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Answer:

42 square units

Step-by-step explanation:

For a better understanding, I add an EFGH Rectangle.

The area of EFGH Rectangle = p x l

The area of EFGH Rectangle = 10 x 8

The area of EFGH Rectangle = 80cm²

The area of BHA Triangle = 1/2 x a x t

The area of BHA Triangle = 1/2 x a x t

The area of BHA Triangle = 1/2 x 6 x 4

The area of BHA Triangle = 2 x 6

The area of BHA Triangle = 12cm²

The area of AGD Triangle = 1/2 x a x t

The area of AGD Triangle = 1/2 x 4 x 2

The area of AGD Triangle = 4cm²

The area of CFD Triangle = 1/2 x a x t

The area of CFD Triangle = 1/2 x 2 x 6

The area of CFD Triangle = 6cm²

The area of BEC Triangle = 1/2 x a x t

The area of BEC Triangle = 1/2 x 8 x 4

The area of BEC Triangle = 4 x 4

The area of BEC Triangle = 16cm²

The area of the trapezoid = EFGH Rectangle - (BHA Triangle + AGD Triangle + CFD Triangle + BEC Triangle)

The area of the trapezoid = 80 - (12 + 4 + 6 + 16)

The area of the trapezoid = 80 - 38

The area of the trapezoid = 42cm²

Conclusion :

The area of the trapezoid is 42 square units

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The area of the considered trapezoid is given by: Option B: 42 square units.

How to find the area of a trapezoid?

The area of a trapezoid is the half of the product of sum of its parallel sides to the height of that trapezoid (the distance between those parallel sides).

Thus, if we have:

Length of its parallel sides = 'a' and 'b' units respectively

Perpendicular distance between its sides = 'h' units,

Then, we get:

[tex]A = \dfrac{1}{2}\times(a+b) \times h \: \rm unit^2[/tex]

What is Pythagoras Theorem?

If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:

[tex]|AC|^2 = |AB|^2 + |BC|^2[/tex]

where |AB| = length of line segment AB. (AB and BC are rest of the two sides of that triangle ABC, AC being the hypotenuse).

Using Pythagoras theorem, we get:

[tex]|AD|^2 = (10-8)^2 + (16-12)^2\\\\|AD| = \sqrt{4 + 16} = \sqrt{20} \: \rm units[/tex]

and

[tex]|BC|^2 = (14-6)^2 + (16-12)^2\\\\|BC| = \sqrt{64 + 16} = 2\sqrt{20} \: \rm units[/tex]

The perpendicular distance between two parallel lines [tex]y = mx + c_1\\y = mx + c_2[/tex] is evaluated by:
[tex]d = \dfrac{|c_2-c_1|}{\sqrt{1+m^2}}[/tex]

The slope of AD = slope of BC = m  = [tex]\dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{10-8}{16-12} = 0.5[/tex]

The equation of line from AD would be:

[tex]y = 0.5x + c_1[/tex]

Putting coordinates of A (12,8) in it, we get:

[tex]8 = 0.5(12) + c_1\\c_1 = 2[/tex]

Thus, equation of line from AD is [tex]y = 0.5x + 2[/tex]

The equation of line from BC would be:

[tex]y = 0.5x + c_2[/tex]

Putting coordinates of B(6, 12) in it, we get:

[tex]12 = 0.5(6) + c_2\\c_2 = 9[/tex]

Thus, we get:

[tex]d = \dfrac{|c_2-c_1|}{\sqrt{1+m^2}} = \dfrac{9-2}{\sqrt{1+(0.5)^2}} = 7/\sqrt{1.25}[/tex]

Thus, the area of the trapezoid is:

A = [tex]\dfrac{1}{2} \times ( 2\sqrt{20} + \sqrt{20})\times \dfrac{7}{\sqrt{1.25}} = \dfrac{21\sqrt{20}}{2\sqrt{1.25}} = 42 \: \rm unit^2[/tex]

Thus, the area of the considered trapezoid is given by: Option B: 42 square units.

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