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.

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.

It is not possible to prove if it is a deltoid or not

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: