Question

Can you please help about this ..Find the perimeter of triangle with vertices A=4-1,B=3,1,C=4,-2 Draw the triangle .P=A+B+C ..can you plz help me...

Answer 1

Question #32075

Can you please help about this ..Find the perimeter of triangle with

vertices $A = 4 - 1, B = 3, 1, C = 4, -2$ Draw the triangle $P = A + B + C$ ..can you plz help

me...

Solution:

The perimeter of triangle is

$$P = | A B | + | A C | + | B C |$$

According to the Pythagorean theorem

$$| A B | = \sqrt {\left(X _ {A} - X _ {B}\right) ^ {2} + \left(Y _ {A} - Y _ {B}\right) ^ {2}}$$ $$| A C | = \sqrt {\left(X _ {A} - X _ {C}\right) ^ {2} + \left(Y _ {A} - Y _ {C}\right) ^ {2}}$$ $$| B C | = \sqrt {\left(X _ {B} - X _ {C}\right) ^ {2} + \left(Y _ {B} - Y _ {C}\right) ^ {2}}$$ $$P = \sqrt {\left(X _ {A} - X _ {B}\right) ^ {2} + \left(Y _ {A} - Y _ {B}\right) ^ {2}} + \sqrt {\left(X _ {A} - X _ {C}\right) ^ {2} + \left(Y _ {A} - Y _ {C}\right) ^ {2}} + \sqrt {\left(X _ {B} - X _ {C}\right) ^ {2} + \left(Y _ {B} - Y _ {C}\right) ^ {2}}$$ $$\begin{array}{l} P = \sqrt {(4 - 3) ^ {2} + (- 1 - 1) ^ {2}} + \sqrt {(4 - 4) ^ {2} + (- 1 - (- 2)) ^ {2}} + \sqrt {(3 - 4) ^ {2} + (1 - (- 2)) ^ {2}} = \\ \sqrt {5} + 1 + \sqrt {1 0} = 6. 4 \\ \end{array}$$

Answer

The perimeter of triangle is 6.4 units.