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anonymous

  • one year ago

The figures below are similar. What are a) the ratio of the perimeters and b) the ratios of the areas of the larger figure to the smaller figure? The figures are not drawn to scale. http://gyazo.com/53b9d9f3f176ae4b195f5cd8c431505f My answer: I know that a part of it is 5/12 but I don't know what else. It's either going to be 25/4 or 7 over 4.

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  1. anonymous
    • one year ago
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    I just need to know.

  2. anonymous
    • one year ago
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    @Hero

  3. anonymous
    • one year ago
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    2.) Find the area of the shaded region. Leave your answer in terms of and in simplest radical form. http://assets.openstudy.com/updates/attachments/4fced66ee4b0c6963ad9ec74-kaitlinlikesmusic-1338955382935-shaded.png My answer: 120 pi -36sqrt3. Is that correct?

  4. anonymous
    • one year ago
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    @phi @amistre64

  5. anonymous
    • one year ago
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    @mathmate

  6. anonymous
    • one year ago
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    @radar

  7. anonymous
    • one year ago
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    @EclipsedStar

  8. phi
    • one year ago
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    for the similar figures with length 30 and 12 the ratio of perimeters would be 30/12 or 5/2

  9. mathmate
    • one year ago
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    @Lollygirl217 For the first question, remember that the area of two similar polygons is proportional to the squares of the ratio between them.

  10. phi
    • one year ago
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    the areas will be the in the square of the lengths 25/4 (an amazing property that holds for similar shapes)

  11. anonymous
    • one year ago
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    Oh mk.

  12. phi
    • one year ago
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    area of circle - area of segment area of segment is area of sector - area of triangle so A(circle) - (A(sector) - A(tri))= A(circle) - A(sector) + A(triangle)

  13. phi
    • one year ago
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    the area of the sector is 1/6 of the whole circle so A(circle) - A(sector) is A(circle) - A(circle)/6 = 5 A(circle)/6 so we want 5 A(circle)/6+ A(triangle)

  14. phi
    • one year ago
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    area of the circle is 144 pi 5/6 *144 pi is 120 pi A(tri) is 36 sqr(3) so the area we want is 120 pi + 36 sqr(3)

  15. anonymous
    • one year ago
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    Thanks again. xD

  16. anonymous
    • one year ago
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