Respuesta :
Answer:
7 square units.
Step-by-step explanation:
On the coordinate plane, triangle ABC has vertices at (1,1), (4,0) and (3,5).
So, the area of the triangle ABC is given by
[tex]\frac{1}{2} | 1(0 - 5) + 4(5 - 1) + 3(1 - 0) | = 7[/tex] square units. (Answer)
We know the formula that area of triangle having vertices at [tex](x_{1}, y_{1})[/tex], [tex](x_{2}, y_{2})[/tex], and [tex](x_{3}, y_{3})[/tex] is given by
Δ = [tex]\frac{1}{2} | x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})|[/tex]
Answer:
Area of Triangle ABC = [tex]\frac{13\sqrt{858} }{4}[/tex] unit² or 7.32 unit²
Step-by-step explanation:
Given coordinates of Triangle are
A = ( 1 , 1)
B = ( 4 , 0)
C = ( 3 , 5)
So , The measure of side AB is
AB = [tex]\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}[/tex]
or, AB = [tex]\sqrt{(4-1)^{2}+(0-1)^{2}}[/tex]
Or, AB = [tex]\sqrt{(3)^{2}+(-1)^{2}}[/tex]
∴ AB = [tex]\sqrt{10}[/tex]
Again ,
The measure of side BC is
BC = [tex]\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}[/tex]
or, BC = [tex]\sqrt{(3-4)^{2}+(5-0)^{2}}[/tex]
Or, BC = [tex]\sqrt{(-1)^{2}+(5)^{2}}[/tex]
∴ BC = [tex]\sqrt{26}[/tex]
Similarly ,
The measure of side CA is
CA = [tex]\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}[/tex]
or, CA = [tex]\sqrt{(3-1)^{2}+(5-1)^{2}}[/tex]
Or, CA = [tex]\sqrt{(2)^{2}+(4)^{2}}[/tex]
∴ CA = [tex]\sqrt{20}[/tex]
Now, let D be the mid points of side BC
So, Points ( D ) = [tex]\frac{(x_1+x_2)}{2}[/tex] , [tex]\frac{(y_1+y_2)}{2}[/tex]
I.e points d = [tex]\frac{(4+3)}{2}[/tex] , [tex]\frac{(0+5)}{2}[/tex]
Or, points D = [tex]\frac{(7)}{2}[/tex] , [tex]\frac{(5)}{2}[/tex]
Now Measure of line AD is
AD = [tex]\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}[/tex]
AD = [tex](\sqrt{\frac{7}{2}-1})^{2}[/tex] + [tex](\sqrt{\frac{5}{2}-1})^{2}[/tex]
Or, AD =( [tex]\sqrt{\frac{5}{2} }[/tex] )² + ( [tex]\sqrt{\frac{3}{2} }[/tex] )²
Or, AD = [tex]\sqrt{\frac{33}{4} }[/tex]
Now, Area of Triangle ABC = [tex]\frac{1}{2}[/tex] × length × base
or, Area of Triangle ABC = [tex]\frac{1}{2}[/tex] × AD × BC
or, Area of Triangle ABC = [tex]\frac{1}{2}[/tex] × [tex]\sqrt{\frac{33}{4} }[/tex] × [tex]\sqrt{26}[/tex]
Or, Area of Triangle ABC = [tex]\frac{13\sqrt{858} }{4}[/tex] unit² or , 7.32 unit² Answer