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 2 isosceles triangles that share a common base.

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

A quadrilateral that has 2 pairs of equal adjacent sides

Example:

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

Then: ABCD 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° 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.
A1 - Deltoid

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True or false:

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

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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 2 isosceles triangles in a kite is called the secondary diagonal.

2 - Kite Main diagonal and 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)

3 - Convex Kite


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Concave Kite

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

4 - Concave kite


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.

new Deltoid_in_kite_shape


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.

3- Kite Rhombus


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).

5 - Kite Square


And, of course, the deltoid itself:

2 pairs of equal adjacent sides.

Deltoid examples 1


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 AD=AB , and also CD=CB CD=CB .

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

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

Therefore: ABC=ADC ∢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: ABC \triangle ABC with ADC \triangle ADC

We would obtain:  

AB=AD AB=AD (given)

BC=DC BC=DC (given)

AC=AC AC=AC (common side)

Therefore, we can conclude:

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

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

Kite practice part 2

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.

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

Therefore: DAC=BAC ∢DAC=∢BAC

Also: BCA=DCA ∢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 AD=AB After all, triangle ADB ADB is an isosceles triangle.

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

Therefore: ACDB AC\perp DB and also: DM=BM DM=BM

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


ABCD ABCD is a kite,

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

ABCD is a kite

X=AB=AD X=AB=AD

X=5cm X=5\operatorname{cm}

According to the definition of a kite.

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

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

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


Check your understanding

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

5+5+4+4=18cm 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 6cm=3+3=BD

And to calculate the length of the main diagonal AC 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 ABO \triangle ABO we obtain:

AO2+32=52 AO^2+3^2=5^2

AO2+9=25 AO^2+9=25

AO2=16 AO^2=16 and we apply the \sqrt{}

AO=4cm AO=4\operatorname{cm}

And in the triangle CBO \triangle CBO we obtain:

CO2+32=42 CO^2+3^2=4^2

9+CO2=16 9+CO^2=16

CO2=7 CO^2=7

2.645cm=CO 2.645cm=CO

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

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

We can calculate the area of the deltoid:

6.645×62=19.935cm2 \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:

This is a kite Answer: not necessarily


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 DO=BO

ACDB AC\perp DB

Is it accepted as a kite?

Is it accepted as a kite?

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

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

According to this, ABCD 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:

A1=A2,C1=C2 ∢A_1=∢A_2,∢C_1=∢C_2

Prove: ABCD ABCD is a kite

ABCD is a kite original

Proof:

A1=A2 ∢A_1=∢A_2 (Given)

C1=C2 ∢C_1=∢C_2 (Given)

AC=AC AC=AC (common side)

Therefore:

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

Therefore:

AB=AD AB=AD

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


Test your knowledge

Exercises

Exercise 1

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

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

The diagonal of the deltoid

Crosses point o.

6 - Deltoid ABCD

Task

Calculate the side of CD CD .

Solution:

Data: Deltoid ABCD ABCD .

Data: Area of ABCD=48cm2 ABCD=48\operatorname{cm}² .

Data: OC=4 OC=4 .

Data: AO=12 AO=12 .

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

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

Calculation: AC=16 AC=16 .

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

Solving for variables: 16×BD2=48 \frac{16\times BD}{2}=48 .

Calculation: 96=16×BD 96=16\times BD .

BD=6cm BD=6\operatorname{cm} .

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

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

Solving for variables DO+OB=6 DO+OB=6 .

Image 7 - Calculate the side of CD

DO2=6 DO²=6 .

DO=3 DO=3 .

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

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

Let's look at the right triangle COD COD :

DO2+CO2=CD2 DO²+CO²=CD² Pythagorean Theorem.

DO=3,CO=4 DO=3,CO=4 Proof.

32+42=CD2 3²+4²=CD² Solving for variable.

9+16=CD2 9+16=CD² .

CD2=25 CD²=25 .

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

Answer:

CD=5cm CD=5\operatorname{cm} .


