Pythagorean Theorem

**Veronica1999** · June 26th, 2010, 06:30 PM

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.

**Also sprach Zarathustra** · June 26th, 2010, 06:39 PM

Look here, what do you see?

**Veronica1999** · June 26th, 2010, 06:48 PM

Are problems 7. 8. 9. 10. correct?

**earboth** · June 26th, 2010, 11:29 PM

Originally Posted by Veronica1999

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 $|\overline{CD}|=\sqrt{30^2+5^2}\approx 30.4$

**Veronica1999** · June 27th, 2010, 01:57 PM

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

**earboth** · June 27th, 2010, 10:46 PM

Originally Posted by Veronica1999

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: $d(18)=\sqrt{549}+\sqrt{244}=3 \sqrt{61}+2 \sqrt{61}=5\sqrt{61}\approx 39.05$

**Veronica1999** · December 6th, 2010, 08:15 PM

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?

**earboth** · December 6th, 2010, 11:53 PM

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 $|\overline{DE}| + |\overline{CE}|$

**Veronica1999** · February 19th, 2011, 01:40 PM

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.

**earboth** · February 19th, 2011, 11:44 PM

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.

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