Math, asked by shelkeakash2005, 5 months ago

area of triangle whose vertices are A(0,0) , B(3, 1) C(2, 4) is

Answers

Answered by hukam0685
0

Area of ∆ ABC is 5 sq-units.

Given:

  • A ∆ABC, with vertices A(0,0) , B(3, 1), C(2, 4).

To find:

  • Find area of ∆ ABC.

Solution:

Formula to be used:

If A(x_1,y_1) , B(x_2, y_2),and C(x_3,y_3) are the vertices of triangle then it's area is given by

\bf Ar.( \triangle \: ABC) =  \frac{1}{2}  |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|  \\

Step 1:

Write the coordinates of given vertices.

A(x1,y1)=A(0,0)

B(x2,y2)=B(3,1)

and C(x3,y3)= C(2,4)

Step 2:

Put the values in the formula.

Ar.( \triangle \: ABC) =  \frac{1}{2}  |0(1 - 4) + 3(4 - 0) +2(0 - 1)|  \\

or

Ar.( \triangle \: ABC) =  \frac{1}{2}  |0+ 12  - 2|  \\

or

Ar.( \triangle \: ABC) =  \frac{1}{2}  |10|  \\

or

Ar.(∆ ABC)= 5 sq-units.

Thus,

Area of ∆ ABC is 5 sq-units.

Learn more:

1) Find the area of the triangle whose vertices are (1,0),(6,0)and(4,3)

https://brainly.in/question/16827510

2) Find the area of triangle abc whose vertices are a(1,2) b(-2,3) c(-3,-4)

https://brainly.in/question/8488575

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