Respuesta :
we know that
A method for calculating the area of a triangle when you know the lengths of all three sides is the Heron's Formula.
The Heron's Formula states that
[tex]A=\sqrt{p(p-a)(p-b)(p-c)}[/tex]
where
a,b.c are the length sides of the triangle
p is half the perimeter of the triangle
Let
[tex]A(-4,1)\\B(-7,5)\\C(0,1)[/tex]
Step 1
Find the distance AB
we know that
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
we have
[tex]A(-4,1)\\B(-7,5)[/tex]
substitute the values in the formula
[tex]d=\sqrt{(5-1)^{2}+(-7+4)^{2}}[/tex]
[tex]d=\sqrt{(4)^{2}+(-3)^{2}}[/tex]
[tex]d=\sqrt{25}[/tex]
[tex]dAB=5\ units[/tex]
Step 2
Find the distance BC
we know that
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
we have
[tex]B(-7,5)\\C(0,1)[/tex]
substitute the values in the formula
[tex]d=\sqrt{(1-5)^{2}+(0+7)^{2}}[/tex]
[tex]d=\sqrt{(-4)^{2}+(7)^{2}}[/tex]
[tex]d=\sqrt{65}[/tex]
[tex]dBC=8.06\ units[/tex]
Step 3
Find the distance AC
we know that
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
we have
[tex]A(-4,1)\\C(0,1)[/tex]
substitute the values in the formula
[tex]d=\sqrt{(1-1)^{2}+(0+4)^{2}}[/tex]
[tex]d=\sqrt{(0)^{2}+(4)^{2}}[/tex]
[tex]d=\sqrt{16}[/tex]
[tex]dAC=4\ units[/tex]
Step 4
Find half the perimeter
[tex]p=\frac{1}{2}(AB+BC+AC)[/tex]
substitute the values
[tex]p=\frac{1}{2}(5+8.06+4)[/tex]
[tex]p=8.53\ units[/tex]
Step 5
Find the area
Applying the Heron's Formula
[tex]A=\sqrt{8.53(8.53-5)(8.53-8.06)(8.53-4)}[/tex]
[tex]A=\sqrt{8.53(3.53)(0.47)(4.53)}[/tex]
[tex]A=\sqrt{64.11}[/tex]
[tex]A=8\ units^{2}[/tex]
therefore
the answer is
the area of the triangle is [tex]8\ units^{2}[/tex]
The area of a shape is the amount of space it occupies.
The area of the triangle is 8 unit square.
The vertices are given as:
[tex]\mathbf{A = (-4,1)}[/tex]
[tex]\mathbf{B = (-7,5)}[/tex]
[tex]\mathbf{C = (0,1)}[/tex]
The area of the triangle is:
[tex]\mathbf{A= \frac{1}{2} \times |A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)|}[/tex]
So, we have:
[tex]\mathbf{A= \frac{1}{2} \times |-4(5 - 1) -7(1 - 1) + 0(1 - 5)|}[/tex]
[tex]\mathbf{A= \frac{1}{2} \times |-4(4) -7(0) + 0|}[/tex]
[tex]\mathbf{A= \frac{1}{2} \times |-4(4) -0 + 0|}[/tex]
[tex]\mathbf{A= \frac{1}{2} \times |-16|}[/tex]
Remove absolute brackets
[tex]\mathbf{A= \frac{1}{2} \times 16}[/tex]
[tex]\mathbf{A= 8}[/tex]
Hence, the area of the triangle is 8 unit square.
Read more about areas at:
https://brainly.com/question/12074976