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