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Math Help - Total Surface Area

  1. #1
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    Total Surface Area

    Hello, I do not understand this geometry question about toal sruface area. I think the answer they gave me is wrong.:

    25. The net of a right triangular prism is shown below. Use the ruler on the Mathematics Chart to measure the dimensions of the right triangular prism to the nearest centimeter.



    Which is closest to the total surface area of this right triangular prism?
    A 18 cm^2
    B 60 cm^2
    C 48 cm^2
    D 36 cm^2

    The answer is C. I keep getting around 33 so, I'm not sure what I'm doing wrong here. How do you find the total surface area of this?
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  2. #2
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    Quote Originally Posted by Mariolee View Post
    Hello, I do not understand this geometry question about toal sruface area. I think the answer they gave me is wrong.:

    25. The net of a right triangular prism is shown below. Use the ruler on the Mathematics Chart to measure the dimensions of the right triangular prism to the nearest centimeter.

    Which is closest to the total surface area of this right triangular prism?
    A 18 cm^2
    B 60 cm^2
    C 48 cm^2
    D 36 cm^2

    The answer is C. I keep getting around 33 so, I'm not sure what I'm doing wrong here. How do you find the total surface area of this?
    1. The total surface area A exists of:
    2 congruent right triangles which form a rectangle.(yellow)
    2 rectangles (lightblue, green).

    2. Unfortunately you didn't give us the dimension you have obtained by measuring . So I called one side a and estimated the missing length x as x = \frac83 \cdot a.
    Then the hypotenuse of the right triangle is x - a

    3. Use Pythagorian theorem to calculate the length of y:

    a^2+y^2=\left(\frac53 a\right)^2~\implies~y=\frac43 a

    4. Calculate the different areas:

    A_{blue} = a \cdot \frac83 a = \frac83 a^2

    A_{green} = a \cdot \frac43 a = \frac43 a^2

    A_{yellow} = 2\cdot \frac12 \cdot a \cdot \frac43 a = \frac43 a^2

    The sum of these 3 areas yields the total surface area A=\frac{16}3 \cdot a^2

    5. If and only if you have measured a = 3 cm then the total area is indeed 48 cm².
    Attached Thumbnails Attached Thumbnails Total Surface Area-netz_prisma.png  
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  3. #3
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    Quote Originally Posted by earboth View Post
    1. The total surface area A exists of:
    2 congruent right triangles which form a rectangle.(yellow)
    2 rectangles (lightblue, green).

    2. Unfortunately you didn't give us the dimension you have obtained by measuring . So I called one side a and estimated the missing length x as x = \frac83 \cdot a.
    Then the hypotenuse of the right triangle is x - a

    3. Use Pythagorian theorem to calculate the length of y:

    a^2+y^2=\left(\frac53 a\right)^2~\implies~y=\frac43 a

    4. Calculate the different areas:

    A_{blue} = a \cdot \frac83 a = \frac83 a^2

    A_{green} = a \cdot \frac43 a = \frac43 a^2

    A_{yellow} = 2\cdot \frac12 \cdot a \cdot \frac43 a = \frac43 a^2

    The sum of these 3 areas yields the total surface area A=\frac{16}3 \cdot a^2

    5. If and only if you have measured a = 3 cm then the total area is indeed 48 cm².
    But that's a lot of estimation. The reason I've measured it in the 33 cm range is because I measured it with an actual ruler for they didn't give me an actual ruler on their Mathematical chart as promised. But this brings up a few questions:

    How did you estimate x = \frac83 \cdot a? Or any of those fractions. Where did all those numbers come from?
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  4. #4
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    Quote Originally Posted by Mariolee View Post
    But that's a lot of estimation. <=== correct but I had to because you didn't post the results of your measuring.
    The reason I've measured it in the 33 cm range is because I measured it with an actual ruler <=== how did you do that?
    for they didn't give me an actual ruler on their Mathematical chart as promised. But this brings up a few questions:

    How did you estimate x = \frac83 \cdot a? <=== actually I printed your sketch and did some measuring. Then I used the ratios between the sides of the areas which form the surface area.
    Or any of those fractions. Where did all those numbers come from? <=== Which numbers do you mean?
    By the way: Could it be that you measured the circumference of the given area? I only ask because the circumference is 32 cm and you say that you have measured 33 cm. But cm is not the unit of an area.
    Last edited by earboth; July 10th 2010 at 12:33 PM. Reason: additional question
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  5. #5
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    I actually put my ruler up against the screen of my comp and measured it in centimeters, as stupid as that sounds.

    I got 33 by measuring all of the regular ares of the different shapes and adding them together.

    Oh, ok. So does the 8 represent the area of the rectangle plus square?

    Quote Originally Posted by earboth View Post
    By the way: Could it be that you measured the circumference of the given area? I only ask because the circumference is 32 cm and you say that you have measured 33 cm. But cm is not the unit of an area.
    I might have on accident.
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  6. #6
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    I don't see how one can "accidently" measure perimeter rather than area. In fact, I don't see how one can measure area with a ruler, to begin with. Are you clear on the distinction between "perimeter" and "area"?
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    I don't see how one can "accidently" measure perimeter rather than area. In fact, I don't see how one can measure area with a ruler, to begin with. Are you clear on the distinction between "perimeter" and "area"?
    To be honest I wouldn't have asked this question if I didn't come across this thread:

    http://www.mathhelpforum.com/math-he...ml?pagenumber=

    So I only wanted to make sure that it was actually the area which should be determined.
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  8. #8
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    Quote Originally Posted by earboth View Post
    To be honest I wouldn't have asked this question if I didn't come across this thread:

    http://www.mathhelpforum.com/math-he...ml?pagenumber=

    So I only wanted to make sure that it was actually the area which should be determined.
    It is indeed.
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