Answer:
D. 100 square units
Explanation:
The area of the quadrilateral can be found by finding the area of the bounding rectangle, and subtracting the areas of the triangles that are not included.
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bounding rectangle
The given points tell us the horizontal extent of the figure is from the minimum x-value (3) to the maximum (17), or 14 units. The vertical extent is from the minimum y-value (2) to the maximum y-value (15), or 13 units. Then the area of the bounding rectangle is ...
A = bh = (14)(13) = 182 . . . square units
corner triangles
The upper-left and lower-right triangles together make a rectangle that is 4 units wide and 8 units high. Its area is ...
A = bh = (4)(8) = 32 . . . square units
The lower-left and upper-right triangles together make a rectangle that is 10 units wide and 5 units high. Its area is ...
A = bh = (10)(5) = 50 . . . square units
quadrilateral area
The area of the quadrilateral is the difference between the area of the bounding rectangle and the total area of the corner triangles:
A = (182 square units) - (32 +50 square units)
A = 100 square units
The area of the quadrilateral is 100 square units.
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Additional comment
If we were to go to the trouble to find the quadrilateral's dimensions, we would find the short side is ...
√(4² +8²) = 4√5 . . . . . . . using the Pythagorean theorem
and the long side is ...
√(5² +10²) = 5√5
Then the area is ...
A = bh = (4√5)(5√5) = 20(√5)² = 100 . . . square units
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We judge the bounding rectangle solution to be simpler, because it involves only 3 multiplications and 2 additions (once we have determined the dimensions of the various parts of the figure). On the other hand, the straightforward exact computation of the quadrilateral's area involve squaring 4 numbers, 2 additions, 2 roots, and three more multiplications. That seems like more work.
The figure was created in the GeoGebra graphing application. It can tell the area directly from the polygon that connects the plotted points. No computation is involved.