Mathematics, 30.11.2019 07:31 juandguardado7045

What is the area of a rectangle with vertices at (−6, 3) , (−3, 6) , (1, 2) , and (−2, −1) ?

enter your answer in the box. do not round any side lengths.

what is the area of a triangle with vertices at (0, −2) , (8, −2) , and (9, 1) ?

enter your answer in the box.

what is the perimeter of a polygon with vertices at (−2, 1) , (−2, 4) , (2, 7) , (6, 4) , and (6, 1) ?

enter your answer in the box. do not round any side lengths.

what is the perimeter of the rectangle shown on the coordinate plane, to the nearest tenth of a unit?

15.3 units

20.4 units

30.6 units

52.0 units

What is the area of a rectangle with vertices at (−6, 3) , (−3, 6) , (1, 2) , and (−2, −1) ?

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Answer from: pc103919

(1) the area of a rectangle with vertices at (−6, 3) , (−3, 6) , (1, 2) , and (−2, −1)

To find area of rectangle we need to find the length and width

Length = distance between (−6, 3) and (−2, −1)

Distance = $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

= $\sqrt{(-2-(-6))^2 + (-1-3)^2}$

= $\sqrt{(4)^2 + (-4)^2}$ = $\sqrt{(16) + 16}$ = $\sqrt{32}$

width = Distance between (−6, 3) , (−3, 6)

Distance = $\sqrt{(-3-(-6))^2 + (6-3)^2}$

= $\sqrt{3^2+3^2}= \sqrt{18}$

Area = Length * width = $\sqrt{32} *\sqrt{18} = \sqrt{576}= 24$

(2) the area of a triangle with vertices at (0, −2) , (8, −2) , and (9, 1)

Area of triangle = $\frac{1}{2} * base * height$

base is the distance between (0,-2) and (8,-2)

Distance = $\sqrt{(8-0)^2 + (-2-(-2))^2}$ = 8

To find out height we take two vertices (8,-2) and (9,1)

Height is the change in y values = 1- (-2) = 3

base = 8 and height = 3

So area of triangle = $\frac{1}{2} * 8 * 3 = 12$

(3) the perimeter of a polygon with vertices at (−2, 1) , (−2, 4) , (2, 7) , (6, 4) , and (6, 1)

To find perimeter we add the length of all the sides

Distance between (−2, 1) and (−2, 4) = $\sqrt{(-2+2)^2 + (4-1)^2}$ = 3

Distance between(−2, 4) and (2, 7) = $\sqrt{(2+2)^2 + (7-4)^2}$ = 5

Distance between (2, 7) and (6, 4) = $\sqrt{(6 - 2)^2 + (4-7)^2}$ = 5

Distance between (6, 4) and (6, 1) = $\sqrt{(6 - 6)^2 + (1-4)^2}$ = 3

Distance between (6, 1) and (−2, 1) = $\sqrt{(-2-6)^2 + (1-1)^2}$ = 8

Perimeter = 3 + 5 + 5 + 3 + 8 = 24

(4) four coordinates are (-7,-1) (-6,4) (3,-3) and (4,2)

Length = Distance between (3,-3) and (4,2) = $\sqrt{(4-3)^2 + (2+3)^2}$ = $\sqrt{26}$

Width = Distance between (-6,4) and (4,2) = $\sqrt{(4+6)^2 + (2-4)^2}$ = $\sqrt{104}$

Perimeter = 2(lenght + width) = 2*( $\sqrt{26}$ + $\sqrt{104}$ )

= 30.6 units

Answer from: qvezne7683

The area of the rectangle is $24\;square\;units$ .

1. According to the question, the vertices of the rectagle are at (-6, 3) ;(-3, 6) ;(1, 2) , and (-2, -1) .

The distance between two vertices on the coordinate plane is

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$Length=D_1\\D_1=\sqrt{(6-3)^2+(-3+6)^2}\\D_1=\sqrt{9+9}\\D_1=3\sqrt 2 units$

Similarly,

$Width=D_2\\D_2=\sqrt{(-2+6)^2+(-1-3)^2}\\D_2=\sqrt{16+16}\\D_2=16\sqrt 2\;units$

The area of the rectangle is equals to product of length and width of the rectabgle.

So,

$Area=length\times width\\Area=3\sqrt{2}\times 4\sqrt{2}\\Area=24\;square\;units$

Hence, the area of the rectangle is $24\;square\;units$ .

Learn more about rectangles here:

link

Answer from: Quest

4038 is the answer lol just use a calculator

Answer from: Quest

why are you giving so many marks make your marks less and make a question list so that i can understand your question very nicely

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What is the area of a rectangle with vertices at (−6, 3) , (−3, 6) , (1, 2) , and (−2, −1) ?

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