SOLUTION: I need help with this question and sorry if it is kind of confusing since I cant get a picture (there is a equilateral triangle ABC i think i might be wrong but the two legs or sid

Question 199614: I need help with this question and sorry if it is kind of confusing since I cant get a picture (there is a equilateral triangle ABC i think i might be wrong but the two legs or sides are both 16 and i am supposed find the range that gives all possible values for b, and another one but same is that if the area of triangle ABC is 114 square units and h = 12, then calculate the value of b.)
thanks I am really sorry that this is really long,
Craig
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

If ABC is indeed an equilateral triangle, then all three sides must be equal (equi means equal, lateral means side). Since your problem implies that there is a range to the value of the measure of the unspecified side, that means that you have an isosceles triangle, one that has two equal sides.

So, we know the sum of the lengths of the two sides you know is 16 + 16 = 32. Now if your side

were any larger than 32, the two given sides would not stretch that far, and you would not have a triangle. Also, if your side

measured exactly 32, then you would just have two straight lines on top of each other, and again you would not have a triangle. But as long as

by any amount however small you have a triangle. When

is very close to 32, you have a very wide but very short triangle. As

gets smaller, the triangle gets narrower but taller.

Now, on the other end,

has to measure something. That is to say it can't be zero. It can be very, very small giving you a very narrow but tall triangle. So we can say that

Putting all of that together, your range of values for

.

As to your second problem, you need to consider the formula for the area of a triangle given the base and height measures:

But you are given the area and the height and wish to know the measure of the base, so:

Just do the arithmetic.

John

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