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anonymous

  • 5 years ago

A square with sides of length s is inscribed in an equilateral triangle with sides of length t. Find the exact ratio of the area of the equilateral triangle to the area of the square.

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  1. watchmath
    • 5 years ago
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    \(\frac{6}{2+\sqrt{3}}\)

  2. anonymous
    • 5 years ago
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    how did you get that?

  3. anonymous
    • 5 years ago
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    cow, remember me? i need helped, wit hthe problem you helped me yesterday

  4. dumbcow
    • 5 years ago
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    yeah

  5. anonymous
    • 5 years ago
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    hey dumbcow lol can u help me lout with my question

  6. anonymous
    • 5 years ago
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    http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4dd9e9930ada8b0b66f142c7

  7. dumbcow
    • 5 years ago
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    Area of square is s^2 Area of equilateral triangle is \[\frac{\sqrt{3}}{4} t ^{2}\] to get ratio we need t in terms of s if we use the right triangle to left of square with side opposite of 60 degree angle the edge of the square of length s and adjacent side length (t-s)/2 tan 60 = opp/adj = 2s/(t-s) = sqrt(3) solve for t \[t = \frac{2+\sqrt{3}}{\sqrt{3}} s\] substitute this into Area equation \[A = \frac{\sqrt{3}}{4}*\frac{(7+4\sqrt{3})}{3} s ^{2}\] \[A = \frac{12 + 7\sqrt{3}}{12} s ^{2}\] divide by area of square s^2 to get ratio \[=\frac{12+7\sqrt{3}}{12}\]

  8. watchmath
    • 5 years ago
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    dumbcow right, I forgot to multiplu by two here :tan 60 = 2s/(t-s) (I did s/(t-s) )

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