Question

What is the area of a rectangle with vertices (-8, -2), (-3,-2), (-3,-6), and (-8. -6)?

Answer 1

Answer:

31 square units

Answer 2

Answer:

The area of the rectangle is 20 sq .units.

Given:

(-8, -2), (-3,-2), (-3,-6), and (-8, -6)

Solution:

The area of the rectangle ‘A’ is given by the formula:

Area = Length × Width

Now, we have to find the sides of the rectangle.

The sides of the rectangle include s1, s2, s3, and s4.

Let’s now assume the points as:

$(x1, y1) = (-8, -2)$

$(x2, y2) = (-3,-2)$

$(x3, y3) = (-3,-6)$

$(x4, y4) = (-8, -6)$

The side s1 is:

$s 1=\sqrt{(x 2-x 1)^{2}+(y 2-y 1)^{2}}$

On substituting the values,

$\Rightarrow s 1=\sqrt{(-3+8)^{2}+(-2+2)^{2}}$

$\Rightarrow s 1=\sqrt{(5)^{2}+(0)^{2}}$

$\Rightarrow s 1=\sqrt{25}$

$\therefore s 1=5 \text { units }$

The side s2 is:

$s 2=\sqrt{(x 3-x 2)^{2}+(y 3-y 2)^{2}}$

On substituting the values,

$\Rightarrow s 2=\sqrt{(-3+3)^{2}+(-6+2)^{2}}$

$\Rightarrow s 2=\sqrt{(0)^{2}+(4)^{2}}$

$\Rightarrow s 2=\sqrt{16}$

$\therefore s 2=4 \text { units }$

The side s3 is:

$s 3=\sqrt{(x 4-x 3)^{2}+(y 4-y 3)^{2}}$

On substituting the values,

$\Rightarrow s 3=\sqrt{(-8+3)^{2}+(-6+6)^{2}}$

$\Rightarrow s 3=\sqrt{(5)^{2}+(0)^{2}}$

$\Rightarrow s 3=\sqrt{25}$

$\therefore s 3=5 \text { units }$

The side s4 is:

$s 4=\sqrt{(x 1-x 4)^{2}+(y 1-y 4)^{2}}$

On substituting the values,

$\Rightarrow s 4=\sqrt{(-8+8)^{2}+(-2+6)^{2}}$

$\Rightarrow s 4=\sqrt{(0)^{2}+(4)^{2}}$

$\Rightarrow s 4=\sqrt{16}$

$\therefore s 4=4 \text { units }$

Now, the length of the given rectangle is 5 units and width of the given rectangle is 4 units.

The area of the rectangle is:

$\Rightarrow A r e a=4 \times 5$

$\therefore A r e a=20 \ s q . \text { units }$