Question

make a sketch for each problem. label the lengths that are known. then use the pythagorean theorem to solve the problem.
a baseball diamond is a square 90 feet on each side. how long is a throw from home plate to second base (to the nearest foot)?
the bottom of a 25 - foot ladder is placed 7 feet from a wall. how far up the wall will the ladder reach?
a path leads across the park from one corner to the opposite corner. the park is 150 meters wide and 200 meters long. how far would you walk if you took the path instead of walking around? how many meters would you save?
sam wants to brace a bookcase by nailing a strip of wood from the lower left corner to the upper right corner. the bookcase is 1 meter wide and 2 meters high. how long should the brace be (to the nearest hundredth of a meter)?

Explanation:

1. Throw from home plate to second base in a baseball diamond

Step1: Identify the right - triangle sides

A baseball diamond is a square with side length \(a = b=90\) feet. The throw from home plate to second base is the hypotenuse \(c\) of a right - triangle.

Step2: Apply the Pythagorean Theorem

The Pythagorean Theorem is \(c^{2}=a^{2}+b^{2}\). Substituting \(a = 90\) and \(b = 90\), we get \(c^{2}=90^{2}+90^{2}=8100 + 8100=16200\).

Step3: Solve for \(c\)

\(c=\sqrt{16200}\approx127.28\approx127\) feet.

2. Height the ladder reaches on the wall

Step1: Identify the right - triangle sides

The length of the ladder is the hypotenuse \(c = 25\) feet and the distance from the wall to the bottom of the ladder is \(a = 7\) feet. Let the height up the wall be \(b\).

Step2: Apply the Pythagorean Theorem

\(b^{2}=c^{2}-a^{2}\). Substituting \(c = 25\) and \(a = 7\), we have \(b^{2}=25^{2}-7^{2}=625 - 49 = 576\).

Step3: Solve for \(b\)

\(b=\sqrt{576}=24\) feet.

3. Path across the park

Step1: Find the length of the path across the park

The park has length \(a = 200\) meters and width \(b = 150\) meters. The path across the park is the hypotenuse \(c\) of a right - triangle. Using the Pythagorean Theorem \(c^{2}=a^{2}+b^{2}\), so \(c^{2}=200^{2}+150^{2}=40000+22500 = 62500\). Then \(c=\sqrt{62500}=250\) meters.

Step2: Find the distance walking around

The distance walking around is \(200 + 150=350\) meters.

Step3: Calculate the distance saved

The distance saved is \(350 - 250 = 100\) meters.

4. Brace for the bookcase

Answer:

127 feet