Question

What is the area of a parallelogram whose verticles are A(-12, 2), B (6, 2), C(-2, -3), and D (-20, -3)?

Answer 1

The area of the parallelogram will be 108 square units.

What is the area of the parallelogram?

The space occupied by any two-dimensional figure in a plane is called the area. The area of the outer surface of any body is called the surface area.

The vertices of the parallelogram are (x₁, y₁), (x₂, y₂), (x₃, y₃), and ( x₄, y₄).

Then the area of the parallelogram will be calculated as below:-

Area = 1/2 |[(x₁y₂ + x₂y₃ + x₃y₄+x₄y₁) - (y₁x₂ + y₂x₃ + y₃x₄+y₄x₁)]|

We have

(x₁, y₁) ⇒ (-12, 2)

(x₂, y₂) ⇒ (6, 2)

(x₃, y₃) ⇒ (-2, -3)

(x₄, y₄) ⇒ (-20 , -3 )

Then the area will be

Area = 1/2[{(-12) x (2) + (6) x (-3) + (-2) x (-3) + (-20 x 2 )} – {(2) x (6) + (2) x (-2) + (-3) x (-20) + ( -3 ) ( -12 )}]

Area = 108 / 2  

Area = 54 square units.

Therefore, the area of the parallelogram will be 108 square units.

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