algebra {4 problems} i am STUCK!?

Question

1. What is the length of each diagnol of a cube that is 45 cm on each side?

2.What is the length of each diagnol of a rectangular box with length 55 cm, width 48 cm, and height 70 cm? Would a meter stick fit in the box?

3.Gary is standing on a dock 2.0 m above the water. He is pulling in a boat that is attached to the end of a 5.2 m rope. If he pulls in 2.3 m of rope, how far did he move the boat?

please explain so i'll learn lol. cuz i don't want to JUST get the answer. these problems involve the pythagoream thyrom {sp?} and tringles. just FYI. :]] thanks

Answer 1

d^2=(a^2+a^2) +a^2

d=45 sqrt(3)

2)
d=sqrt(55^2+ 48^2+ 70^2)
=44,26

3)
5,2-2=3,2

Answer 2

Yes, all of these problems need the pythagorean theorem and the trick is knowing which are the sides and which is the diagonal.

1) Picture a right triangle whose sides are any two sides of the cube that start from a common point. The diagonal you're looking for is the triangle they form when you join their endpoints, I guess you will know how to use pythagoras there.

2) Same as one. Do it with 55 and 48 as right sides and obtain the length of diagonal of the side they form. Then do it with 55 and 70, last with 48 and 70. Make a drawing so its easier.

This is an interesting question, in 1 and 2 you obtained the diagonals in the SIDES of the boxes (the cube in 1 and the rectangle in 2). Now, say you small and can stand inside any of those boxes in one of the corners with your back against that corner. Then imagine a line going from the corner close to your ankles all the way to the TOP corner farther away from you. If that line is more than 1 meter long (in problem 2) you can fit the ruler in the box.

To obtain the length of such line you have to make a new triangle: one of height 70 and of length equal to the diagonal formed by the other two sides ( easy the root of the sum of 48 squared plus 55 squared ). Math is about writing the formulas and making drawings all you need to solve the problems is here, be patient.

3) This one is trickier, picture your right angled triangle. Now, the height is 2 and the DIAGONAL is 5.2 . Obtain the other side -answer below-

(5.2 squared) minus (2 squared) equals how far the boat is from the dock, name this answer as "A"

Now, the answer is "A" minus what follows:

Make a new triangle of height 2, diagonal 2.3 and obtain the other side ( you do this by 2.3 squared minus 2 squared, name this "B" )

The answer to 3 is B minus A

Sorry if this was too direct, hope it was of assistance. For more info check on the source (below I guess). Good luck.
Please learn your trigonometry right, it'll save you a lot from problems later.

Answer 3

1. this is tricky because it is a 3D object; since each side of the cube is the same length, 45 cm, the by the Pythagorean theorem of 3D
45^2 + 45^2 + 45^2 = d^2 where d is the length of diagonal

2. Use the same method here as above:
55^2 + 48^2 + 70^2 = d^2

3. We have to assume that the boat is at the water line and that Gary is standing on the dock but has no height ie he pulls on his belly, maybe; anyway, the length of rope attached to the boat is 5.2 m, this is the hypotenuse so that the length of the boat from the dock initially is (by old Pythag) given by:
5.2^2 = l^2 + 2^2 or l^2 = 5.2^2 - 2^2
Now let x be the distance required to be found, then by similar triangles:
2.3/5.2 = x/l where l is given above
So just solve for x and you are done

Answer 4

I'm gonna try and do this...but I haven't done geometry in quite a while.

1) So, squares (and cubes) have only 90 degrees angles...so any diagonal across a square would create a 45'45'90 triangle. So then you could use the pythagorean theorem to find the diagonal because the two short sides are both 45cm. Try that.

Answer 5

1. Use Pythag. Thm. : a^2 + b^2 = c^2
45^2 + 45^2 = c^2
2025 + 2025 = c^2
4050 = c^2
sq.rt. 4050 = c
c = 2 sqrt 45

Answer 6

Yuck.. Reminds me of why I hate maths.. Sorry!