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12. two joggers run 8 miles north and then 5 miles west. what is the sh…

Question

  1. two joggers run 8 miles north and then 5 miles west. what is the shortest distance, to the nearest tenth of a mile, they must travel to return to their starting point?
  2. oscars dog house is shaped like a tent. the slanted sides are both 5 feet long and the bottom of the house is 6 feet across. what is the height of his dog house, in feet, at its tallest point?
  3. to get from point a to point b you must avoid walking through a pond. to avoid the pond, you must walk 34 meters south and 41 meters east. to the nearest meter, how many meters would be saved if it were possible to walk through the pond?
  4. a suitcase measures 24 inches long and the diagonal is 30 inches long. how much material is needed to cover one side of the suitcase?

Explanation:

Response
Question 12

Step1: Identify the triangle type

The joggers' path forms a right triangle, with legs \(a = 8\) miles (north) and \(b = 5\) miles (west). The shortest return distance is the hypotenuse \(c\).

Step2: Apply the Pythagorean theorem

The Pythagorean theorem states \(c=\sqrt{a^{2}+b^{2}}\). Substitute \(a = 8\) and \(b = 5\):
\(c=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4\) (rounded to the nearest tenth).

Step1: Analyze the tent shape

The dog house (tent) is an isosceles triangle. The base is \(6\) feet, so half the base is \(\frac{6}{2}=3\) feet. The slanted side (hypotenuse) is \(5\) feet. Let the height be \(h\).

Step2: Apply the Pythagorean theorem

Using \(a = 3\) (half - base), \(c = 5\) (hypotenuse), and \(b=h\) (height) in \(a^{2}+h^{2}=c^{2}\):
\(3^{2}+h^{2}=5^{2}\)
\(9 + h^{2}=25\)
\(h^{2}=25 - 9=16\)
\(h=\sqrt{16}=4\)

Step1: Find the direct distance (hypotenuse)

The path around the pond forms a right triangle with legs \(a = 34\) meters (south) and \(b = 41\) meters (east). The direct distance \(c\) is \(\sqrt{a^{2}+b^{2}}\).
\(c=\sqrt{34^{2}+41^{2}}=\sqrt{1156+1681}=\sqrt{2837}\approx53.3\) meters.

Step2: Find the distance saved

The distance around the pond is \(34 + 41=75\) meters. The distance saved is \(75-53.3 = 21.7\approx22\) meters (to the nearest meter).

Answer:

\(9.4\) miles

Question 13