construct an isoceles triangle with an area of 100. what is the length of the base of the triangle which has the smallest perimeter?
Prepping for a test, explanations please
Sorry Ignore
To construct an isosceles triangle with an area of 100, we need to find the dimensions of the triangle that will satisfy this condition.
Let's denote the base of the triangle as 'b' and the height as 'h'. The area of a triangle is given by the formula: Area = (base * height) / 2. In this case, the area is 100, so we have the equation:
100 = (b * h) / 2
Simplifying the equation, we get:
200 = b * h
Now, since it is an isosceles triangle, the base and height will be equal. So we can rewrite the equation as:
200 = b^2
To find the length of the base of the triangle with the smallest perimeter, we need to minimize the sum of the three sides of the triangle.
Let's call the lengths of the equal sides 's' and the base 'b'. The perimeter is given by the formula: Perimeter = 2s + b.
We can substitute the value of 's' to find the perimeter in terms of 'b':
Perimeter = 2(b/2) + b
Perimeter = b + b
Perimeter = 2b
Since we want to find the length of the base with the smallest perimeter, we need to minimize the value of 'b'. We already have an equation for 'b' from earlier:
200 = b^2
Now, to find the minimum value of 'b', we can take the derivative of the equation with respect to 'b' and set it to zero. Differentiating the equation:
d(200)/db = d(b^2)/db
0 = 2b
Solving for 'b', we get b = 0.
However, since we are looking for a positive value for 'b', we discard the solution b = 0.
Therefore, the minimum value for the base 'b' is when it is as close to zero as possible. However, in this case, the base cannot be zero because that would result in a degenerate triangle.
Hence, the answer is that the length of the base with the smallest perimeter is "as close to zero as possible" without being zero.