# Kite

🏆Practice deltoid

## The Deltoid and Everything You Need to Know to Verify It

### What is a Kite or Deltoid?

In geometry, a deltoid is defined as a quadrilateral consisting of $2$ isosceles triangles that share a common base.

#### So, what is the identification of a deltoid in the family of quadrilaterals?

Example:

If given : $AD=AB,DC=BC$

Then: $ABCD$ is a deltoid by definition.

• 2 isosceles triangles with a common base form a deltoid.
• The sum of the angles in the deltoid is $360°$ degrees.
• The area of the deltoid contains the number of quadrilaterals that cover the selected parts of the plane.
• The perimeter of the deltoid is the length of the thread with which we border the outline of the deltoid and is measured in units of length in meters or cm.

## Test yourself on deltoid!

True or false:

A deltoid is composed of an isosceles triangle and a right triangle.

## 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.

Do you know what the answer is?

### 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:

#### 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?

See example:

Do you think you will be able to solve it?

### 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)

## 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.

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.

$CD=5\operatorname{cm}$.

### 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 $a$

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)

$a=5\operatorname{cm}$

$a=5\operatorname{cm}$

Do you know what the answer is?

### Exercise 3

Given a kite $ABCD$

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

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

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.

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 $BD=14$

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

$AO=3\operatorname{cm}$

$AO=3\operatorname{cm}$

### 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 $3\operatorname{cm}$

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.

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

$MN=AC=8$ (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

$48-24=24\operatorname{cm}²$

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 $108X \operatorname{cm}^2$.

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)

$X=288$

$288\operatorname{cm}$

Do you think you will be able to solve it?

## A Brief Visual Summary of the Article

A quick visual summary of the deltoid article

## examples with solutions for deltoid

### Exercise #1

Shown below is the deltoid ABCD.

The diagonal AC is 8 cm long.

The area of the deltoid is 32 cm².

Calculate the diagonal DB.

### Step-by-Step Solution

First, we recall the formula for the area of a kite: multiply the lengths of the diagonals by each other and divide this product by 2.

We substitute the known data into the formula:

$\frac{8\cdot DB}{2}=32$

We will reduce the 8 and the 2:

$4DB=32$

Divide by 4

$DB=8$

8 cm

### Exercise #2

Look at the deltoid in the figure:

What is its area?

### Step-by-Step Solution

Initially, let's remember the formula for the area of a kite

$\frac{Diagonal1\times Diagonal2}{2}$

Both pieces of information already exist, so we can place them in the formula:

(4*7)/2

28/2

14

14

### Exercise #3

The kite ABCD shown below has an area of 42 cm².

AB = BC

BD = 14

The diagonals of the kite intersect at point 0.

Calculate the length of side AO.

### Step-by-Step Solution

We substitute the data we have into the formula for the area of the kite:

$S=\frac{AC\times BD}{2}$

$42=\frac{AC\times14}{2}$

We multiply by 2 to remove the denominator:

$14AC=84$

Then divide by 14:

$AC=6$

In a rhombus, the main diagonal crosses the second diagonal, therefore:

$AO=\frac{AC}{2}=\frac{6}{2}=3$

3 cm

### Exercise #4

True or false:

A deltoid is composed of an isosceles triangle and a right triangle.

False.

### Exercise #5

True or false?

In a concave deltoid, the sum of the angles is 180.