1. answer the three questions below about the quadrilateral:

a. How could we find the perimeter below about the quadrilateral?

b. How might we calculate the area of this figure?

c. It looks like it might be a rectangle. What can we do to help us decide if it's a rectangle or a parallelogram?​

1 answer the three questions below about the quadrilaterala How could we find the perimeter below about the quadrilateralb How might we calculate the area of th class=

Respuesta :

Before we begin, note that we need to use the distanse formula for everything to reach our conclusion. With that being said, here we go:

[tex]\displaystyle \sqrt{[-x_1 + x_2]^2 + [-y_1 + y_2]^2} = d[/tex]

Using the ordered pairs [tex]\displaystyle [3, -3][/tex] and [tex]\displaystyle [4, -2][/tex] for instanse:

[tex]\displaystyle \sqrt{[-3 + 4]^2 + [3 - 2]^2} = \sqrt{1^2 + 1^2} = \sqrt{2}[/tex]

Now, sinse the distanse between [tex]\displaystyle [3, -3][/tex] and [tex]\displaystyle [4, -2][/tex] is [tex]\displaystyle \sqrt{2}[/tex] units, then the distanse between [tex]\displaystyle [1, 1][/tex] and [tex]\displaystyle [0, 0][/tex] ALSO has to be [tex]\displaystyle \sqrt{2}[/tex] units. By definition, lettre c has already been answered for you because sinse it is a rectangle, if the two short sides are congruent, then the two long sides ALSO have to be congruent, but just in case you want to be sure [you do not trust your instincts], just simply re-use the distanse formula.

Using the ordered pairs [tex]\displaystyle [1, 1][/tex] and [tex]\displaystyle [4, -2][/tex] for instanse:

[tex]\displaystyle \sqrt{[-1 + 4]^2 + [-1 - 2]^2} = \sqrt{3^2 + [-3]^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}[/tex]

So there you have it. The length of both of the long sides is [tex]\displaystyle 3\sqrt{2}[/tex] units.

Now that we cleared all of that up, we can now find the perimetre and area of this rectangle:

[tex]\displaystyle 2w + 2l = P[/tex]

[tex]\displaystyle 2[\sqrt{2}] + 2[3\sqrt{2}] = 2\sqrt{2} + 6\sqrt{2} = 8\sqrt{2}[/tex]

The perimetre of this rectangle is [tex]\displaystyle 8\sqrt{2}[/tex] units.

[tex]\displaystyle wl = A[/tex]

[tex]\displaystyle [\sqrt{2}][3\sqrt{2}] = [3][2] = 6[/tex]

The area of this rectangle is [tex]\displaystyle 6[/tex] squared units.

You have now found what you were looking for.

** All rectangles are parallelograms because they both have two pairs of parallel and congruent sides, while vice versa is falce because a parallelogram does not have four congruent right angles, so it is safe to say that this is both a rectangle AND parallelogram, sinse the picture displayed is a rectangle.

I am joyous to assist you at any time. ☺️

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Answer:

  a) compute and add up the side lengths

  b) multiply length by width

  c) compare the slopes of adjacent sides

Step-by-step explanation:

a) The perimeter of any figure is the sum of the lengths of its sides. It can be computed for a rectangle or parallelogram by adding the lengths of adjacent sides and multiplying the sum by 2.

In this graph, each grid square is 1/2 unit. The width of the rectangle is the diagonal of a square that is 1 unit on a side, so is √2 units. The length is 3 times that, so the perimeter is ...

  P = 2(L+W)

  P = 2(3√2 +√2) = 8√2 . . . . perimeter of the rectangle in units

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b) The area is the product of the length and width.

  A = LW

  A = (3√2)(√2) = 6 . . . . area of the rectangle in square units

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c) Each side lies on the diagonal of a unit square. The diagonals of a square are perpendicular, so the sides of this parallelogram are perpendicular. That means the figure is a rectangle.

The basic idea is to look at adjacent sides to see if they are perpendicular. Here, we have determined that using the properties of a square. One could compute "rise"/"run" for each side to see of the values are opposite reciprocals.

  origin to B: rise/run = 1/1 = 1

  B to C: rise/run = -3/3 = -1

Each of these values is the opposite reciprocal of the other: 1 = -1/-1.

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Additional comment

This question seems to be about methods, not about numbers. We have given numbers anyway.