# Pythagorean Theorem

• Jun 26th 2010, 07:30 PM
Veronica1999
Pythagorean Theorem
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.
• Jun 26th 2010, 07:39 PM
Also sprach Zarathustra
For 6.
Look here, what do you see?
• Jun 26th 2010, 07:48 PM
Veronica1999
Are problems 7. 8. 9. 10. correct?
• Jun 27th 2010, 12:29 AM
earboth
Quote:

Originally Posted by Veronica1999
6. I couldn't solve.
7. 14.3<<< 0.04 percent = 0.0004
8. 40.4<<<< 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 $|\overline{CD}|=\sqrt{30^2+5^2}\approx 30.4$
• Jun 27th 2010, 02:57 PM
Veronica1999

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
• Jun 27th 2010, 11:46 PM
earboth
Quote:

Originally Posted by Veronica1999

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: $d(18)=\sqrt{549}+\sqrt{244}=3 \sqrt{61}+2 \sqrt{61}=5\sqrt{61}\approx 39.05$
• Dec 6th 2010, 09:15 PM
Veronica1999
Quote:

Originally Posted by earboth
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?
• Dec 7th 2010, 12:53 AM
earboth
Quote:

Originally Posted by Veronica1999
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 $|\overline{DE}| + |\overline{CE}|$
• Feb 19th 2011, 02:40 PM
Veronica1999
Quote:

Originally Posted by earboth
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.
• Feb 20th 2011, 12:44 AM
earboth
Quote:

Originally Posted by Veronica1999
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.