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Math Help - Pythagorean Theorem

  1. #1
    Member Veronica1999's Avatar
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    Pythagorean Theorem

    I have five problems attached. I wasn't sure about my answers.
    6. I couldn't solve.
    7. 14.3
    8. 40.4
    9. 39.2
    10. y(squared) = 2(x-15)squared + 775
    The shortest distance is 28.

    The last part of problem 10 is Draw an accurate picture of this path, and make a conjecture about angles AED and BEC. Use your protractor to test your conjecture.
    Attached Thumbnails Attached Thumbnails Pythagorean Theorem-geo.pdf  
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    For 6.

    Look here, what do you see?
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  3. #3
    Member Veronica1999's Avatar
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    Are problems 7. 8. 9. 10. correct?
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    Quote Originally Posted by Veronica1999 View Post
    I have five problems attached. I wasn't sure about my answers.
    6. I couldn't solve.
    7. 14.3<<< 0.04 percent = 0.0004
    8. 40.4<<<< OK
    9. 39.2<<<< the answer is: No. (Your result is OK!)
    10. y(squared) = 2(x-15)squared + 775
    The shortest distance is 28.

    The last part of problem 10 is Draw an accurate picture of this path, and make a conjecture about angles AED and BEC. Use your protractor to test your conjecture.
    to #10: I'm very curious to learn how you got your result. It can't be correct because the straight line segment CD has a length of |\overline{CD}|=\sqrt{30^2+5^2}\approx 30.4
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  5. #5
    Member Veronica1999's Avatar
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    7. new answer 1.4

    10. I set up the equation y to equal the value of DE + EC
    y =  \[15^{2}+x^{2}+ (30-x)^{2} + 100\]

    If I get the minimum value for this equation, I thought it would be the answer.
    I got 775, so root 775 = 27.8
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  6. #6
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    Quote Originally Posted by Veronica1999 View Post
    7. new answer 1.4

    10. I set up the equation y to equal the value of DE + EC
    y =  \[15^{2}+x^{2}+ (30-x)^{2} + 100\]

    If I get the minimum value for this equation, I thought it would be the answer.
    I got 775, so root 775 = 27.8
    1. This method is only valid if the distance is calculated by a single square-root.

    2. d(x)=|\overline{DE}+\overline{EC}|= \sqrt{15^{2}+x^{2}}+ \sqrt{(30-x)^{2} + 100}

    3. Differentiate d wrt x. You'll get:

    d'(x)=\dfrac{x \sqrt{x^2-60x+1000} + (30-x) \sqrt{225+x^2}}{\sqrt{225+x^2} \cdot \sqrt{x^2-60x+1000}}

    Now solve for x: d'(x) = 0
    Caution: To solve this equation you need a lot of paper and even more patience. Finally you should come out with x = 18.

    4. Thus the minimum distance is: d(18)=\sqrt{549}+\sqrt{244}=3 \sqrt{61}+2 \sqrt{61}=5\sqrt{61}\approx 39.05
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  7. #7
    Member Veronica1999's Avatar
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    Quote Originally Posted by earboth View Post
    1. This method is only valid if the distance is calculated by a single square-root.

    2. d(x)=|\overline{DE}+\overline{EC}|= \sqrt{15^{2}+x^{2}}+ \sqrt{(30-x)^{2} + 100}

    3. Differentiate d wrt x. You'll get:

    d'(x)=\dfrac{x \sqrt{x^2-60x+1000} + (30-x) \sqrt{225+x^2}}{\sqrt{225+x^2} \cdot \sqrt{x^2-60x+1000}}

    Now solve for x: d'(x) = 0
    Caution: To solve this equation you need a lot of paper and even more patience. Finally you should come out with x = 18.

    4. Thus the minimum distance is: d(18)=\sqrt{549}+\sqrt{244}=3 \sqrt{61}+2 \sqrt{61}=5\sqrt{61}\approx 39.05
    I was studying reflection and figured out that this could have been easily solved if I reflected the 15 and used the pythagorean theorem. Then, I would have got 30squared +25squared = 1525 which is 39.0512

    Is my logic correct?
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  8. #8
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    Quote Originally Posted by Veronica1999 View Post
    I was studying reflection and figured out that this could have been easily solved if I reflected the 15 and used the pythagorean theorem. Then, I would have got 30squared +25squared = 1525 which is 39.0512

    Is my logic correct?
    1. This way to solve the problem is only valid if the angle \angle(CED) = 90^\circ

    2. If \angle(CED) \ne 90^\circ you calculate the length of the red line in my sketch which hasn't the same length as |\overline{DE}| + |\overline{CE}|
    Attached Thumbnails Attached Thumbnails Pythagorean Theorem-min_distance.png  
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  9. #9
    Member Veronica1999's Avatar
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    Quote Originally Posted by earboth View Post
    1. This way to solve the problem is only valid if the angle \angle(CED) = 90^\circ

    2. If \angle(CED) \ne 90^\circ you calculate the length of the red line in my sketch which hasn't the same length as |\overline{DE}| + |\overline{CE}|
    Could you please take a look at my attached work.

    I am concluding that you can get the shortest length when angle AED and BEC is equal.
    Attached Thumbnails Attached Thumbnails Pythagorean Theorem-mathhelpp.pdf  
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  10. #10
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    Quote Originally Posted by Veronica1999 View Post
    Could you please take a look at my attached work.

    I am concluding that you can get the shortest length when angle AED and BEC is equal.
    That's correct.

    But instead of using the values of the angles you should see when this situation will occur: If you reflect D to D' (or C to C' as you've proposed) then you'll have similar triangles (greyed).

    Now you can use proportions:

    \dfrac{10}{15}=\dfrac x{30-x}

    Solve for x.
    Attached Thumbnails Attached Thumbnails Pythagorean Theorem-min_dist2.png  
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