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3 years ago

What is the most precise name for quadrilateral ABCD with vertices A(-2, -2), B(-2, 0), C(1, 0), and D(1, -2)

Mathematics

2 answers:

True [87]3 years ago

8 0

D.Trapezoid is the answer

Ad libitum [116K]3 years ago

7 0

Answer: The correct option is (D) Trapezoid.

Step-by-step explanation: Given that the co-ordinates of the vertices of quadrilateral ABCD are A(-2, -2), B(-2, 0), C(1, 0), and D(1, -2).

We are to select the most precise name of the quadrilateral ABCD from the given options.

The lengths of the sides of the quadrilateral ABCD are calculated using distance formula as follows:

$AB=\sqrt{(-2+2)^2+(-2-0)^2}=\sqrt{0+4}=2~\textup{units},\\\\BC=\sqrt{(-2-1)^2+(0-0)^2}=\sqrt{9+0}=3~\textup{units},\\\\CD=\sqrt{(1-1)^2+(0+2)^2}=\sqrt{0+4}=2~\textup{units},\\\\DA=\sqrt{(1+2)^2+(-2+2)^2}=\sqrt{9+0}=3~\textup{units}.$

The slopes of the sides are calculated as follows:

$\textup{slope of AB, }m=\dfrac{0+2}{-2+2}=\textup{cannot be determined},\\\\\textup{slope of BC, }n=\dfrac{0-0}{1+2}=0,\\\\\textup{slope of CD, }o=\dfrac{-2-0}{1-1}=\textup{cannot be determined},\\\\\textup{slope of DA, }p=\dfrac{-2+2}{-2-1}=0.$

Since, AB = CD, AD = BC and slope of BC (n) = slope of DA (p), so we have

the opposite sides are congruent and one pair of opposite sides are parallel.

Therefore, the most precise name for the quadrilateral ABCD is a TRAPEZOID.

Thus, (D) is the correct option.

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3 years ago

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Ksju [112]

Answer:

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Step-by-step explanation:

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Now , let $\frac{1}{y}$ = u

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pashok25 [27]

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Read 2 more answers

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aev [14]

Check the picture below.

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after that, they both walked for 15 minutes more or 900 seconds, and for her that means 5400 feet, whilst for him going slower means 4500 feet, as you see in the picture.

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$\bf r^2=x^2+y^2\stackrel{chain~rule}{2r\cfrac{dr}{dt}}=0+\stackrel{chain~rule}{2y\cfrac{dy}{dt}}\implies \cfrac{dr}{dt}=\cfrac{2y}{2r}\cdot \cfrac{dy}{dt}
\\\\\\
\cfrac{dr}{dt}=\cfrac{y}{r}\cdot \cfrac{dy}{dt}$

so, what is "r" value 15 minutes later anyway?

$\bf r=\sqrt{(1500+4500+5400)^2+500^2}\implies r=\sqrt{11400^2+500^2}
\\\\\\
r=\sqrt{130210000}\implies r=100\sqrt{13021}$

then we just plug those folks in the derivative,

$\bf \cfrac{dr}{dt}=\cfrac{y}{r}\cdot \cfrac{dy}{dt}\qquad 
\begin{cases}
r=100\sqrt{13021}\\
\frac{dy}{dt}=11\\
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\end{cases}\implies \cfrac{dr}{dt}=\cfrac{11400}{100\sqrt{13021}}\cdot 11
\\\\\\
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7 0

4 years ago