Question

GeometryFind the area of the of the following shape. Show all work.

Answer 1

Solution:

The area of the shape is the sum of the areas of the triangles ABE and BCD as shown below:

$\text{Area of shape = area of triangle ABE + area of triangle BCD}$

Area of triangle ABE:

$\begin{gathered} \text{Area}_{\text{ABE}}=\frac{1}{2}\times\text{base}\times\text{height} \\ =\frac{1}{2}\times AE\times BE \end{gathered}$

To evalutae the area of triangle ABE, we need the dimensions of the triangle ABE.

AE: To evaluate AE, we determine the distance between the points A and E.

Where the respective coordinates of A and E are (-4,0) and (0,0).

The distance between any two points is expressed as

$\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \text{where} \\ (x_1,y_1)\text{ and }(x_2,y_2)\text{ are the endpoints of the line} \end{gathered}$

Thus, the distance AE is evaluated as

$\begin{gathered} AE=\sqrt[]{(0_{}-(-4)_{})^2+(0-0)^2} \\ \Rightarrow\sqrt[]{16} \\ \therefore AE=4\text{ units} \end{gathered}$

similarly, BE has endpoints B and E whose respective coordinates are (0,4) and (0,0).

Thus,

$\begin{gathered} BE=\sqrt[]{(0_{}-0_{})^2+(0-4)^2} \\ \Rightarrow\sqrt[]{16} \\ \therefore BE=4\text{ units} \end{gathered}$

Hence, the area of the triangle ABE becomes

$\begin{gathered} =\frac{1}{2}\times AE\times BE \\ \Rightarrow\frac{1}{2}\times4\text{ units}\times4\text{ units} \\ \therefore\text{Area}_{\text{ABE}}=\text{ 8 square units} \end{gathered}$

Area of triangle BCD:

$\begin{gathered} \text{Area}_{\text{BCD}}=\frac{1}{2}\times\text{base}\times\text{height} \\ \Rightarrow\frac{1}{2}\times DC\times BD \end{gathered}$

DC has endpoints at D and C whose respective coordinates are (0,2) and (2,2).

Thus,

$\begin{gathered} DC=\sqrt[]{(2-0)^2+(2-2)^2} \\ =\sqrt[]{4} \\ \Rightarrow DC\text{ = 2 units} \end{gathered}$

BD has endpoints at B and D whose respective coordinates are (0,4) and (0,2).

Thus,

$\begin{gathered} BD=\sqrt[]{(0-0)^2+(2-4)^2} \\ =\sqrt[]{4} \\ \Rightarrow BD=\text{ 2 units} \end{gathered}$

Hence, area of the triangle BCD becomes

$\begin{gathered} \text{Area}_{\text{BCD}}=\frac{1}{2}\times DC\times BD \\ =\frac{1}{2}\times2\text{ units}\times2\text{ units} \\ \Rightarrow\text{Area}_{\text{BCD}}=\text{ 2 square units} \end{gathered}$

Recall that:

$\text{Area of shape = area of triangle ABE + area of triangle BCD}$

Thus,

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