SOLUTION: a fifteen long piece of wire is cut and formed into a circle and an equilateral triangle. graph the function for total area of the shapes.
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Question 1161857: a fifteen long piece of wire is cut and formed into a circle and an equilateral triangle. graph the function for total area of the shapes.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Assume the 15 unit piece of wire is cut at C units from one end where C is the circumference of the circle that is formed from that piece. That leaves 15 - C units with which to form the equilateral triangle. Since the perimeter of an equilateral triangle is 3s where s is the measure of one side of the triangle, we know that 3s = 15 - C, which is to say
The radius of a circle as a function of the circumference is:
The area of an equilateral triangle in terms of the measure of a side is derived from the facts that an altitude to one side is a perpendicular bisector of that side and it bisects the apex angle so that the altitude is the long leg of a 30-60-90 right triangle. Since the hypotenuse measures 
, the short leg is 
, and the long leg measures 
. The area of the 30-60-90 triangle is the measure of the short leg times the measure of the long leg divided by 2. But since the whole equilateral triangle is two of these right triangles, the area of the whole equilateral triangle is:
Substituting 
^2\sqrt{3}}{36})
Since the area of a circle is
Substituting 
^2}\ =\ \frac{C^2}{4\pi})
Then the total area of the two shapes as a function of the circumference of the circle shape is:
\ =\ \frac{C^2}{4\pi}\ +\ \frac{(15\,-\,C)^2\sqrt{3}}{36})
The graph is a parabola, but I'll let you work out the details.
John
My calculator said it, I believe it, that settles it
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