Each square in the grid below has area 1. Find the area of the irregular quadrilateral below.

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.
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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