Question

What is the area of the trapezoid shown? trapezoid A = ______ units²?

Answer 1

Answer:

74.1 units²

Step-by-step explanation:

Area of the trapezoid shown = ½×(AD + BC)×BE

Use distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to find AD, BC, and BE.

✍️Distance between A(-9, 4) and D(9, 1):

$AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(9 -(-9))^2 + (1 - 4)^2}$

$= \sqrt{(18)^2 + (-3)^2}$

$= \sqrt{(324 + 9} = \sqrt{333}$

$AD = 18.2 units$ (nearest tenth)

✍️Distance between B(-4, -3) and C(2, -4):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(2 -(-4))^2 + (-4 -(-3))^2}$

$= \sqrt{(6)^2 + (-1)^2}$

$= \sqrt{(36 + 1} = \sqrt{37}$

$BC = 6.1 units$ (nearest tenth)

✍️Distance between B(-4, -3) and E(-3, 3):

$BE = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-3 -(-4))^2 + (3 -(-3))^2}$

$= \sqrt{(1)^2 + (6)^2}$

$= \sqrt{(1 + 36} = \sqrt{37}$

$BE = 6.1 units$ (nearest tenth)

✔️Area of the trapezoid:

Area = ½×(AD + BC)×BE

Plug in the values into the formula

Area = ½×(18.2 + 6.1)×6.1

Area = ½×(24.3)×6.1

Area = 74.115 ≈ 74.1 units² (nearest tenth)