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lyzkhalifaa

  • 3 years ago

if an equillateral triangke has the perimeter of 66, what is the altitude? i just wanna know how to solve it, like the formula.

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  1. Hero
    • 3 years ago
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    Well, if given the perimeter of an equilateral triangle, you can find the lengths of the sides by using the formula \[S_{\text{Triangle}} = \frac{P}{3}\] |dw:1357829776184:dw| Afterwards, you'll have to draw the altitude which divides the base of the triangle in half. |dw:1357829954412:dw| Furthermore, you'd have to use Pythagorean Theorem to find the the length of the altitude

  2. lyzkhalifaa
    • 3 years ago
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    |dw:1357830348662:dw| so, how would i make this into a pythagoream theorom?

  3. Hero
    • 3 years ago
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    Split the bigger triangle into two smaller ones. Then just solve for one of the missing sides. The missing side is the altitude of course.

  4. Hero
    • 3 years ago
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    |dw:1357830687072:dw| a^2 + b^2 = c^2 a^2 + 11^2 = 22^2

  5. lyzkhalifaa
    • 3 years ago
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    i got a=24.6 .. is that right?

  6. Hero
    • 3 years ago
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    c = 22 is the hypotenuse, right? So either leg has to be less than that. In other words a < c

  7. Hero
    • 3 years ago
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    So there's no way that a = 24. 6

  8. lyzkhalifaa
    • 3 years ago
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    11x11=121 and 22x22=484, then i added them together and 605 and the squareroot of 605 was 24.6 :c

  9. Hero
    • 3 years ago
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    a^2 + 11^2 = 22^2 a^2 = 22^2 - 11^2 a^2 = 484 - 121

  10. lyzkhalifaa
    • 3 years ago
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    i got 11\[\sqrt{3}\]

  11. lyzkhalifaa
    • 3 years ago
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    with the 11 infront

  12. lyzkhalifaa
    • 3 years ago
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    thank you Hero (:

  13. Hero
    • 3 years ago
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    a^2 + 11^2 = 22^2 a^2 = 22^2 - 11^2 a^2 = 484 - 121 a^2 =121(4 - 1) a^2 = 121(3) a = sqrt(121*3) a = sqrt(11^2)sqrt(3) a = 11sqrt(3)

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