Question

If A(-3, 5), B(0, -4), C(6, -2), and D(3, 7), find the area of the rectangle ABCD.

• A. 20 square units
• B. 30 square units
• C. 40 square units
• D. 60 square units

Answer 1

The correct option is D 60 square units

Given: ABCD is a rectangle with A(-3, 5), B(0, -4), C(6, -2), and D(3, 7).

The distance between two points $(x_1,y_1)\text{and}(x_2,y_2)$ is $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}.$

AB = $\sqrt{(0-(-3){)}^2+(-4-5{)}^2}$

AB = $\sqrt{(3{)}^2+(-9{)}^2}$

AB = $\sqrt{9+81}$

AB = $\sqrt{90}$

AB = 3$\sqrt{10}$ units

BC = $\sqrt{(6-0{)}^2+(-2-(-4){)}^2}$

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

BC = $\sqrt{36+4}$

BC = $\sqrt{40}$

BC = 2$\sqrt{10}$ units

Area of the rectangle = Length $\times$ Breadth
Area of rectangle = 3$\sqrt{10}$ $\times$ 2$\sqrt{10}$
Area of rectangle = 60 square units