A sequence of two transformations that can be used to show that polygon ABCDE is congruent to polygon GFJIH
1 answer:
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Answer:
a) 0 m
b) 16.8 m
Step-by-step explanation:
A piece of wire, 30 m long, is cut in two sections: a and b. Then, the relation between a and b is:
The section "a" is used to make a square and the section "b" is used to make a circle.
The section "a" will be the perimeter of the square, so the square side will be:
Then, the area of the square is:
The section "b" will be the perimeter of the circle. Then, the radius of the circle will be:
The area of the circle will be:
The total area enclosed in this two figures is:
To calculate the extreme values of the total area, we derive and equal to 0:
We obtain one value for the extreme value, that is a=16.8.
We can derive again and calculate the value of the second derivative at a=16.8 in order to know if the extreme value is a minimum (the second derivative has a positive value) or is a maximum (the second derivative has a negative value):
As the second derivative is positive at a=16.8, this value is a minimum.
In order to find the maximum area, we analyze the function. It is a parabola, which decreases until a=16.8, and then increases.
Then, the maximum value has to be at a=0 or a=30, that are the extremes of the range of valid solutions.
When a=0 (and therefore, b=30), all the wire is used for the circle, so the total area is a circle, which surface is:
When a=30, all the wire is used for the square, so the total area is:
The maximum value happens for a=0.