Question

1. walt disney railroad to winnie the pooh adventure $10^{2}+3^{2}=c^{2}$ 1044.0
2. tom sawyer island to the hall of presidents $1^{2}+2^{2}=c^{2}$ 224.0
3. buzz lightyear spin to mad tea $1^{2}+6^{2}=c^{2}$ 608.0
4. jungle cruise to the barnstormer $6^{2}+13^{2}=c^{2}$ 1432.0
5. peter pans flight to astro orbiter $3^{2}+11^{2}=c^{2}$ 916.0
6. legend of the indy speedway $1^{2}+7^{2}=c^{2}$ 707.0
7. the timekeeper to big thunder mountain railroad $2^{2}+13^{2}=c^{2}$ 1393.0
8. the enchanted tiki room to the haunted mansion $2^{2}+5^{2}=c^{2}$ c = 539.0
9. the alien encounter to splash mountain $2^{2}+14^{2}=c^{2}$
10. juan and michael are supposed to meet at space mountain. michael has to walk from country bear jamboree and juan has to walk from small world. michael has to walk from country bear jamboree. who has to walk further? by how much. $4^{2}+11^{2}=c^{2}$ 447.0 $1^{2}+15^{2}=c^{2}$

Explanation:

Step1: Recall Pythagorean theorem

For a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(c=\sqrt{a^{2}+b^{2}}\). And we need to multiply the result by 100.

Step2: Solve for problem 1

Given \(a = 10\), \(b = 3\). First calculate \(a^{2}+b^{2}=10^{2}+3^{2}=100 + 9=109\), then \(c=\sqrt{109}\), and the value after multiplying by 100 is \(100\sqrt{109}\approx1044.0\).

Step3: Solve for problem 2

Given \(a = 1\), \(b = 2\). Calculate \(a^{2}+b^{2}=1^{2}+2^{2}=1 + 4 = 5\), then \(c=\sqrt{5}\), and the value after multiplying by 100 is \(100\sqrt{5}\approx223.6\approx224.0\).

Step4: Solve for problem 3

Given \(a = 1\), \(b = 6\). Calculate \(a^{2}+b^{2}=1^{2}+6^{2}=1+36 = 37\), then \(c=\sqrt{37}\), and the value after multiplying by 100 is \(100\sqrt{37}\approx608.3\approx608.0\).

Step5: Solve for problem 4

Given \(a = 6\), \(b = 13\). Calculate \(a^{2}+b^{2}=6^{2}+13^{2}=36 + 169=205\), then \(c=\sqrt{205}\), and the value after multiplying by 100 is \(100\sqrt{205}\approx1431.8\approx1432.0\).

Step6: Solve for problem 5

Given \(a = 3\), \(b = 11\). Calculate \(a^{2}+b^{2}=3^{2}+11^{2}=9 + 121 = 130\), then \(c=\sqrt{130}\), and the value after multiplying by 100 is \(100\sqrt{130}\approx1140.2\approx1140.0\) (there seems to be a calculation error in the original for this one).

Step7: Solve for problem 6

Given \(a = 1\), \(b = 7\). Calculate \(a^{2}+b^{2}=1^{2}+7^{2}=1 + 49 = 50\), then \(c=\sqrt{50}\), and the value after multiplying by 100 is \(100\sqrt{50}\approx707.1\approx707.0\).

Step8: Solve for problem 7

Given \(a = 1\), \(b = 13\). Calculate \(a^{2}+b^{2}=1^{2}+13^{2}=1+169 = 170\), then \(c=\sqrt{170}\), and the value after multiplying by 100 is \(100\sqrt{170}\approx1303.8\approx1304.0\) (there seems to be a calculation error in the original for this one).

Step9: Solve for problem 8

Given \(a = 2\), \(b = 5\). Calculate \(a^{2}+b^{2}=2^{2}+5^{2}=4 + 25 = 29\), then \(c=\sqrt{29}\), and the value after multiplying by 100 is \(100\sqrt{29}\approx538.5\approx539.0\).

Step10: Solve for problem 9

Given \(a = 2\), \(b = 14\). Calculate \(a^{2}+b^{2}=2^{2}+14^{2}=4 + 196 = 200\), then \(c=\sqrt{200}\), and the value after multiplying by 100 is \(100\sqrt{200}\approx1414.2\).

Step11: Solve for problem 10

For Juan: \(a = 4\), \(b = 11\), \(a^{2}+b^{2}=4^{2}+11^{2}=16 + 121 = 137\), \(c_{1}=100\sqrt{137}\approx1170.5\).
For Michael: \(a = 1\), \(b = 15\), \(a^{2}+b^{2}=1^{2}+15^{2}=1+225 = 226\), \(c_{2}=100\sqrt{226}\approx1503.3\).
The difference is \(100(\sqrt{226}-\sqrt{137})\approx1503.3 - 1170.5=332.8\). Michael walks further by approximately \(332.8\).

Answer:

1. \(1044.0\)
2. \(224.0\)
3. \(608.0\)
4. \(1432.0\)
5. \(1140.0\)
6. \(707.0\)
7. \(1304.0\)
8. \(539.0\)
9. \(1414.2\)
10. Michael walks further by approximately \(332.8\)