Respuesta :

Answer:

31.275 units²

Step-by-step explanation:

We'll find the total area by breaking up this given area into two separate parts, finding the area of each part separately, and then adding these two sub-areas together.

First, draw a horizontal line through the rightmost vertex, which creates a triangle of base 7.5 and height 5.  

The formula for the area of a triangle of base b and height h is A = (1/2)(b)(h).  Thus, the area of the triangle we have created is

A = (1/2)(7.5)(5) = 18.75 units²

The bottom half of the given irregular quadrilateral is a trapezoid of width 3 units.  We must use the Pythagorean Theorem twice to find the lengths of the two legs.  On the far left we have a triangle with longer side 3 and shorter side 1.5; the hypotenuse of this triangle is

√(1.5² + 3²), or √11.25.  This is the measure of the shorter leg of the trapezoid.

On the far right we have another triangle with loger leg 4 and shorter leg 3.  The length of the hypotenuse is also the length of the longer leg of the trapezoid.  It is √(3°2 + 4°) = 5.

Now we'll use the formula for the area of a trapezoid:

A = (average length of legs of trapezoid)(width of trapezoid)

        √11.25 + 5                8.35              25.06

   = ------------------- * 3  = ----------- * 3 = ------------- = 12.53 units²

                 2                          2                     2

Our last step is to add together the triangle area found earlier and the trapezoid area just found here:

Total area = 18.75 units² + 12.53 units² =  31.275 units²

The total area of the irregular quadrilateral is 31.275 units²

A quadrilateral is a polygon with four sides. The area of the quadrilateral is 37.5 units².

What is a quadrilateral?

A quadrilateral is a polygon with four sides.

The area of the quadrilateral is the sum of all the areas of triangles A, B, C, and rectangle D.

1.  Area of triangle A

[tex]\rm Area\ \triangle A=\dfrac{1}{2} \times Perpendicular \times Base[/tex]

                [tex]\rm =\dfrac{1}{2} \times 8\times 4\\\\=16\ units^2[/tex]

2. Area of triangle B

[tex]\rm Area\ \triangle B=\dfrac{1}{2} \times Perpendicular \times Base[/tex]

                [tex]\rm =\dfrac{1}{2} \times 5\times 5\\\\=12.5\ units^2[/tex]

3.  Area of triangle C

[tex]\rm Area\ \triangle C=\dfrac{1}{2} \times Perpendicular \times Base[/tex]

                [tex]\rm =\dfrac{1}{2} \times 4\times 3\\\\=6\ units^2[/tex]

4. Area of rectangle D,

[tex]\rm Area\ Rectangle\ D= Perpendicular \times Base[/tex]

                            [tex]\rm =1\times 3\\\\=3\ units^2[/tex]

[tex]\text{The area of the quadrilateral} = \triangle A+\triangle B+\triangle C+\rm Rectangle\ D[/tex]

                                             [tex]=16+12.5+6+3\\\\=37.5\rm\ units^2[/tex]

Hence, the area of the quadrilateral is 37.5 units².

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