# Thread: triangle?

1. ## triangle?

Tom has two sticks, one 17 inches long and the other 10 inches long. Joan is looking for another stick so they can build a triangle. What is the maximum range that the length of the third stick can fall within in order to be able to build a triangle with the three sticks as sides?

2. Originally Posted by igottaquestion
Tom has two sticks, one 17 inches long and the other 10 inches long. Joan is looking for another stick so they can build a triangle. What is the maximum range that the length of the third stick can fall within in order to be able to build a triangle with the three sticks as sides?
If you join the 2 sticks at a point, then the maximum length of the third stick is the sum of the lengths of the 2 sticks that Tom has.
All 3 sticks would make a line 27 inches long.
If the 3rd stick is less than that sum, they can make a triangle.

In a triangle, all 3 sides touch.

3. ## triangle

yeah but what if the stick is one inch long...then they wouldnt touch

4. Oh yeah!
sorry, thanks for pointing that out.

If Tom's 2 sticks lie one on top of the other,
then the minimum length of Joan's stick is the difference between the lengths of Tom's sticks.

Therefore the range of the length of Joan's stick is

$(17-10)

You could draw a diagram to illustrate.

Draw the 17 inch side (make it 17cm or so).
Then, with one end as centre, draw a circle of radius 10 inches (10cm).
All points on the circle circumference give the 3rd apex.

5. Hello, igottaquestion!

Tom has two sticks, one 17 inches long and the other 10 inches long.
Joan is looking for another stick so they can build a triangle.
What is the maximum range that the length of the third stick can fall within
in order to be able to build a triangle with the three sticks as sides?
We have this triangle:
Code:
            *
*  *
*     *
10 *        * 17
*           *
*              *
*  *  *  *  *  *  *
x
The Triangle Inequality says:
. . The sum of any two sides must be greater than the third side.

We have: . $x + 10 \:>\:17 \quad\Rightarrow\quad x \:>\:7$

. . . .And: . $10+17 \:>\:x \quad\Rightarrow\quad 27 \:>\:x \quad\Rightarrow\quad x \:<\:27$

Therefore: . $7 \:<\:x\:<\:27$

6. for any triangle, with 3 sides a,b,c the sum of 2 sides is always greater than the third side
ie. a+b>c,a+c>b,b+c>a