Exercise 2

Given: The deltoid ABCD ABCD

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

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

The secondary diagonal is equal to a a

Given The deltoid ABCD

Task

Calculate the value of: a a

Solution:

Given the area of the deltoid:

A=AC×DB2=6a1 A=\frac{AC\times DB}{2}=\frac{6a}{1}

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

DB=a DB=a (Secondary diagonal)

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

2a2+2a=12a 2a^2+2a=12a

2a2=10a 2a^2=10a /: (Divide by a)

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

a=5cm a=5\operatorname{cm}

Answer:

a=5cm a=5\operatorname{cm}


Do you know what the answer is?

Exercise 3

Given a kite ABCD ABCD

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

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

Exercise 3 Given the kite ABCD

Task

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=AC×DB2 A=\frac{AC\times DB}{2}

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

AD=4cm AD=4\operatorname{cm}

The formula cannot be applied because the diagonal AC 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 ABCD

Given that Area ABCD=42cm2 ABCD=42\operatorname{cm}²

Given that BD=14 BD=14

Task:

Calculate the value of AO AO

Solution:

Given that ABCD ABCD is a kite

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

BD=14 BD=14

The formula to calculate the area of the kite is:

AC×BD2= A \frac{AC\times BD}{2}=\text{ A}

AC×142= 42 \frac{AC\times 14}{2}=\text{ 42}

84=14×AC 84=14\times AC

AC=6 AC=6

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

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

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

2AO=6 2AO=6 Q.E.D

AO=3cm AO=3\operatorname{cm}

Answer:

AO=3cm AO=3\operatorname{cm}


Check your understanding

Exercise 5

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

Side AC=8 AC=8

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

Calculate the value of the white area

Task:

Calculate the value of the white area

Solution:

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

Given that DO=3cm 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= MN×KM A=\text{ MN}\times KM

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

DO=OB=3 DO=OB=3

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

Therefore: DO+OB=DB=6 DO+OB=DB=6

DB=KM (Equal parallel sides in a rectangle)

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

Area of the deltoid:

A=ABCD=AC×DB2=8×62=24cm2 A=\text{ABCD=}\frac{AC\times DB}{2}=\frac{8\times 6}{2}=24cm²

Area of rectangle - Area of deltoid = Dotted area

4824=24cm2 48-24=24\operatorname{cm}²

Answer:

The answer is 24cm2 24\operatorname{cm}²


Exercise 6

Given the concave kite ABCD ABCD

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

The area of the kite is equal to 108Xcm2 108X \operatorname{cm}^2 .

Given the concave kite ABCD

Task:

Calculate the side DB DB

DB=X DB=X

Solution:

Given the area of the kite =108X =108X

Given: DB=X DB=X

Given:

AC=75%X=34X AC=\frac{75\%}{X}=\frac{3}{4}X

This is because AC AC is equal to 75% 75\% of DB DB which is equal to 34 \frac{3}{4} ,DB DB

Formula to calculate the area of the kite =

A=AC×DB2=108X A=\frac{AC\times DB}{2}=108X

AC×DB=216X AC\times DB=216X

X×34X=216X X\times\frac{3}{4}X=216X

34X2=216X \frac{3}{4}X^2=216X : divide by 34 \frac{3}{4}

X2=288X X^2=288X : (divided by X)

X=288 X=288

Answer:

288cm 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


Test your knowledge

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.

S=32S=32S=32888AAABBBCCCDDD

Video Solution

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:

 8DB2=32 \frac{8\cdot DB}{2}=32

We will reduce the 8 and the 2:

4DB=32 4DB=32

Divide by 4

DB=8 DB=8

Answer

8 cm

Exercise #2

Look at the deltoid in the figure:

777444

What is its area?

Video Solution

Step-by-Step Solution

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

Diagonal1×Diagonal22 \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

Answer

14

Exercise #3

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

AB = BC

DC = AD

BD = 14

The diagonals of the kite intersect at point 0.

Calculate the length of side AO.

S=42S=42S=42141414DDDAAABBBCCCOOO

Video Solution

Step-by-Step Solution

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

S=AC×BD2 S=\frac{AC\times BD}{2}

42=AC×142 42=\frac{AC\times14}{2}

We multiply by 2 to remove the denominator:

 14AC=84 14AC=84

Then divide by 14:

AC=6 AC=6

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

AO=AC2=62=3 AO=\frac{AC}{2}=\frac{6}{2}=3

Answer

3 cm

Exercise #4

True or false:

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

Video Solution

Answer

False.

Exercise #5

True or false?

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

Video Solution

Answer

False.

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