Respuesta :
Answer:
1) The area of the composite figure whose vertices have the coordinates (−8,3) , (−4,4) , (−1,1) , (−4,−2) , (−8,−2) is (20 + 4 + 3) = 27 unit²
Step-by-step explanation:
1) Given : coordinate of a composite figure (−8,3) , (−4,4) , (−1,1) , (−4,−2) , (−8,−2)
We have to find the area of the composite figure whose vertices have the given coordinates.
We first plot the given coordinate and obtained the figure as shown in attachment below.
Join points (−8,3) and (−4,−2)
and (−4,4) and (−4,−2)
Thus, we have obtained three triangles.
Area of triangle having coordinates [tex](a_x,a_y),(b_x,b_y),(c_x,c_y)[/tex] is given by ,
[tex]\left | \frac{a_x(b_y-c_y)+b_x(c_y-a_y)+c_x(a_y-b_y}{2} \right |[/tex]
Thus, for triangle with coordinate (−8,3) ,(-8, -2)and (−4,−2)
Area is given by
[tex]\left | \frac{-8(-2+2)+(-8)(-2-3)+(-4)(3-(-2)}{2} \right |[/tex]
thus area for triangle with coordinate (−8,3) ,(-8, -2)and (−4,−2) is 20 unit²
Similarly, for triangle with coordinate (−8,3) ,(-4,-4)and (−4,−2)
Area is given by
[tex]\left | \frac{-8(-4+2)+(-4)(-2-3)+(-4)(3-(-4)}{2} \right |[/tex]
thus area for triangle with coordinate (−8,3) ,(-8, -2)and (−4,−2) is 4 unit²
Similarly, for triangle with coordinate (−1,1) ,(-4,-4)and (−4,−2)
Area is given by
[tex]\left | \frac{-1(-4+2)+(-4)(-2-1)+(-4)(1-(-4)}{2} \right |[/tex]
thus area for triangle with coordinate (−8,3) ,(-8, -2)and (−4,−2) is 3 unit²
Thus, the area of the composite figure whose vertices have the coordinates (−8,3) , (−4,4) , (−1,1) , (−4,−2) , (−8,−2) is (20 + 4 + 3) = 27 unit²
2)
Given : The coordinates of the vertices of a polygon are (−2, −1) , (2, 3) , (4, 3) , (4, −3) , and (−1, −4) .
We have to find the perimeter of the polygon.
We first plot these point to obtain the polygon.
Using distance formula [tex]\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/tex]
thus, we find the distance between each two points using above formula,
we have
a) [tex]\mathrm{Distance\:between\:}\left(-2,\:-1\right)\mathrm{\:and\:}\left(2,\:3\right)[/tex]
[tex]=\sqrt{\left(2-\left(-2\right)\right)^2+\left(3-\left(-1\right)\right)^2}=4\sqrt{2}=5.7[/tex]
b)
[tex]\mathrm{The\:distance\:between\:}\left(2,\:3\right)\mathrm{\:and\:}\left(4,\:3\right)\mathrm{\:is\:}[/tex]
[tex]=\sqrt{\left(4-2\right)^2+\left(3-3\right)^2}=2[/tex]
c)
[tex]\mathrm{The\:distance\:between\:}\left(4,\:3\right)\mathrm{\:and\:}\left(4,\:-3\right)\mathrm{\:is\:}[/tex]
[tex]=\sqrt{\left(4-4\right)^2+\left(-3-3\right)^2}=6[/tex]
d)
[tex]\mathrm{The\:distance\:between\:}\left(4,\:-3\right)\mathrm{\:and\:}\left(-1,\:-4\right)\mathrm{\:is\:}[/tex]
[tex]=\sqrt{\left(-1-4\right)^2+\left(-4-\left(-3\right)\right)^2}=\sqrt{26}=5.1[/tex]
e)
[tex]\mathrm{The\:distance\:between\:}\left(-1,\:-4\right)\mathrm{\:and\:}\left(-2,\:-1\right)\mathrm{\:is\:}[/tex]
[tex]=\sqrt{\left(-2-\left(-1\right)\right)^2+\left(-1-\left(-4\right)\right)^2}=\sqrt{10}=3.2[/tex]
thus, perimeter is sum of all distance is 22 units.
Thus, the perimeter of the coordinates of the vertices of a polygon are (−2, −1) , (2, 3) , (4, 3) , (4, −3) , and (−1, −4) is 22 units.
