Model each scenario with an equation and a sketch. Solve for the missing value and u. Use complete sentences to interpret the solution. In your final answer, include your equation, sketch, interpretation, and all calculations necessary for the solution. To get home from school, Bob walks four blocks north and three blocks east. What is the straight line distance between Bob’s house and his school? One of the requirements of your summer window-washing job is to provide yourself with all of the necessary supplies, including a fourteen foot ladder. When you arrive at your first job, you place your ladder on the ground six feet from the base of the house and lean it towards a second story window, only to realize that the ladder doesn’t reach the window. Given the length of the ladder and its current position, what is one possible height of the second story window? (Hint: There is more than one correct answer.) In a softball diamond, each of the bases, including home plate, are equidistant from each other. Although the name implies differently, a softball diamond is in the shape of a square. Given that the distance between the bases is unknown, determine an expression for the straight line distance between first and third bases. A sailboat drifts 600 meters west, makes a turn and sails 800 meters south. How far is the sailboat from its original position?

1. Since bob walks north and then east, he creates a right triangle in which the legs are the distances that Bob walks, and the hypotenuse is the straight line distance between Bob’s house and his school. To relate all the distances we are going to use the Pythagorean theorem: [tex]c= \sqrt{a^{2}+b^{2}} [/tex]
where 
[tex]c[/tex] is the straight line distance between Bob’s house and his school
[tex]a[/tex] is the distance that Bob walks north
[tex]b[/tex] is the distance that Bob walks east
We know for our problem that Bob walks four blocks north, so [tex]a=4[/tex]. We also know that Bob walks three blocks east, so [tex]b=3[/tex]. Lets replace those values in our equation to find the distance between Bob's house and his school:
[tex]c= \sqrt{a^{2}+b^{2}} [/tex]
[tex]c= \sqrt{4^{2}+3^{2}} [/tex]
[tex]c= \sqrt{16+9} [/tex]
[tex]c= \sqrt{25} [/tex]
[tex]c=5[/tex]
We can conclude that the straight line distance between Bob’s house and his school is 5 blocks.

2. Here we are going to use the Pythagorean theorem as well, but this time the length of the ladder will be the hypotenuse of our triangle, and the distance between the house and the ladder will be one of the legs of our triangle. So to model the situation we are going to use the equation: [tex]a= \sqrt{c^{2}-b^{2} } [/tex]
where
[tex]a[/tex] is the distance that the leader reach
[tex]b[/tex] is the distance between the house and the ladder
[tex]c[/tex] is the length of the ladder
For our problem we know that [tex]b=6[/tex] and [tex]c=14[/tex], so lets replace those values in our equation to find [tex]a[/tex]:
[tex]a= \sqrt{c^{2}-b^{2}} [/tex]
[tex]a= \sqrt{14^{2}-6^{2}} [/tex]
[tex]a= \sqrt{196-36} [/tex]
[tex]a= \sqrt{160} [/tex]
[tex]a=12.6[/tex]
We now know that the current position of our leader in the wall of the house is 12.6 feet. Since we know that our leader does not reach the window, we can conclude that the possible height of the second store window is greater than 12.6 feet.

3. We know that a softball diamond is in the shape of a square, and all the bases are equidistant from each other; therefore the straight line distance between first and third bases is the diagonal of the square. We also now that the distance between the bases are the sides of the square. Since that distance is unknown, we are going to represent it with [tex]x[/tex]. We are also going to represent the straight line distance between first and third bases with [tex]d[/tex]. So now, we have a right triangle of hypotenuse [tex]d[/tex] and legs [tex]x[/tex]. Lets use the Pythagorean theorem to set up our equation:
[tex]d= \sqrt{ x^{2} + x^{2} } [/tex]
[tex]d= \sqrt{2 x^{2} } [/tex]
[tex]d=x \sqrt{2} [/tex]
We can conclude that the expression that represents the straight line distance between first and third bases is [tex]d=x \sqrt{2} [/tex].

4. When our sailboat drifts west and then makes a turn south it creates a right triangle. The hypotenuse of our right triangle will be the distance of the sailboat from its original position. So lets use the Pythagorean theorem: [tex]c= \sqrt{a^{2}+b^{2}} [/tex]
where 
[tex]a[/tex] is the distance covered by our sailboat in the west direction
[tex]b[/tex] is the distance covered by our sailboat in the south direction
[tex]c[/tex]  is the distance of the sailboat from its original position
From our problem we know that [tex]a=600[/tex] and [tex]b=800[/tex], so lets replace those values in our equation to find [tex]c[/tex]:
[tex]c= \sqrt{a^{2}+b^{2}} [/tex]
[tex]c= \sqrt{600^{2}+800^{2}} [/tex]
[tex]c= \sqrt{360000+640000}[/tex]
[tex]c= \sqrt{1000000} [/tex]
[tex]c=1000[/tex]
We can conclude that our sailboat is 1000 meters away from its original position.