 20×14\begin{align*}20 \times 14\end{align*} (units vary depending on the size, so we will not use any here). The vertices of the yellow rhombus in the middle are 1.7 units from the midpoint of each side.

Find the total area of the flag and the area of the rhombus (including the circle). Do not round your answers.

### Area and Perimeter of Rhombuses and Kites

Recall that a rhombus is an equilateral quadrilateral and a kite has 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. In the rhombus, all 4 triangles are congruent and in the kite there are two sets of congruent 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 A=12d1d2\begin{align*}A=\frac{1}{2} d_1 d_2\end{align*} if the diagonals of the rhombus or kite are d1\begin{align*}d_1\end{align*} and d2\begin{align*}d_2\end{align*}. You could also say that the area of a kite and rhombus are half the product of the diagonals.

#### Calculating the Perimeter

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.

122+82144+64sideP=side2=side2=208=413=4(413)=1613A=121624A=192\begin{align*}12^2+8^2 &= side^2 && A=\frac{1}{2} \cdot 16 \cdot 24\\ 144+64 &= side^2 && A=192\\ side &= \sqrt{208}=4 \sqrt{13}\\ P &= 4 \left(4 \sqrt{13} \right)=16 \sqrt{13}\end{align*}

#### Calculating the Perimeter and Area

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, we can determine the short leg is 7 and the long leg is 73\begin{align*}7 \sqrt{3}\end{align*}. Each diagonal is two times the length of the leg, so the lengths of the two diagonals are 14 and 143\begin{align*}14 \sqrt{3}\end{align*}.

P=414=56A=1214143=19632169.74\begin{align*}P= 4 \cdot 14=56 && A=\frac{1}{2} \cdot 14 \cdot 14 \sqrt{3}=\frac{196 \sqrt{3}}{2} \approx 169.74\end{align*}

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

For this problem, it might be helpful to plot the points. From the graph we can see this is probably a kite. Upon further review of the sides, AB=AD\begin{align*}AB = AD\end{align*} and BC=DC\begin{align*}BC = DC\end{align*} (you can do the distance formula to verify). Let’s see if the diagonals are perpendicular by calculating their slopes.

\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*}

Yes, the diagonals are perpendicular because the slopes are opposite signs and reciprocals. \begin{align*}ABCD\end{align*} is a kite. To find the area, we need to find the length of the diagonals. Use the distance formula.

\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*}

Now, 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*}

#### Flag Problem Revisited

The total area of the Brazilian flag is \begin{align*}A=14 \cdot 20=280 \ units^2\end{align*}. To find the area of the rhombus, we need to find the length of the diagonals. One diagonal is \begin{align*}20-1.7-1.7=16.6 \ units\end{align*} and the other is \begin{align*}14-1.7-1.7=10.6 \ units\end{align*}. The area is \begin{align*}A=\frac{1}{2} (16.6)(10.6)=87.98 \ units^2\end{align*}.

### Examples

Find the perimeter and area of the kites below.

In a kite, there are two pairs of congruent triangles. You will need to use the Pythagorean Theorem in both problems to find the length of sides or diagonals.

#### Example 1

'

\begin{align*}& \text{Shorter sides of kite} && \text{Longer sides of kite}\\ & 6^2+5^2=s_1^2 && 12^2+5^2=s_2^2\\ & 36+25=s_1^2 && 144+25=s_2^2\\ & \qquad \ s_1=\sqrt{61} && \qquad \quad s_2=\sqrt{169}=13\\ & P=2 \left( \sqrt{61} \right)+2(13)=2 \sqrt{61} +26 \approx 41.6\\ & A= \frac{1}{2} (10)(18)=90\end{align*}

#### Example 2

\begin{align*}& \text{Smaller diagonal portion} && \text{Larger diagonal portion}\\ & 20^2+d_s^2=25^2 && 20^2+d_l^2=35^2\\ & \qquad \ \ d_s^2=225 && \qquad \ \ d_l^2=825\\ & \qquad \ \ d_s=15 && \qquad \quad d_l=5 \sqrt{33}\\ & P=2(25)+2(35)=120\\ & A=\frac{1}{2} \left(15+5 \sqrt{33} \right)(40) \approx 874.5\end{align*}

#### Example 3

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*}

### Review

1. Do you think all rhombi and kites with the same diagonal lengths have the same area? Explain your answer.
2. Use this picture of a rhombus to show that the area of a rhombus is equal to the sum of the areas of the four congruent triangles. Write a formula and reduce it to equal \begin{align*}\frac{1}{2} d_1 d_2\end{align*}.
3. Use this picture of a kite to show that the area of a kite is equal to the sum of the areas of the two pairs of congruent triangles. Recall that \begin{align*}d_1\end{align*} is bisected by \begin{align*}d_2\end{align*}. Write a formula and reduce it to equal \begin{align*}\frac{1}{2} d_1 d_2\end{align*}.
4. The area of a kite is \begin{align*}54 \ units^2\end{align*}. What are two possibilities for the lengths of the diagonals?
5. Sherry designed the logo for a new company. She used three congruent kites. What is the area of the entire logo?

For problems 6-8, determine what kind of quadrilateral \begin{align*}ABCD\end{align*} is and find its area.

1. \begin{align*}A(-2, 2), B(5, 6), C(6, -2), D(-1, -6)\end{align*}
2. Given that the lengths of the diagonals of a kite are in the ratio 4:7 and the area of the kite is 56 square units, find the lengths of the diagonals.
3. Given that the lengths of the diagonals of a rhombus are in the ratio 3:4 and the area of the rhombus is 54 square units, find the lengths of the diagonals.
4. Sasha drew this plan for a wood inlay he is making. 10 is the length of the slanted side and 16 is the length of the horizontal line segment as shown in the diagram. Each shaded section is a rhombus. What is the total area of the shaded sections?
5. In the figure to the right, \begin{align*}ABCD\end{align*} is a square. \begin{align*}AP = PB = BQ\end{align*} and \begin{align*}DC = 20 \ ft\end{align*}.
1. What is the area of \begin{align*}PBQD\end{align*}?
2. What is the area of \begin{align*}ABCD\end{align*}?
3. What fractional part of the area of \begin{align*}ABCD\end{align*} is \begin{align*}PBQD\end{align*}?

6. In the figure to the right, \begin{align*}ABCD\end{align*} is a square. \begin{align*}AP = 20 \ ft\end{align*} and \begin{align*}PB = BQ = 10 \ ft\end{align*}.
1. What is the area of \begin{align*}PBQD\end{align*}?
2. What is the area of \begin{align*}ABCD\end{align*}?
3. What fractional part of the area of \begin{align*}ABCD\end{align*} is \begin{align*}PBQD\end{align*}?

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.

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### Vocabulary Language: English

Congruent

Congruent figures are identical in size, shape and measure.

Legs of a Right Triangle

The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.