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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:

The area of a rectangle is 42 units ².

Step-by-step 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)

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

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

\(AB = \sqrt{49}\)

\(\sqrt{49} =7\)

AB = 7 units

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

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

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

\(BC = \sqrt{36}\)

\(\sqrt{36} =6\)

BC = 6 units

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

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

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

\(CD = \sqrt{49}\)

\(\sqrt{49} =7\)

CD =  7 units

As take  D (4, 9) to  A (4, 3)

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

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

\(DA = \sqrt{36}\)

\(\sqrt{36} =6\)

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.