For both a rhombus and a kite, area can be found if you know the lengths of the diagonals. What is special about the diagonals of these shapes?
Area or Perimeter of Triangles and Quadrilaterals
Perimeter is the distance around a shape. To find the perimeter of any two dimensional shape, find the sum of the lengths of all the sides.
Area is the number of square units it takes to cover a two dimensional shape. The most basic shape to find the area of is a rectangle. The area of a rectangle is base times height.
\begin{align*}Area_{Rectangle}=bh\end{align*}
By dissection, a parallelogram can be turned into a rectangle.
Therefore, the area of a parallelogram is also base times height.
\begin{align*}Area_{Parallelogram}=bh\end{align*}
You can think of any triangle as half a parallelogram. If you rotate a triangle \begin{align*}180^\circ\end{align*}
Therefore, the area of a triangle is base times height divided by two. Remember that any of the three sides can be the base. Also remember that the height must be perpendicular to the base and extend to the highest point of the triangle.
\begin{align*}{Area}_{Triangle}=\frac{bh}{2}=\frac{1}{2}bh\end{align*}
These and other area formulas that are good to know are shown below.
Shape  Area Formula  Picture 
Rectangle  \begin{align*}A=bh\end{align*} 

Parallelogram  \begin{align*}A=bh\end{align*} 

Triangle  \begin{align*}A=\frac{bh}{2}\end{align*} 

Trapezoid  \begin{align*}A=\frac{(b_1+b_2)h}{2}\end{align*} 

Rhombus  \begin{align*}A=bh\end{align*} 

Kite  \begin{align*}A=\frac{d_1d_2}{2}\end{align*} 

Square  \begin{align*}A=s^2\end{align*} 
Deriving the Formula for the Area of a Trapezoid
Derive the formula for the area of a trapezoid.
A trapezoid can be thought of as half a parallelogram by rotating it \begin{align*}180^\circ\end{align*}
The base of this parallelogram is \begin{align*}b_1+b_2\end{align*}
Finding the Area and Perimeter
1. Find the area of the trapezoid.
\begin{align*}b_1=6\end{align*}, \begin{align*}b_2=10\end{align*}, \begin{align*}h=8\end{align*}.
\begin{align*}Area=\frac{(6+10)8}{2}=\frac{(16)8}{2}=64 \ units^2\end{align*}
2. Find the area and perimeter of a square with side length 10 inches.
Perimeter is the distance around the shape. A square has four congruent sides, so each side is 10 inches. The perimeter is 40 in.
Area is the number of square units it takes to cover the shape. The area is \begin{align*}10^2=100 \ in^2\end{align*}.
Examples
Example 1
Earlier, you were asked what is special about the diagonals of a rhombus and a kite.
The diagonals of both rhombuses and kites are perpendicular. This means that when the shapes are broken down into triangles, one diagonal can be the base of the triangle and a portion of the other diagonal will be the height.
Example 2
The diagonals of a rhombus bisect each other. This means that they cut each other in half. The diagonals are also perpendicular. Derive the area formula for a rhombus \begin{align*}A=\frac{d_1d_2}{2}\end{align*}.
The diagonals divide the rhombus into four triangles. For each of the triangles, the base and height are \begin{align*}\frac{d_1}{2}\end{align*} and \begin{align*}\frac{d_2}{2}\end{align*}. Therefore, the area of each triangle is \begin{align*}\frac{\left(\frac{d_1}{2} \right) \left(\frac{d_2}{2} \right)}{2}=\frac{d_1d_2}{8}\end{align*}. The rhombus is made up of four triangles with the same area. The area of the rhombus is \begin{align*}4 \left(\frac{d_1d_2}{8} \right)=\frac{d_1d_2}{2}\end{align*}.
Example 3
True or false: To find the area of a parallelogram you just multiply the lengths of two adjacent sides.
False. The base and height need to be perpendicular. In a generic parallelogram, two adjacent sides will not be perpendicular.
Example 4
Explain why the formula \begin{align*}A=bh\end{align*} works to find the area of a rhombus.
\begin{align*}A=bh\end{align*} is the formula for the area for any parallelogram. Since a rhombus is a parallelogram, this area formula works for a rhombus.
Review
1. Find the area and perimeter of a rectangle with a length of 12 inches and a width of 15 inches.
2. Find the area and perimeter of a right triangle with legs 3 cm and 4 cm and hypotenuse 5 cm.
3. Find the area of a trapezoid with bases 4 cm and 12 cm and height 9 cm.
4. The perimeter of a rectangle is 150 cm. The length is 4 times more than the width. What is the length of the rectangle?
5. The area of a triangle is \begin{align*}30 \ cm^2 \end{align*}. The base is twice as long as the height. What is the height of the triangle?
6. The perimeter of a right triangle is 24 in. The area is \begin{align*}24 \ in^2\end{align*}. The hypotenuse is 4 inches longer than the base. What are the lengths of the sides of the triangle?
7. Why is it necessary to use square units (such as \begin{align*}in^2\end{align*}) when referring to area?
8. Why don't you use square units when referring to perimeter?
9. When finding the area of a trapezoid, does it matter which sides you label as \begin{align*}b_1\end{align*} and \begin{align*}b_2\end{align*}?
10. Find the area of a square with a diagonal of 12 in.
11. Explain in your own words why you divide by two in the formula for the area of a triangle.
12. Explain in your own words why you divide by two in the formula for the area of a trapezoid.
13. Find the area of a kite with diagonals 10 cm and 18 cm.
14. Derive the formula \begin{align*}A=s^2\end{align*} for the area of a square.
15. The diagonals of a kite are perpendicular and one diagonal bisects the other diagonal. Derive the formula \begin{align*}A=\frac{d_1d_2}{2}\end{align*} for the area of a kite.
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.6.