anonymous
  • anonymous
Find the perimeter of the following polygon.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1362244727495:dw|
anonymous
  • anonymous
26 29 36 39
anonymous
  • anonymous
@sami-21

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anonymous
  • anonymous
@ZeHanz
anonymous
  • anonymous
help quick please
anonymous
  • anonymous
i thougth times but that isnt right
anonymous
  • anonymous
Look at just the left half of the diagram for now. You have a right triangle and you have the measures of 2 sides, so you can use the Pythagorean Theorem: x^2 + 12^2 = 13^2 You can get x^2 from that and then get "x" Your perimeter will be easy from there.
anonymous
  • anonymous
so i add 13+14+12+12
anonymous
  • anonymous
?
anonymous
  • anonymous
i mean 13+13
anonymous
  • anonymous
+12+12
anonymous
  • anonymous
Well, first, you have to solve: x^2 + 12^2 = 13^2
anonymous
  • anonymous
12*12?
anonymous
  • anonymous
12+12 = 24
anonymous
  • anonymous
Hint: subtract 12^2 from each side.
anonymous
  • anonymous
ugh
anonymous
  • anonymous
36 is the answer
anonymous
  • anonymous
am i correct
anonymous
  • anonymous
Don't worry, I'll help you: (x^2 + 12^2) - 12^2 = 13^2 - 12^2 x^2 + (12^2 - 12^2) = 13^2 - 12^2
anonymous
  • anonymous
Yes, 36 is the answer! You got it!
anonymous
  • anonymous
Good luck to you in all of your studies and thx for the recognition! @rebecca1233
anonymous
  • anonymous
your welcome :)

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