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lisabon 2012 [21]
3 years ago
15

Figure ABCD has vertices A(−3, 2), B(2, 2), C(2, −4), and D(−3, −2). What is the area of figure ABCD?

Mathematics
2 answers:
Sidana [21]3 years ago
8 0

Answer:

The area of the figure ABCD is 25\ units^{2}

Step-by-step explanation:

we have

A(-3, 2), B(2, 2), C(2, -4), D(-3, -2)

using a graphing tool

Plot the figure

see the attached figure

The figure is a trapezoid

The area of the trapezoid is equal to

A=\frac{1}{2}(AD+BC)AB

Find the distance between the points in the graph

we have

AD=4\ units

BC=6\ units

AB=5\ units

substitute

A=\frac{1}{2}(4+6)5=25\ units^{2}

FromTheMoon [43]3 years ago
5 0
25 tried so pls give a thnx if right if wrong just give me a heads up! : )
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the scale is x4

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Wittaler [7]

Answer:

See below

Step-by-step explanation:

Use the formulae directly

For a cone, with base radius = r and height = h, here are the related formula

\textrm{Slant height  l} = \sqrt{r^2 + h^2}  (1)

\textrm{Lateral surface area} = \pi r l

\textrm {Base area} = \pi r^2

\textrm { Total Surface Area SA} = \textrm{Base Area + Lateral Area} = \pi r^2 + \pi rl = \pi r(r + l) (2)

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Therefore directly plugging in the numbers in the above equations:

Note:

l = slant height in cm
SA = total surface area in sqcm
V = Volume in cubic cm

Figure(a)

r = 4, h = 8

\textrm{l} = \sqrt{4^2 + 8^2} =\sqrt{80} = 8.944 \\\textrm{SA}  = 4\pi(4 + 8.944) =  4\pi(12.944) = 162.66\\\textrm{V} = (1/3)\pi(4^2)(8) = 134.04

Figure(b)

r = 7, h =15

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\textrm{SA}  = 7\pi(7 + 16.55) =  517.89

Figure (c)

r = 5, l = 8

h = \sqrt{l^2 - r^2} = \sqrt{8^2 - 5^2} = \sqrt{39} = 6.245\\SA = \textrm{SA}  = 5\pi(5 + 8) =  204.2\\V = (1/3)\pi(5^2)(6.245) = 163.5


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2 years ago
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