Question

what is the area of pentagon ABCDE with vertices A(-5, 6), B(1, -7), C(6, 0), D(1, 3), E(1, 6)

Answer 1

What is the area of pentagon ABCDE with vertices A(-5, 6), B(1, -7), C(6, 0), D(1, 3), E(1, 6)

Solution:

1). $x_{E} = x_{D} = x_{B} = 1$ , so the points E, D and B are collinear points.

Hence

Area ABCDE = Area ABE + Area BCD.

2). $y_{A} = y_{E} = 6$ , so AE is parallel y-axis

$x_{B} = x_{E} = 1$ , so BE is parallel x-axis

Hence the triangle ABE is a right triangle and its area is

Area ABE = $\frac{1}{2} (AE \times EB)$

$AE = |x_{A} - x_{E}| = |-5 - 1| = 6$

$EB = |y_{E} - y_{B}| = |6 + 7| = 13$

Area ABE = $\frac{1}{2} (6 \times 13) = 39$

3). The points D, H and B are collinear points, so $x_{H} = x_{D} = 1$

Area BCD = $\frac{1}{2} (BD \times CH)$

$CH = |x_{C} - x_{H}| = |6 - 1| = 5$

$BD = |y_{B} - y_{D}| = |-7 - 3| = 10$

Area ABE = $\frac{1}{2} (10 \times 5) = 25$

Hence Area ABCDE = 25 + 39 = 64