Area of a Kite: Definition, Formula & Example

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As a matter of fact, there is! In this article, we will discuss a formula that calculates the area of kites and observe several worked examples that employ this technique.

Recap. Defining a Kite

Before we begin, let us begin by refreshing our memories of kites. A kite is a type of quadrilateral that has two pairs of equal adjacent sides. Like all other quadrilaterals, it contains 4 sides, 4 angles, and 2 diagonals.

The structure of a kite satisfies the characteristics of a cyclic quadrilateral. A cyclic quadrilateral is a quadrilateral where all four of its vertices lie on a circle. It is sometimes referred to as an inscribed quadrilateral. The circle that holds all four of these vertices on its circumference is called the circumcircle or circumscribed circle. Here is a diagram of a kite within a circle.

A kite inscribed inside a circle. An example of a cyclic quadrilateral.

Example of a cyclic quadrilateral

Properties of a Kite

Let us now recall the fundamental properties of a kite. Here we have a kite denoted by ABCD. M is the point at which the diagonals intersect.

A kite denoted by ABCD and M is the point at which the diagonals intersect.

Diagram of a kite

The following table is a list of its features.

Properties of a Kite	Description
It has two pairs of equal adjacent sides	AB = BC and AD = DC
It has one pair of equal opposite angles that are obtuse	∠BAD = ∠BCD > 90^o
No parallel lines
It has two non-equal diagonals	AC ≠ BD
Diagonals are perpendicular and bisect each other	AC ⊥ BD and AM = MC and BM = MD

We are now ready to learn more about the area of a Kite.

The Area Formula for a Kite

The area of a kite is the space bounded by its sides. Referring back to our previous diagram of a kite, the area formula is given by

\[A=\frac{1}{2}\times d_1 \times d_2\]

where $d_1$ and $d_2$ are the lengths of the vertical diagonal and horizontal diagonal, respectively.

$Diagram of a kite with two diagonals, one horizontal called $d_1$ and one vertical called $d_2$$

Area of a kite

Deriving the Area of a Kite

Now we have an explicit recipe for finding the area of a kite. But how did it come about? This segment will discuss a step of step derivation of how this formula actually satisfies the area of a given kite. Again, let's turn our attention back to our previous kite, shown below.

A kite denoted by ABCD and M is the point at which the diagonals intersect.

Area of a kite

For our kite ABCD above, let's call the length of the shorter diagonal $AC=x$ and the length of the longer diagonal $BD=y$. From the properties of a kite, both these diagonals are perpendicular (at right angles) and bisect each other.

With that in mind, we have

\[AM=MC=\frac{AC}{2}=\frac{x}{2}\]

The area of kite ABCD is made up of the sum of two areas: triangle ABD and triangle BCD. Writing this as an expression, we have

Area of kite ABCD = Area of ΔABD + Area of ΔBCD

Let's called this Equation 1.

The area of a triangle is the product of its base and height multiplied by half, that is,

\[\text{Area of a Triangle}=\frac{1}{2}\times b \times h\]

where $b$ is the base and $h$ is the height. Using this formula, let's determine the areas of triangle ABD and triangle BCD.

\[\text{Area of Triangle ABD}=\frac{1}{2}\times AM\times BD\]

\[\text{Area of Triangle BCD}=\frac{1}{2}\times MC\times BD\]

Now replacing AM, BD and MC by $x$ and $y$, we have

\[\text{Area of Triangle ABD}=\frac{1}{2}\times\frac{x}{2}\times y=\frac{xy}{4}\]

\[\text{Area of Triangle BCD}=\frac{1}{2}\times\frac{x}{2}\times y=\frac{xy}{4}\]

Now, using Equation 1, we obtain

\[\text{Area of kite ABCD}=\frac{xy}{4}+\frac{xy}{4}=\frac{xy}{2}\]

Finally, substituting the values of $x$ and $y$, we have the required formula for the area of a kite.

\[\text{Area of a kite}=\frac{1}{2}\times AC \times BD\]

The Area of a Kite and a Rhombus

The area formula of a kite happens to follow the same idea as the area of a rhombus. Let's recall the structure of a rhombus. Here we have a rhombus denoted by ABCD. M is the point at which the diagonals intersect.

