Area of Trapezoids and Kites Formulas
The area is the extent of the region occupied by a shape on a plane. The area of a shape tells us how many unit squares can fit within the boundary of the shape. Area is measured in square units such as square inch, square centimeter, square mile, etc. We use different formulas to find the area of different shapes. Here we will focus on the formula used to find the area of trapezoids and kites.
List of formulas:
- Area of trapezoid \( =\frac{1}{2}h(a+b) \)
Here, a and b are the bases of the trapezoid, and h is its height.
- Area of kite \( =\frac{pq}{2} \)
Here p and q are the diagonals of the kite.
Trapezoids
Trapezoids are quadrilaterals with at least one set of parallel sides, known as bases. The non-parallel sides of trapezoids are known as legs. We come across various real-life examples of trapezoids quite frequently. Trapezoids can be found in the popcorn tubs that have a wide opening and a narrow base, Chinese takeout boxes, typical bedside lamps, flower pots, handbags and much more. In all of these shapes, one set of sides is parallel to the other.
Area of Trapezoids
We can find the area of a trapezoid by dividing the shape into triangles and rectangles and adding up the areas of the individual shapes. But the easiest way to find the area of a trapezoid is to use the direct formula. To find the area of a trapezoid, we multiply its height by the sum of the length of its bases and divide the result by 2.
Kites
Kites are a different set of quadrilaterals that have two pairs of sides adjacent to each other that are equal in length. Kites are not as common as trapezoids but this shape can be observed very easily in kites.
Area of Kites
We can calculate the area of a kite by dividing the shape into two triangles. But, we have a direct formula that we can use to find the area of a kite. We just need to find the product of the length of the diagonals and divide the result by 2. This formula is derived from the idea of splitting a kite into two triangles in order to find its area.
Solved Examples
Example 1: Find the area of the trapezoid.
Solution:
Area of a trapezoid \( =\frac{1}{2}h(a+b) \)
\( ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2}\times 4(7+10) \)
\( ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\times 17 \)
\( ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=34 \) square meters
So, the area of the trapezoid is 34 square meters.
Example 2: Find the area of the kite. (sides are in inches).
Solution:
Here, AO = OC = 3 in
So,
AC = p
= 3 + 3
= 6 in
BO = 8 in
OD = 4 in
BD = BO + OD
So, q = 12 in
Area of a kite \( =\frac{pq}{2} \)
\( ~~~~~~~~~~~~~~~~~~~~~~=\frac{6\times 12}{2} \)
\( ~~~~~~~~~~~~~~~~~~~~~~=6\times 6 \)
\( ~~~~~~~~~~~~~~~~~~~~~~=36 \) sq. inches
So, the area of the kite is 36 square inches.
Example 3: The bases of a trapezoidal window are 26 inches and 18 inches long. Find the distance between the parallel sides if the area of the window is 528 square inches.
Solution:
We know the area and the length of the bases of the trapezoid. We need to find the height of the trapezoid.
Area of a trapezoid \( =\frac{1}{2}h(a+b) \)
\( 528=\frac{1}{2}\times h\times (26+18) \)
\( 528=\frac{1}{2}\times h\times 44 \)
\( 528= h\times 22 \) Divide both sides by 22,
\( \frac{528}{22}=h \)
\( h=24 \) inches
Therefore, the window is 24 inches tall.
Example 4: A shepherd herds a flock of sheep in a land that has the shape of a trapezoid. If the perpendicular distance between the parallel sides is 2 miles and the length of the parallel sides is 3 miles and 5 miles, find the area of land available for the flock of sheep.
Solution:
The field should look like this:
One base of the trapezoid is 3 miles long and the other base is 5 miles long. The distance between the two bases is 2 miles.
So, a = 3 miles, b = 5 miles and h = 2 miles
Area of a trapezoid \( =\frac{1}{2}h(a+b) \)
\( ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2}\times 2\times (5+2) \)
\( ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=7 \) square miles
Therefore, the area of the field available for the flock of sheep is 7 square miles.
Example 5: Penny and Leo reached the final round of a kite flying competition. The diagonals of Penny’s kite were 3 feet and 6 feet long, and the diagonals of Leo’s kite were 4 feet 5 feet long. Who has the bigger kite?
Solution:
For Penny’s kite:
p = 3 feet and q = 6 feet
Area of a kite \( =\frac{pq}{2} \)
\( ~~~~~~~~~~~~~~~~~~~~~~=\frac{3\times 6}{2} \)
\( ~~~~~~~~~~~~~~~~~~~~~~=3\times 3 \)
\( ~~~~~~~~~~~~~~~~~~~~~~=9 \) square feet
For Leo’s kite:
p = 4 feet and q = 5 feet
Area of a kite \( =\frac{pq}{2} \)
\( ~~~~~~~~~~~~~~~~~~~~~~=\frac{4\times 5}{2} \)
\( ~~~~~~~~~~~~~~~~~~~~~~=2\times 5 \)
\( ~~~~~~~~~~~~~~~~~~~~~~=10 \) square feet
Therefore, Leo had a bigger kite.
Trapezoids are quadrilaterals that have at least one pair of parallel sides. On the other hand, kites are quadrilaterals that have two pairs of adjacent sides of equal length.
A kite can be a trapezoid, and a trapezoid can be a kite only if it is a rhombus. So, a pair of adjacent sides need to be equal and the opposite sides have to be parallel.
Squares and rhombuses can be considered as kites as a pair of adjacent sides of both squares and rhombuses are equal.
Parallelograms can be considered a trapezoid because at least one pair of lines are parallel in the case of a parallelogram.
The diagonals of a kite are perpendicular bisectors. That means they create 90-degree angles at the center. If we know the length of the sides of a kite, we can use the Pythagorean theorem to find the length of the diagonals.