Answer:
Area of triangle is 9.88 units^2
Step-by-step explanation:
We need to find the area of triangle
Given E(5,1), F(0,4), D(0,8)
We will use formula:
[tex]Area\,\,of\,\,triangle =\sqrt{s(s-a)(s-b)s-c)} \\where\,\, s = \frac{a+b+c}{2}[/tex]
We need to find the lengths of side DE, EF and FD
Length of side DE = a = [tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
Length of side DE = a = [tex]=\sqrt{(5-0)^2+(1-8)^2}\\=\sqrt{(5)^2+(-7)^2}\\=\sqrt{25+49}\\=\sqrt{74}\\=8.60[/tex]
Length of side EF = b = [tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
Length of side EF = b = [tex]=\sqrt{(0-5)^2+(4-1)^2}\\=\sqrt{(-5)^2+(3)^2}\\=\sqrt{25+9}\\=\sqrt{34}\\=5.8[/tex]
Length of side FD = c = [tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
Length of side FD = c = [tex]=\sqrt{(0-0)^2+(8-4)^2}\\=\sqrt{(0)^2+(4)^2}\\=\sqrt{0+16}\\=\sqrt{16}\\=4[/tex]
so, a= 8.60, b= 5.8 and c = 4
s = a+b+c/2
s= 8.6+5.8+4/2
s= 9.2
Area of triangle=[tex]=\sqrt{s(s-a)(s-b)s-c)}\\=\sqrt{9.2(9.2-8.6)(9.2-5.8)(9.2-4)}\\=\sqrt{9.2(0.6)(3.4)(5.2)}\\=\sqrt{97.5936}\\=9.88[/tex]
So, area of triangle is 9.88 units^2
