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Math Help - Urgent! Help! Similar Solids!

  1. #1
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    Post Urgent! Help! Similar Solids!

    Two solids are similar, with volumes of 54cm^3 and 16cm^3. If the larger solid has a surface area of 27cm^2, what is the surface area of the smaller solid?

    So its a proportion...

    16cm^3 X
    _______ = ______
    54cm^3 27cm^2

    But how do i solve it? Cross multiply? do i have to turn them into cubed instead of squared?
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by sarahgasoline View Post
    Two solids are similar, with volumes of 54cm^3 and 16cm^3. If the larger solid has a surface area of 27cm^2, what is the surface area of the smaller solid?

    So its a proportion...

    16cm^3 X
    _______ = ______
    54cm^3 27cm^2

    But how do i solve it? Cross multiply? do i have to turn them into cubed instead of squared?
    Let's see what happens if we have a 2x2x2 cube (volume=8, s.a.=24) and a 4x4x4 cube (volume=64, s.a.=92)

    does \frac{8}{64}=\frac{24}{96}?

    The answer: no...

    However, does: \frac{8}{24}=\frac{64}{96}

    The answer: no...

    in fact, I think you need to know what kind of object it is...

    but you should wait for someone else's opinion.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sarahgasoline View Post
    Two solids are similar, with volumes of 54cm^3 and 16cm^3. If the larger solid has a surface area of 27cm^2, what is the surface area of the smaller solid?

    So its a proportion...

    16cm^3 X
    _______ = ______
    54cm^3 27cm^2

    But how do i solve it? Cross multiply? do i have to turn them into cubed instead of squared?
    Say there is a size factor of x between the lengths of the respective sides L_1 and L_2 of the solid. That is to say: L_2 = x L_1. (This ratio obviously will hold for any chosen linear measurement between the two solids. This is the definition of similar solids.) Then the volumes will be in the cube of that ratio:
    V_2 = x^3V_1

    In this case:
    54 \, cm^3 = x^3(16 \, cm^3)

    x^3 = \frac{54 \, cm^3}{16 \, cm^3} = \frac{27}{8}

    x = \frac{3}{2}

    Now, surface areas will be in the square of the ratio:
    SA_2 = x^2SA_1

    So
    27 \, cm^2 = \left ( \frac{3}{2} \right ) ^2SA_1

    27 \, cm^2 = \frac{9}{4} \cdot SA_1

    SA_1 = \frac{4}{9} ( 27 \, cm^2 )

    SA_1 = 12 \, cm^2

    -Dan
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  4. #4
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    Quote Originally Posted by sarahgasoline View Post
    Two solids are similar, with volumes of 54cm^3 and 16cm^3. If the larger solid has a surface area of 27cm^2, what is the surface area of the smaller solid?

    So its a proportion...

    16cm^3 X
    _______ = ______
    54cm^3 27cm^2

    But how do i solve it? Cross multiply? do i have to turn them into cubed instead of squared?
    Here is one way.

    You cannot equate ratios that are not in the same power. Example, you cannot equate a ratio of cubes to a ratio of squares. They have to be in the same power. A ratio of cubes to another ratio of cubes, ....

    (16 cu.cm)/(54 cu.cm) is a ratio of cubes.
    (x sq.cm)/(27 sq.cm) is a ratio of squares.
    The two cannot be equal.

    So, either (a) reduce the (16/54) into a ratio of squares, or (b) raise the (x/27) into a ratio of cubes.

    a) All in ratios of cubes:
    (16/54) = (x/27)^(3/2)
    In the righthand side, we took first the square roots of x and 27, then we cubed those roots. Now both ratios are in cubes.
    To continue,
    Clear the radical, raise both sides to their (2/3) powers,
    (16/54)^(2/3) = (x/27)
    x = 27*[(16/54)^(2/3)]
    x = 12 -----------------answer.

    b) All in ratios of squares:
    (16/54)^(2/3) = (x/27)
    In the lefthand side, we got first the cuberoots of 16 and 54, then we squared those cuberoots. Now both ratios are in squares.
    Hey, we have the same equation as the cleared one in part (a) above!
    So,
    x = 12 ---------------answer.
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