How do you find the perimeter of triangle ABC with vertices A(-5,-2), B(-2,-2), and C(-5,2) ?
Answer
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Hint: The given question need to find the perimeter of the triangle, we should know that perimeter of any figure is the sum of the length of the boundaries of the given shape, here for the triangle we are provided with the vertices of the triangle with the help of which we can get side length to get the perimeter.
Formulae Used:
Length between two points =\[ \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
Complete step by step solution:
The question needs to find the perimeter of the given triangle whose vertices are known to us, let’s draw the given triangle. On drawing we get:
Now for finding the length of the sides by using vertices we have to use the standard formulae for line segment for given two points, on solving we get:
\[ \Rightarrow length\,d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
For the side AB we get: A(-5,-2) and B(-2,-2)
\[\Rightarrow length\,AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\
\Rightarrow AB = \sqrt {{{( - 2 - ( - 5))}^2} + {{( - 2 - ( - 2))}^2}} = \sqrt {{{(3)}^2} + {{(0)}^2}} = 3\,units \\ \]
For the side BC we get: B(-2,-2) and C(-5,2)
\[\Rightarrow length\,BC = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\
\Rightarrow AB = \sqrt {{{( - 5 - ( - 2))}^2} + {{( - 2 - (2))}^2}} = \sqrt {{{( - 3)}^2} + {{(0)}^2}} = 3\,units \\ \]
For the side CA we get: C(-5,2) and A(-5,-2)
\[\Rightarrow length\,CA = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\
\Rightarrow AB = \sqrt {{{( - 5 - ( - 5))}^2} + {{( - 2 - (2))}^2}} = \sqrt {{{(0)}^2} + {{(4)}^2}} = 4\,units \\ \]
Now perimeter of the given triangle is the sum of all sides, on solving we get:
\[ \therefore Perimeter\,of\,the\,given\,triangle = length(AB + BC + CA) = 3 + 3 + 4 = 10units\]
Hence we obtain the perimeter of the given figure which is ten units.
Note: The above question can also be solved by plotting the graph on the coordinate axis and then you can easily get the vertices on graph, with the help of these vertices, we can find the length of the sides and then we can easily find the perimeter of the triangle by adding the sides.
Formulae Used:
Length between two points =\[ \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
Complete step by step solution:
The question needs to find the perimeter of the given triangle whose vertices are known to us, let’s draw the given triangle. On drawing we get:
Now for finding the length of the sides by using vertices we have to use the standard formulae for line segment for given two points, on solving we get:
\[ \Rightarrow length\,d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
For the side AB we get: A(-5,-2) and B(-2,-2)
\[\Rightarrow length\,AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\
\Rightarrow AB = \sqrt {{{( - 2 - ( - 5))}^2} + {{( - 2 - ( - 2))}^2}} = \sqrt {{{(3)}^2} + {{(0)}^2}} = 3\,units \\ \]
For the side BC we get: B(-2,-2) and C(-5,2)
\[\Rightarrow length\,BC = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\
\Rightarrow AB = \sqrt {{{( - 5 - ( - 2))}^2} + {{( - 2 - (2))}^2}} = \sqrt {{{( - 3)}^2} + {{(0)}^2}} = 3\,units \\ \]
For the side CA we get: C(-5,2) and A(-5,-2)
\[\Rightarrow length\,CA = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\
\Rightarrow AB = \sqrt {{{( - 5 - ( - 5))}^2} + {{( - 2 - (2))}^2}} = \sqrt {{{(0)}^2} + {{(4)}^2}} = 4\,units \\ \]
Now perimeter of the given triangle is the sum of all sides, on solving we get:
\[ \therefore Perimeter\,of\,the\,given\,triangle = length(AB + BC + CA) = 3 + 3 + 4 = 10units\]
Hence we obtain the perimeter of the given figure which is ten units.
Note: The above question can also be solved by plotting the graph on the coordinate axis and then you can easily get the vertices on graph, with the help of these vertices, we can find the length of the sides and then we can easily find the perimeter of the triangle by adding the sides.
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