## Some Basic Concepts of the Kite

**Main diagonal:** The diagonal that passes between the identical sides in a kite.

**Secondary diagonal:** The common base of $2$ isosceles triangles in a kite is called the secondary diagonal.

**Vertex angles:** The angles between the equal sides in a kite.

**Base angles:** The angles through which the common base passes.

## Types of Kites

### Convex Kite

**Convex Kite:** A kite with diagonals on the inside (as in the images of the kites above)

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### Concave Kite

**Concave Kite:** A kite with one of its diagonals outside (like a kind of bowl).

On many occasions, when we sit on the beach facing the sea, we observe a good number of kites. Have you looked at their shape? This is a deltoid shape. The deltoid has a somewhat complicated form. It's a quadrilateral but not a square, and it has a shape similar to a rhombus and a parallelogram, but their definitions are different. In this article, we will learn what a deltoid is and how to identify it.

## Who Else Belongs to the Kite Family?

### Diamond Shape

**Rhombus:** All sides are equal vertical diagonals, diagonals that cross each other and bisect the angles, from each side we look at the quadrilateral of the kite. The rhombus is actually an equilateral kite.

### Square

**Square****:** The most elaborate of the group: its diagonals are perpendicular and intersect; they cross the angles as in a rhombus, but in a square, the lengths of the diagonals are equal as in a rectangle. Also, from every side we look, we'll notice 2 isosceles triangles with a common base, so the characteristics of the kite will also be present in it. The square is a kite with equal sides and angles (all angles are right angles).

**And, of course, the deltoid itself:**

2 pairs of equal adjacent sides.

## Deltoid Test

### Why are the base angles equal in a kite?

**We will use the definition of a Kite:** 2 equilateral triangles with a common base

**Therefore:** $AD=AB$, and also $CD=CB$.

**According to this:** $∢ABD=∢ADB$ Because the base angles in an equilateral triangle are equal

**Also:** $∢BDC=∢DBC$ Base angles in an isosceles triangle are equal

**Therefore:** $∢ABC=∢ADC$ We combine equal angles with equal angles so that the sum of the angles is equal (the total amount)

Even if we overlaid the triangles: $\triangle ABC$ with $\triangle ADC$

**We would obtain: **

$AB=AD$ (given)

$BC=DC$ (given)

$AC=AC$ (common side)

**Therefore, we can conclude**:

$\triangle ABC≅\triangle ADC$ (according to the superposition theorem: side, side, side)

$∢ABC=∢ADC$ (Corresponding angles in equal overlaid triangles)

**As a result of the overlay, the kite principle can be deduced:**

- The main diagonal in the kite intersects the angles, crosses a secondary diagonal, and is perpendicular to it.

$\triangle ABC≅\triangle ADC$ (according to the superposition theorem: side, side, side) Proven

**Therefore:** $∢DAC=∢BAC$

**Also:** $∢BCA=∢DCA$, Corresponding angles in equal overlaid triangles

- The main diagonal in the kite intersects a secondary diagonal and is perpendicular to it.

**According to the data:** $AD=AB$ After all, triangle $ADB$ is an isosceles triangle.

In an isosceles triangle the vertex angle is perpendicular to the base and bisects it.

**Therefore:** $AC\perp DB$ and also: $DM=BM$

**From this, we can calculate the missing sides and the missing angles in the given kite:**

**$ABCD$**** is a kite,**

**Find** **$X,Y,α,β$**** in the given kite**

$X=AB=AD$

$X=5\operatorname{cm}$

According to the definition of a kite.

$∢BAC=α=40°$ The main diagonal of the kite intersects the angles.

$∢ACD=β=50°$ The main diagonal of the kite intersects the angles.

$Y=3\operatorname{cm}$, the main diagonal in the kite intersects the secondary diagonal.

