In this mini-lesson, we will explore the world of kites.

You will get to learn about the interesting concepts related to the definition of a kite, its properties, and the area of a kite.

A closed figure made with 4 line segments forms the shape of a kite.

We will explore everything about kites, which are commonly seen around us.

You can observe the shape of a kite in the kites flown by kids in the sky, and an Indian dessert - Kaju Katli.

So let's get started!

**Lesson Plan**

**What Is a Kite and its Properties?**

A kite is a quadrilateral in which two pairs of adjacent sides are equal.

The elements of a kite are its 4 angles, its 4 sides, and 2 diagonals.

Let's look at the kite ABCD.

The elements of this kite are:

- \(\angle ABC\)
- \(\angle BCD\)
- \(\angle CDA\)
- \(\angle DAB\)
- Side AB
- Side BC
- Side CD
- Side AD
- Diagonal AC
- Diagonal BD

**Properties of Kite**

A kite has:

- Two pairs of adjacent equal sides

\(AC = BC \text { and } AD = BD\)

- One pair of opposite angles (which are obtuse) that are equal

\(\angle A = \angle B\)

- Diagonals that are perpendicular to each other

\(AB \perp CD\)

- A longer diagonal that bisects the shorter diagonal

\(AO=OB\)

- a longer diagonal that bisects the pair of opposite angles

\(\angle ACD = \angle DCB\)

\(\angle ADC = \angle CDB\)

**What Is the Area of a Kite?**

The area of a kite is half the product of the lengths of its diagonals.

The formula to determine the area of a kite is:

Area = \(\dfrac{1}{2}\times d_1 \times d_2\) |

The area of kite ABCD given below is \(\dfrac{1}{2}\times AC \times BD\)

## Unlock the Derivation of the Formula

Consider a kite ABCD as shown above.

Assume the lengths of the diagonals of ABCD to be \(AC = p, BD = q\)

We know that the longer diagonal of a kite bisects the shorter diagonal at right angles, i.e., BD bisects AC and \(\angle AOB = 90^\circ, \angle BOC = 90^\circ\)

Therefore,

\(AO=OC=\dfrac{AC}{2}=\dfrac{p}{2}\)

Area of kite \(ABCD\) = Area of \(\Delta ABD\) + Area of \(\Delta BCD\ \ \ \ \cdots (1)\)

We know that,

Area of a triangle = \(\dfrac{1}{2}\times \text{Base}\times \text{Height}\)

Now, we will calculate the areas of triangles ABD and BCD

Area of \(\Delta ABD = \dfrac{1}{2}\times AO \times BD\) = \(\dfrac{1}{2}\times \dfrac{p}{2}\times q=\dfrac{pq}{4}\)

Area of \(\Delta BCD = \dfrac{1}{2}\times OC \times BD\) =\( \dfrac{1}{2}\times \dfrac{p}{2}\times q=\dfrac{pq}{4}\)

Therefore, using (1)

\(\begin{align} \text{Area of kite ABCD }&= \dfrac{pq}{4}+\dfrac{pq}{4}\\[0.2 cm]&=\dfrac{pq}{2}\\[0.2 cm]&=\dfrac{1}{2}\times BD\times AC \end{align}\)

**How Do you Calculate the Area of a Kite?**

The area of a kite is calculated by plugging in the lengths of its diagonals into the formula of the area of the kite.

Enter the lengths of the diagonals of a kite and hit the 'Calculate' button to determine its area.

- The perimeter of a kite is \(2(Side_1 + Side_2)\)
- The area of a kite is \(\dfrac{1}{2}\times d_1\times d_2\)
- A kite has two pairs of adjacent equal sides.
- A kite is a cyclic quadrilateral, hence, satisfies all the properties of a cyclic quadrilateral.

**Solved Examples**

Example 1 |

Four friends are flying kites of the same size in a park.

The lengths of diagonals of each kite are 12 in and 15 in.

Determine the sum of areas of all the four kites.

**Solution**

Lengths of diagonals are:

\(d_1=12 \text{ in}\)

\(d_2=15 \text{ in}\)

The area of each kite is:

\(\begin{align} A&=\dfrac{1}{2}\times d_1\times d_2\\&=\dfrac{1}{2}\times 12 \times 15\\&=90 \text{ in}^2\end{align}\)

Since each kite is of the same size, therefore the total area of all the four kites is:

\(4\times 90 = 360 \text{ in}^2\)

\(\therefore\) Area of the four kites is 360 \(in^2\) |

Example 2 |

Kate wants to give a kite-shaped chocolate box to her friend

She wants to paste a picture of herself with her friend to cover the top of the box.

Determine the area of the top of the box if the diagonals of the lid of the box are 9 in and 12 in.

**Solution**

\(d_1=9 \text{ in}\)

\(d_2=12 \text{ in}\)

Since the box is kite-shaped, therefore the area of the top of the box is:

\(\begin{align} A&=\dfrac{1}{2}\times d_1\times d_2\\&=\dfrac{1}{2}\times 9\times 12\\&=54 \text{ in}^2\end{align}\)

\(\therefore\) The area of the top of the box is \(54 \text{ in}^2\) |

- Can a kite be called a parallelogram?
- Is the area of a rhombus the same as the area of a kite? Why or Why not?

**Interactive Questions**

**Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.**

**Let's Summarize**

We hope you enjoyed learning about the area of a kite with the simulations and interactive questions. Now, you will be able to easily solve problems in the area of a kite and its properties in math**.**

**About Cuemath**

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Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

**Frequently Asked Questions (FAQs)**

## 1. How to find the area of a kite?

The area of a kite can be calculated using the formula \(A=\dfrac{1}{2}\times d_1\times d_2\)

## 2. How to find the diagonals of a kite?

The length of one diagonal of a kite can be found using the Pythagorean theorem.

The length of the other diagonal can be found by substituting the length of the first diagonal into the area of a kite formula if area is known.