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## Geometry (all content)

### Course: Geometry (all content)>Unit 9

Lesson 2: Pythagorean theorem application

# Pythagorean theorem word problem: fishing boat

If we assume that the mast of a boat is at a right angle to the deck, then we can model the length of a rope stretched between the mast and deck as the hypotenuse of a right triangle. Then we can use the Pythagorean theorem to relate the lengths. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• Does the Pythagorean theorem only work on right triangles?
• Good question!

Draw a right-angled triangle, with a square drawn on each of the sides. The areas of the two smaller squares, added together, is equal to the area of the largest square.

Now draw two more triangles. In the first one, keep the longest side the same, but make the two shorter sides a bit shorter than in your first triangle. They will meet at an angle greater than 90°. Can you see that the areas of the squares drawn on their sides will add up to less than before? In the next triangle, keep the longest side the same, but make the two shorter sides a bit longer than in your first triangle. They will meet at an angle less than 90°. Can you see that the areas of the squares drawn on their sides will add up to more than before?

It is only when the two shorter sides meet at 90° do you get the relationship between the areas of the squares.

• Does it matter what order you pit the a2 or b2
• In the equation, no it does not because the terms are being added together. This is called the commutative property. In math we write the terms in alphabetical order according to the letter. That is why you see a2 written first, because a is before b in the alphabet.
On the triangle, as long as you are using the smaller sides of the triangle, called legs, it does not matter which you label as a or b.
• What if the mast isn't absolutely upright? How would we find the length of the rope then?
• If it isn't a right triangle you will need to use the Law of Cosines, or the Law of Sines.
• how is the Pythagorean useful in word problems?
• It you need to find the length of a ladder on the side of a house, which makes a right triangle
• Is there another theorem for other types of triangles?
• There are several theorems relating the sides and angles of general triangles. Try the Law of Sines and Law of Cosines in the Trigonometry section.
• can someone help? i have a hard time converting the answers in the modules.
• Try finding the lengths first before you calculate anything and make sure to draw a triangle so it makes it alot easier for yourself!
• lol its art class now
• This might have been asked before, but i couldn't find it.

Why can't you just (15^2)^1/2 + (20^2)^1/2= (c^2)^1/2 ---> 15+20=35. Why doesn't this work?
• This doesn't work because you have to take the square root of the whole left side to undo the "c^2" on the right side and get c by itself. Consider this example:

Let's say we had a right triangle with lengths 3, 4, and 5. We would write the pythagorean theorem like this:
3^2 + 4^2 = 5^2
And this works, because 3^2 + 4^2 is the same as 9 + 16, which is 25, which is 5^2. So the equation works.

However, what would happen if I tried to solve it the way you did in your question? It would look like this:
(3^2)^1/2 + (4^2)^1/2 = (5^2)^1/2
When we take a number and square it, and then take the square root of that, we just get the original number again. So "(3^2)^1/2" would just be "3", and "(4^2)^1/2" would just be 4, and "(5^2)^1/2" would just be 5. So we could re-write the equation like this:
3 + 4 = 5
But this clearly isn't true.

The reason it didn't work is because square roots do not distribute. This means that if you take the square root of both sides to get rid of the "^2" on the "c", you have to take the square root of the WHOLE left side too, not just the individual terms. So the actual equation would be:
a^2 + b^2 = c^2
(a^2 + b^2) ^ 1/2 = c
You can't just throw that "1/2" onto the a^2 and the b^2. That would get you a totally different answer.

I hope this helped.