### Calculating the perimeter of a kite is done by adding up all its sides:

#### $5+5+4+4=18\operatorname{cm}$

#### And the calculation of the area of the deltoid is done using the product of the diagonals divided by two:

**Calculation of the secondary diagonal:** $6cm=3+3=BD$

And to calculate the length of the main diagonal $AC$ **we use the Pythagorean theorem** in right-angled triangles formed by the diagonals (as it has been proven to us that they are perpendicular to each other)

**And therefore, in the triangle** **$\triangle ABO$**** we obtain:**

$AO^2+3^2=5^2$

$AO^2+9=25$

$AO^2=16$ and we apply the $\sqrt{}$

$AO=4\operatorname{cm}$

**And in the triangle** **$\triangle CBO$**** we obtain:**

$CO^2+3^2=4^2$

$9+CO^2=16$

$CO^2=7$

$2.645cm=CO$

**Therefore, the length of the main diagonal is equal to:**

$4+2.645=6.645\operatorname{cm}$

**We can calculate the area of the deltoid:**

$\frac{6.645\times6}{2}=19.935\operatorname{cm}^2$

## Deltoid Test: What is the necessary condition to get a deltoid?

Does every quadrilateral whose diagonals are perpendicular form a kite?

**The answer is:** not necessarily

**See example:**

### If that's the case, what is the additional condition for the vertical diagonals that a kite requires for acceptance?

Let's check, here we have a quadrilateral where one diagonal crosses the other and is perpendicular to it, is it necessarily accepted to be a kite?

**Given:**

$DO=BO$

$AC\perp DB$

**Is it accepted as a kite?**

Since $DO=BO$ and also $AC\perp DB$

Therefore, we can conclude that $AD=AB$ and also $DC=BC$ (in a triangle where the altitude is also the median, it is an isosceles triangle)

According to this, $ABCD$ is a kite according to the definition: **2 isosceles triangles on a common base form a kite**.

### Another condition for a quadrilateral to be a kite: one of the diagonals bisects the angles.

**Given:**

$∢A_1=∢A_2,∢C_1=∢C_2$

**Prove:** $ABCD$ is a kite

**Proof**:

$∢A_1=∢A_2$ (Given)

$∢C_1=∢C_2$(Given)

$AC=AC$ (common side)

**Therefore:**

$\triangle ABC≅\triangle ADC$ (by the Angle-Side-Angle postulate)

**Therefore:**

$AB=AD$

$BC=DC$ (corresponding sides in congruent triangles are equal)

**If you're interested in learning how to calculate areas of other geometric shapes, you can check out one of the following articles:**

- How do you calculate the area of a trapezoid?
- How to calculate the area of a triangle
- The area of a parallelogram: what is it and how is it calculated?
- Circular area
- Surface area of triangular prisms
- How do you calculate the area of a rhombus?
- How to calculate the area of a regular hexagon?
- How to calculate the area of an orthohedron
- Proof by contradiction

**On the** **Tutorela** **blog, you'll find a variety of articles about mathematics.**

## Exercises

### Exercise 1

In the following exercise, it is necessary to know the Pythagorean Theorem.

**Given:** The deltoid $ABCD$ $\left(AB=AD, DC=BC\right)$.

The diagonal of the deltoid

Crosses point o.

**Task**

Calculate the side of $CD$.

**Solution:**

Data: Deltoid $ABCD$.

Data: Area of $ABCD=48\operatorname{cm}²$.

Data: $OC=4$.

Data: $AO=12$.

Sum of the parts $AC=AO+OC$.

Solving for variables: $AC=12+4$.

Calculation: $AC=16$.

Formula to calculate the area of deltoid abcd= $\left(AC\times BD\right):2$.

Solving for variables: $\frac{16\times BD}{2}=48$.

Calculation: $96=16\times BD$.

$BD=6\operatorname{cm}$.

In the deltoid, the main diagonal $\left(AC\right)$ crosses the secondary diagonal $\left(BD\right)$ $OB=DO$.

Sum of the parts $BD=DO+OB$.

Solving for variables $DO+OB=6$.

$DO²=6$.

$DO=3$.

$AC\perp BD$ In the deltoid the diagonals are perpendicular to each other.

Between perpendicular lines there are right angles ($90°$ degrees).

**Let's look at the right triangle \( COD \):**

$DO²+CO²=CD²$ Pythagorean Theorem.

$DO=3,CO=4$ Proof.

$3²+4²=CD²$ Solving for variable.

$9+16=CD²$.

$CD²=25$.

$CD=5\operatorname{cm}$ Q.E.D.

**Answer:**

$CD=5\operatorname{cm}$.

