Correct Answer
56 square units
Explanation
By selecting the horizontal line segment A C \( \overline{AC} \) as the base, we can calculate the distance between B and this line as the height. Using the familiar formula for the area of a triangle,Area equals one half times base times height
$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$
we can find the area to be 56 square units. Since vertex
A is at
coordinates 2, 3 \( (2, 3) \) and vertex
C is at
coordinates 16, 3, \( (16, 3), \) the length of the base
segment A C \( \overline{AC} \) is simply the difference in their
x-coordinates:
A C equals 16 minus 2 equals 14 \( AC = 16 - 2 = 14 \) units.
The height of the triangle is the perpendicular distance from point B to line segment A C. \( \overline{AC}. \) The y-coordinate of B is 11, while line segment A C \( \overline{AC} \) is at y equals 3. \( y = 3. \) Thus, the height is the difference between these y-values: 11 minus 3 equals 8 \( 11 - 3 = 8 \) units.
Area equals one-half times base times height $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$
equals one half times 14 times 8 $$ = \frac{1}{2} \times 14 \times 8 $$
equals one half times 1 hundred 12, which equals 56 $$ = \frac{1}{2} \times 112 = 56 \text{ square units} $$
The Shoelace Method
For this problem, you had a nice horizontal base to work with, but the "Shoelace Formula" (also known as the Polygon Area Formula), is particularly convenient for finding the area of a polygon, including a triangle, given its vertices' coordinates.
The formula for the area of a triangle with vertices at x sub 1, y sub 1 \( (x_1, y_1), \) x sub 2, y sub 2 \( (x_2, y_2), \) and x sub 3, y sub 3 \( (x_3, y_3) \) is:Area equals one half, the absolute value of, x sub 1 y sub 2 plus x sub 2 y sub 3 plus x sub 3 y sub 1 minus, open parenthesis y sub 1 x sub 2 plus y sub 2 x sub 3 plus y sub 3 x sub 1, close parenthesis, end absolute value.
$$ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_1 - (y_1x_2 + y_2x_3 + y_3x_1) \right| $$
Using the vertices given in the problem, we assign coordinates x sub 1 y sub 1 equals coordinates 2, 3, \( (x_1, y_1) = (2, 3), \) coordinates x sub 2 y sub 2 equals coordinates 20, 11, \( (x_2, y_2) = (20, 11), \) and coordinates x sub 3 y sub 3 equals coordinates 16, 3. \( (x_3, y_3) = (16, 3). \) And substituting those into the shoelace formula for a triangle and evaluating, we get:Area equals one half, the absolute value of 2 times 11 plus 20 times 3 plus 16 times 3 - open parenthesis 3 times 20 plus 11 times 16 plus 3 times 2, end absolute value.
$$ A = \frac{1}{2} \left| 2 \cdot 11 + 20 \cdot 3 + 16 \cdot 3 - (3 \cdot 20 + 11 \cdot 16 + 3 \cdot 2) \right| $$
equals one half, the absolute value of 22 plus 60 plus 48 minus open parenthesis 60 plus 176 plus 6, end absolute value $$ = \frac{1}{2} \left| 22 + 60 + 48 - (60 + 176 + 6) \right| $$
equals one half, the absolute value of 130 minus 242, end absolute value $$ = \frac{1}{2} \left| 130 - 242 \right| $$
equals one half times 112, which equals 56. $$ = \frac{1}{2} \times 112 = 56 $$
The Shoelace Formula lets us calculate the area by multiplying and subtracting values from the vertices' coordinates. It's a straightforward way to find the area of any triangle or polygon when plotted on a coordinate plane. By substituting our points carefully and following the formula, we determined that triangle ABC has an area of 56 square units.
