# The coordinates of the vertices of a rectangle are (-3,5), (0,-4), (3,7), and (6,-2). How do you find the area of this figure?

Dec 11, 2017

$60$ square units

#### Explanation:

In this case it is a good idea to draw a quick sketch so that you can see what you are dealing with.

We can use Pythagoras to solve this as appropriately projecting lines from any point will form a triangle. The vertical and horizontal lengths of which can be read of from the axis.

The distance between two points is $\sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

Let the distance between two points be ${d}_{i}$

Distance between points A and B

${d}_{1} = \sqrt{{\left[3 - \left(- 3\right)\right]}^{2} + {\left[7 - 5\right]}^{2}}$

${d}_{1} = \sqrt{36 + 4} = \sqrt{{2}^{2} \times 10} = 2 \sqrt{10}$

Distance between points B and C

${d}_{2} = \sqrt{{\left[6 - 3\right]}^{2} + {\left[7 - \left(- 2\right)\right]}^{2}}$

${d}_{2} = \sqrt{9 + 81} = \sqrt{{3}^{2} \times 10} = 3 \sqrt{10}$
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Area $= {d}_{1} \times {d}_{2} = 2 \sqrt{10} \times 3 \sqrt{10} = 60$

Dec 11, 2017

The area of the rectangle is $108.17$ sq.unit.

#### Explanation:

The four vertices are $A \left(- 3 , 5\right) B \left(0 , - 4\right) , C \left(3 , 7\right) , D \left(6 , - 2\right)$

Distance between two points D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2

Side $A B = \sqrt{{\left(- 3 - 0\right)}^{2} + {\left(5 + 4\right)}^{2}} = \sqrt{90} \approx 9.49$unit

Side $B C = \sqrt{{\left(0 - 3\right)}^{2} + {\left(- 4 - 7\right)}^{2}} = \sqrt{130} \approx 11.40$unit

Side $C D = \sqrt{{\left(3 - 6\right)}^{2} + {\left(7 + 2\right)}^{2}} = \sqrt{90} = 9.49$unit

Side $A D = \sqrt{{\left(6 + 3\right)}^{2} + {\left(- 2 - 5\right)}^{2}} = \sqrt{130} = 11.40$unit

The area of the rectangle is${A}_{r} = A B \cdot B C$

$\therefore {A}_{r} = = \sqrt{90} \cdot \sqrt{130} = \sqrt{117} \cdot 10 \approx 108.17 \left(2 \mathrm{dp}\right)$sq.unit

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