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grigory [225]
3 years ago
10

What is the perimiter of ABC with vertices A (-2,9), B(7-3), and c(-2,-3) in the coordinate plane?

Mathematics
1 answer:
Serggg [28]3 years ago
6 0

Answer:

|AB| = √((7 - (-2))^2 + (-3 - 9)^2) = 15

|BC| = √((-2 - 7)^2 + (-3 - (-3))^2) = 9

|AC| = √((-2 - (-2))^2 + (-3 - 9)^2) = 12

Perimeter = 15 + 9 + 12 = 36

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Answer:

perimeter is  4 sqrt(29) + 4pi  cm

area is 40 + 8pi cm^2

Step-by-step explanation:

We have a semicircle and a triangle

First the semicircle with diameter 8

A = 1/2 pi r^2 for a semicircle

r = d/2 = 8/2 =4

A = 1/2 pi ( 4)^2

  =1/2 pi *16

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Now the triangle with base 8 and height 10

A = 1/2 bh

  =1/2 8*10

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Add the areas together

A = 40 + 8pi cm^2

Now the perimeter

We have 1/2 of the circumference

1/2 C =1/2 pi *d

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        = 4pi

Now we need to find the length of the hypotenuse of the right triangles

using the pythagorean theorem

a^2+b^2 = c^2

The base is 4 ( 1/2 of the diameter) and the height is 10

4^2 + 10 ^2 = c^2

16 + 100 = c^2

116 = c^2

sqrt(116) = c

2 sqrt(29) = c

Each hypotenuse is the same so we have

hypotenuse + hypotenuse + 1/2 circumference

2 sqrt(29) + 2 sqrt(29) + 4 pi

4 sqrt(29) + 4pi  cm

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