Can someone tell me how to solve this Problem?

Question

I'm studying for my Geometry EOC, and I don't know how to answer this problem. Can someone help me with it, or at least tell me the topic to look up so I can solve it?

On a map, Tannersville, Chadwick, and Barkersville form a triangle. Chadwick is 70 miles from Tannersville and Barkersville is 90 miles from Tannersville. Which is a possible distance between Chadwick and Barkersville?

Answer 1

Lets look at what you have.

On the paper in front of you, draw a triangle with angle at the top and angles on the left and right side.

On the lower left angle write Chadwick. On the lower right angle write Barkerville. On the top angle write Tannersville.

On the line between Chadwick and Tannersville write 70.
On the line between Barkersville and Tannersville write 90.

So that represents the known information. Now the question is, what is a POSSIBLE distance between Chadwick and Barkersville.

The reason the question is so hard to think about is because there are lots and lots of answers. What happens if the two roads are almost the opposite direction (east and west).

Since it is a triangle you cannot have a straight line (180 degrees) but you could have almost a straight line which would make the distance almost up to the sum of the two distances. In that case, your triangle would have a LONG base and the top angle would be almost 180 degrees.

What happens if the two roads connecting the cities are almost parallel. In that case it could be almost the difference between the two values. That would make the angle at Chadwick almost 180 degrees (since that is the shortest leg).

So the answer must be any value greater than the difference and less that the sum of the two sides.
another way to write the solution is
20 > x < 160

Answer 2

if the three towns are in a straight line going tannersville then chadwick and then barkersville, then the maximum distance that chadwick can be from barkersville is

90 miles + 70 miles = 160 miles

if the three towns are still in a straight line ,but we change the order of them, say barkersville first then chadwick and then tannersville, then the minimum distance between chadwick and barkersville is

90 miles - 70 miles = 20 miles

so the possible distances between chadwick and barkersville is greater than 20 miles (since they form a triangle and not a straight line) and less than 160 miles. in interval notation your answer would be:

(20,160)

hoped this helped
10 pts best answer?

Answer 3

You do not list the possible distances, but since a triangle is formed, the distance on the third side can not exceed the sum of the other two sides.

Assuming an angle of 180 degrees, the maximum distance of the third side would be 160 miles (70 + 90). Assuming an angle of 0 degrees, the distance would be 20 miles.

If the three cities formed a right angle triangle, then the distance of the third side can be determined, but that was not mentioned.

Answer 4

Just set up a triangle with each city as a corner. I will use T, C, and B for simplicity. With only 2 pieces of information, you can't find the exact distance becuase the angle could vary. What you do know is that one side of a triangle must be less than the sum of the other two sides. If it is equal to the sum then you have a straight line and if it is greater, it cannot exist. Therefore, the distance from B-C must be less than the distance from T-B + the distance from T-C, so it must be less than 70+90=160

Also, it must be greater than the difference of the other 2 sides or the 2 sides would not be able to connect so it must be greater than 90-70=20

So the distance must be more than 20 but less than 160

Unless it says that it is a right triangle or there is a right angle, you are NOT looking for the hypotenuse

Answer 5

if a triangle exist,than if a,b and c are the sides then:
ac
(the sum of the 2 smaller sides > the biggest one)
lets say the third distance(lets call it x) is between 70 and 90-the 2 other values.then the triangle exists with no problems.
lets call distance 70miles=y
lets call distance 90miles=z
if x<70 then x+y>z => x > z - y =>x>20 miles
if x>90 then y+z>x => x < 160 miles
so x is between 20 and 160(not containing these exact values)

Answer 6

Any between 20 and 160 miles but not 20 or 160

Answer 7

Hmm...Does the problem tell you how many miles the three cities are away from each other altogether? I would guess 160 miles...or maybe 110...It depends on what the triangle looks like (if you have a picture of it, because if you do, then maybe you can look at it, and guess which one fits better). Hope I was a little bit of help. Good luck!

Answer 8

in the beginning, i've got not touched the programming long term in the past and easily knowledgeable in C++.. yet right here is a few concept if might help. you could elect the huge variety to circulate into one after the other somewhat of awhole.. as an occasion, 123, you may make it enter a million & 2 & 3 and assign each and each huge variety to a cellular like a million to cellular A, 2 to cellular B, 3 to cellular C.. then you truly can write this methodology to look opposite.. that's show cellular C, cellular B and cellular A. so first this methodology will ask "enter first huge variety" then you truly enter a million. then "enter 2d huge variety" then you truly enter 2. so on.. i'm unsure that's going to paintings however the assumption is there..

Answer 9

The sum of 2 legs of a triangle has to be greater than the value of the third leg - does this help?

Answer 10

sounds like your looking for the hypotenuese. so the square of the hypotenuese is equal to the sum of the square of the other 2 sides. you have the 2 sides so you should be able to plug in the numbers. i hope this is correct and helpful for you.