Question

What is the area of this triangle ?

Answer 1

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:

$Area\,\,of\,\,triangle =\sqrt{s(s-a)(s-b)s-c)} \\where\,\, s = \frac{a+b+c}{2}$

We need to find the lengths of side DE, EF and FD

Length of side DE = a = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side DE = a = $=\sqrt{(5-0)^2+(1-8)^2}\\=\sqrt{(5)^2+(-7)^2}\\=\sqrt{25+49}\\=\sqrt{74}\\=8.60$

Length of side EF = b = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side EF = b = $=\sqrt{(0-5)^2+(4-1)^2}\\=\sqrt{(-5)^2+(3)^2}\\=\sqrt{25+9}\\=\sqrt{34}\\=5.8$

Length of side FD = c = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side FD = c = $=\sqrt{(0-0)^2+(8-4)^2}\\=\sqrt{(0)^2+(4)^2}\\=\sqrt{0+16}\\=\sqrt{16}\\=4$

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=$=\sqrt{s(s-a)(s-b)s-c)}\\=\sqrt{9.2(9.2-8.6)(9.2-5.8)(9.2-4)}\\=\sqrt{9.2(0.6)(3.4)(5.2)}\\=\sqrt{97.5936}\\=9.88$

So, area of triangle is 9.88 units^2