Flagstaff

Kieran

New member
Joined
Nov 16, 2012
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A flagstaff broke a second time. Previously , it broke @ a level 3 feet lower. The top of the staff touches the ground 6 foot from the bottom. Previously , it was 12 feet from the base. How tall is the flagstaff ?
No diagram was given. Should right ang triangles be used ? A problem that contains a 6 and a 12 is usually a subtle hint. Thanks.
 
A flagstaff broke a second time. Previously , it broke @ a level 3 feet lower. The top of the staff touches the ground 6 foot from the bottom. Previously , it was 12 feet from the base. How tall is the flagstaff ?
No diagram was given. Should right ang triangles be used ? A problem that contains a 6 and a 12 is usually a subtle hint. Thanks.

For this problem you need to draw the picture. Yes right angled triangle should be used.

Please share your work with us.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (e.g. "are these correct?")

Hints:

Assume

The length of the pole is = L

The height broken first time = y1

Then hypotenuse (h1) = L - y1

Base = 6

Apply Pythagorean theorem and get equation (1)

Then

The height broken second time = y1 - 3

Then hypotenuse (h1) = L - y1 + 3

Base = 12

Apply Pythagorean theorem and get equation (2)

Using (1) and (2) solve for L.
 
A flagstaff broke a second time. Previously , it broke @ a level 3 feet lower. The top of the staff touches the ground 6 foot from the bottom. Previously , it was 12 feet from the base. How tall is the flagstaff ?
No diagram was given. Should right ang triangles be used?
No picture? Yeesh! They're expecting a lot of assumptions, then. :-?

Yes, right triangles will likely be the way to go. I'd assume that, when they talk about the "top of the staff" "touching the ground" a certain distance from "the bottom", they mean us to gather that the upper portion (above the break) is still kind-of attached, with the broken-off bit clinging to the upright remainder of the pole at the break point, and the tip in the dirt, with the broken-off part being slanty sideways, forming the hypotenuse of the right triangle.

So your triangle for the current break could have "h" for the height (of the remaining upright portion), "6" for the base (along the ground), and, say, "L" for the length of the broken-off bit.

Your triangle for the previous break could have "h - 3" for the height, "12" for the base, and -- because the hypotenuse and the height together form the total length of the unbroken pole -- "L + 3" for the hypotenuse.

This gives you two equations in two unknowns. Solve the system, and back-solve for the value they want. ;)
 
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