Question

What is the area of a rectangle with vertices at (4, 3), (11, 3), (11, 9), and (4, 9)?

Answer 1

This is rectangle ABCD, where coordinates of the vertices are: A ( ( 4, 3 ); B ( 11, 3 ); C ( 11, 9 ) and D ( 4, 9 ). The area of the rectangle is: A = L x W ( L - length and W - width ). Length = AB = 11 - 4 = 7. Width = BC = 9 - 3 = 6. Finally: A = 7 * 6 = 42. Answer: The area of a rectangle is 42. Hope this helps. Let me know if you need additional help!

Answer 2

The area of a rectangle is 42 units ².

Explanation:

As the diagram is given below.

Distance Formula

`Distance\ formula = \sqrt{(x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2)}`

As the rectangle with vertices at A (4, 3), B(11, 3),C (11, 9), and D (4, 9).

Take A (4, 3) to B(11, 3)

`AB = \sqrt{(11-4)^(2)+(3-3)^(2)}`

`AB = \sqrt{(7)^(2)+(0)^(2)}`

`AB = √(49)`

`√(49) =7`

AB = 7 units

As take B(11, 3) to C (11, 9).

`BC = \sqrt{(11-11)^(2)+(9-3)^(2)}`

`BC = \sqrt{(0)^(2)+(6)^(2)}`

`BC = √(36)`

`√(36) =6`

BC = 6 units

As take C (11, 9) toD (4, 9).

`CD = \sqrt{(4-11)^(2)+(9- 9)^(2)}`

`CD = \sqrt{(7)^(2)+(0)^(2)}`

`CD = √(49)`

`√(49) =7`

CD = 7 units

As take D (4, 9) to A (4, 3)

`DA = \sqrt{(4-4)^(2)+(3-9)^(2)}`

`DA = \sqrt{(0)^(2)+(-6)^(2)}`

`DA = √(36)`

`√(36) =6`

DA = 6 units

Thus

AB = DC= 7 units

AD = BC = 6 units

Formula

Area of rectangle = Length × Breadth

Length = 7 units

Breadth = 6 units

Put value in the above

Area of rectangle = 7 × 6

= 42 units ²

Therefore the area of a rectangle is 42 units ².