Question

For #'s 12 - 13, find the area of each figure.

Answer 1

Using the distance(d) formula to obtain the length AB,BC,CA.

The distance formula is,

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

Given

`\begin{gathered} A\rightarrow(5,-6) \\ B\rightarrow(-5,-3) \\ C\rightarrow(5,6) \end{gathered}`

Therefore,

`\begin{gathered} AB=√((-5-5)^2+(-3--6)^2)=√((-10)^2+(-3+6)^2)=√(100+3^2)=√(109) \\ BC=√((5--5)^2+(6--3)^2)=√(10^2+9^2)=√(100+81)=√(181) \\ CA=√((5-5)^2+(6--6)^2)=√(0^2+12^2)=√(144)=12 \end{gathered}`

Therefore,

`\begin{gathered} AB=√(109)=10.44030\approx10.4 \\ BC=√(181)=13.45362\approx13.5 \\ CA=12 \end{gathered}`

Using Heron's formula to solve for the area

`\begin{gathered} Area=√(s(s-a)(s-b)(s-c)) \\ s=(a+b+c)/(2) \end{gathered}`

where,

`\begin{gathered} a=10.4 \\ b=13.5 \\ c=12 \\ \\ s=(10.4+13.5+12)/(2)=17.95 \end{gathered}`

Therefore, the area is