Deltoid - Examples, Exercises and Solutions

First, let's recall the formula for the area of a rhombus:

(Diagonal 1 * Diagonal 2) divided by 2

Now we will substitute the known data into the formula, giving us the answer:

(12*16)/2
192/2=
96

To solve the exercise, we first need to remember how to calculate the area of a rhombus:

(diagonal * diagonal) divided by 2

Let's plug in the data we have from the question

10*6=60

60/2=30

And that's the solution!

To solve the exercise, we first need to know the formula for calculating the area of a kite:

It's also important to know that a concave kite, like the one in the question, has one of its diagonals outside the shape, but it's still its diagonal.

Let's now substitute the data from the question into the formula:

(6*5)/2=
30/2=
15

Let's begin by reminding ourselves of the formula for the area of a kite

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

Both these values are given to us in the figure thus we can insert them directly into the formula:

(4*7)/2

28/2

14

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

We substitute the known data into the formula:

 $\frac{8\cdot DB}{2}=32$

We reduce the 8 and the 2:

$4DB=32$

Divide by 4

$DB=8$

In order to answer the question asked first, we need to recall some properties of the deltoid. For this purpose, let's draw deltoid $ABCD$ where we connect every two non-adjacent vertices (meaning - draw the diagonals) and mark the intersection of the diagonals with the letter $E$:

$$

Let's recall two properties of the deltoid that will help us answer the question (markings from the previous drawing):

a. Definition of a deltoid - A deltoid is a convex quadrilateral with two pairs of adjacent equal sides:

$BA=BC\\ DA=DC$

b. The diagonals in a deltoid are perpendicular to each other:

$AC\perp BD\\ \updownarrow\\ \sphericalangle BEA= \sphericalangle AED= \sphericalangle DEC= \sphericalangle CEB=90\degree$

Now we can clearly answer the question that was asked, and the answer is that the deltoid can indeed be described as composed of two isosceles triangles since triangles: $\triangle ABC,\hspace{6pt}\triangle ADC$ are isosceles - (from property a' mentioned earlier):

Or can be described as composed of four right triangles, since triangles: $\triangle AEB,\hspace{6pt}\triangle CEB,\hspace{6pt}\triangle AED,\hspace{6pt}\triangle CED$ are right triangles (from property b' mentioned earlier):

Therefore, the correct answer is answer a'.

To solve the problem, we need to remember the formula for the area of a rhombus:

The product of the diagonals multiplied together and divided by 2.

Let's plug in the data we have into the formula:

(8*X)=60
2

Note that we can simplify the fraction, thus eliminating the denominator:

4X=60

Let's divide the equation by 4

X=15

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:

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

To solve the question, we first need to remember the formula for the area of a kite:

Diagonal * Diagonal / 2

This means that if we substitute the given data we can see that:

a(2a+2)/2 = area of the kite

Let's remember that we are also given the area, so we'll put that in the equation too

a(2a+2)/2 = 6a

Now we have an equation that we can easily solve.

First, let's get rid of the fraction, so we'll multiply both sides of the equation by 2

a(2a+2)=6a*2
a(2a+2)=12a

Let's expand the parentheses on the left side of the equation

2a²+2a=12a

2a²=10a

Let's divide both sides of the equation by a

2a=10

Let's divide again by 2

a=5

And that's the solution!

We can calculate the area of rectangle ABCD like so:

$20\times30=600$

Now let's divide the deltoid along its length and width and add the following points:

Finally, we can calculate the area of deltoid PMNK as follows:

$PMNK=\frac{PN\times MK}{2}=\frac{20\times30}{2}=\frac{600}{2}=300$