Rhombus ABCD with midpoint D

Diagram of a rhombus

You can already see the resemblances with a kite, just by looking at this diagram. The following table is a list of its features.

Properties of a Rhombus	Description
Has four equal sides	AB = BC = CD = DA
Has opposite angles of equal measures	∠ABC = ∠CDA and ∠BCD = ∠DAB
Has two pairs of parallel sides	AB // DC and AD // BC
Has two non-equal diagonals	AC ≠ BD
Diagonals are perpendicular and bisect each other	AC ⊥ BD and AM = MC and BM = MD

The Area Formula for a Rhombus

\[A=\frac{1}{2}\times d_1 \times d_2\]

where d₁ and d₂ are the lengths of the vertical diagonal and horizontal diagonal, respectively.

$Diagram of a rhombus with two diagonals, one horizontal called $d_1$ and one vertical called $d_2$$

Area of a rhombus

Examples of Area of Kites

In this section, we shall look at several worked examples that make use of this formula that deduces the area of a kite. Here is the first example.

Cathy has 3 identical kite-shaped notecards with diagonals of lengths 5 inches and 17 inches. Determine the sum of the area for these 3 notecards.

Solution

The diagonals of each box are given by $d_1=5$ and $d_2=17$. Using the area formula for a kite, the area of one notecard is

\[A=\frac{1}{2}\times 5 \times 17=\frac{82}{2}=42.5\]

Thus, the area of each kite is 42.5 in². Since we have 3 identical notecards, we can simply multiply this area by 3 to find their total area.

\[42.5\times 3=127.5\]

Thus, the total area of all 3 notecards is 127.5 in².

Let's look at another example.

Mary has a cardboard cutout shaped like a kite. The shorter diagonal measures 3 feet while the longer diagonal measured is 14 feet. What is the area of this cutout?

She then decides to divide this cutout into 7 separate pieces of equal areas. What would the area of each piece be?

Solution

The diagonals of this cutout are given by $d_1=3$ and $d_2=14$. Using the area formula for a kite, the area of this cutout is

\[A=\frac{1}{2}\times 3 \times 14=\frac{42}{2}=21\]

Thus, the area of this cutout is 21 ft². Since Mary wants to divide this cutout into 7 identical segments, we can simply divide this area by 7 to identify the area of each piece.

\[\frac{21}{7}=3\]

Hence, the area of each piece would be 3 ft².

Here is one last example before we end this topic.

David has a kite with an area of 304 square inches. The shorter diagonal is 16 inches long. What is the length of the longer diagonal?

Solution

In this question, we are given the measures of the area and one of the diagonals of this kite, namely $A=304$ and $d_1=16$. In order to find the length of the longer diagonal, $d_2$, we need to rearrange the given formula to make $d_2$ the subject. Given that the formula for the area of a kite is

\[A=\frac{1}{2}\times d_1 \times d_2\]

Rearranging this formula so that $d_2$ becomes the subject yields

\[d_2=\frac{2A}{d_1}\]

Now substituting our known values for $A$ and $d_1$, we have

\[d_2=\frac{2\times 304}{16}=38\]

Thus, the length of the longer diagonal is 38 inches.

Area of Kites - Key takeaways

A kite is a type of quadrilateral with no parallel lines.
A kite has two pairs of equal adjacent sides and one pair of equal opposite angles that are obtuse.
A kite has two non-equal diagonals.
The diagonals of a kite are perpendicular and bisect each other.
The area of a kite is given by \[A=\frac{1}{2}\times d_1 \times d_2\] where $d_1$ and $d_2$ are the lengths of the vertical diagonal and horizontal diagonal, respectively.