### Exercise 2

Given: The deltoid $ABCD$

The area of the deltoid is equal to $6a$

The main diagonal is equal to $2a+2$

The secondary diagonal is equal to $a$

**Task**

**Calculate the value of:** $a$

**Solution:**

**Given the area of the deltoid:**

$A=\frac{AC\times DB}{2}=\frac{6a}{1}$

$AC=2a+2$ (Main diagonal)

$DB=a$ (Secondary diagonal)

$12a=(2a+2)a$ (Expanding the parenthesis)

$2a^2+2a=12a$

$2a^2=10a$ /: (Divide by **a**)

$2a=10$ /: (Divide by 2)

$a=5\operatorname{cm}$

**Answer:**

$a=5\operatorname{cm}$

### Exercise 3

Given a kite $ABCD$

The diagonal $DB$ is equal to $5\operatorname{cm}$.

The side $AD$ is equal to $4\operatorname{cm}$

**Homework**

Is it possible to calculate the area of the kite? If so, calculate its area.

**Solution**

Formula for calculating the area of the kite

$A=\frac{AC\times DB}{2}$

Given that $DB=5\operatorname{cm}$

$AD=4\operatorname{cm}$

The formula cannot be applied because the diagonal $AC$ is not given, and there is no information provided that would help to find it.

**Answer**

It is not possible to calculate the area of the kite

### Exercise 4

Given the kite $ABCD$

Given that Area $ABCD=42\operatorname{cm}²$

Given that $BD=14$

**Task:**

Calculate the value of $AO$

**Solution:**

Given that $ABCD$ is a kite

The area of the kite $ABCD$ is equal to $42\operatorname{cm}$

$BD=14$

**The formula to calculate the area of the kite is:**

$\frac{AC\times BD}{2}=\text{ A}$

$\frac{AC\times 14}{2}=\text{ 42}$

$84=14\times AC$

$AC=6$

In the kite, the main diagonal crosses the secondary diagonal. $AO=OC$

$OC+AO=AC$ (Sum of the parts)

$OC+AO=AC$ variable isolation $\left(AC=6,OC=AO\right)$

$2AO=6$ Q.E.D

$AO=3\operatorname{cm}$

**Answer:**

$AO=3\operatorname{cm}$

### Exercise 5

Given that the deltoid $ABCD$ is enclosed within a rectangle $KMNH$

Side $AC=8$

Height $DO$ of the triangle $ADC$ is equal to $3\operatorname{cm}$

**Task:**

Calculate the value of the white area

**Solution:**

Given that $AC=8\operatorname{cm}$

Given that $DO=3\operatorname{cm}$

To calculate the dotted area, we calculate the area of the rectangle and subtract the area of the deltoid.

We start with the area of the rectangle:

$A=\text{ MN}\times KM$

$MN=AC=8$ (in an equilateral parallel rectangle)

$DO=OB=3$

(The main diagonal in a rhombus is perpendicular to the secondary diagonal and crosses it)

Therefore: $DO+OB=DB=6$

DB=KM (Equal parallel sides in a rectangle)

**Area of the rectangle:** $A\text{ }=6\times 8=48cm²$

**Area of the deltoid:**

$A=\text{ABCD=}\frac{AC\times DB}{2}=\frac{8\times 6}{2}=24cm²$

Area of rectangle - Area of deltoid = Dotted area

$48-24=24\operatorname{cm}²$

**Answer:**

The answer is $24\operatorname{cm}²$

### Exercise 6

Given the concave kite $ABCD$

Given that the diagonal $AC$ is equal to $75\%$ of the diagonal $DB$

The area of the kite is equal to $108X \operatorname{cm}^2$.

**Task:**

Calculate the side $DB$

$DB=X$

**Solution:**

Given the area of the kite $=108X$

Given: $DB=X$

Given:

$AC=\frac{75\%}{X}=\frac{3}{4}X$

This is because $AC$ is equal to $75\%$ of $DB$ which is equal to $\frac{3}{4}$,$DB$

Formula to calculate the area of the kite =

$A=\frac{AC\times DB}{2}=108X$

$AC\times DB=216X$

$X\times\frac{3}{4}X=216X$

$\frac{3}{4}X^2=216X$ : divide by $\frac{3}{4}$

$X^2=288X$ : (divided by X)

$X=288$

**Answer:**

$288\operatorname{cm}$

## A Brief Visual Summary of the Article

A quick visual summary of the deltoid article