A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?
A. pir^2
B. pir^2 + 10
C. pir^2 + 1/4pi^2r^2
D. pir^2 + (40 - 2\pir)^2E. pir^2 + (10 - 1/2pir)^2

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DarcySea DarcySea · Answer 1 · 2020-03-12T02:55:53+01:00

Answer:

E. pi*r^2 + (10 - 1/2*pi*r)^2

Step-by-step explanation:

Let the first cut piece be of length 'x'.

Let the second piece square be of length 'a'.

So, second piece length = Total length - first piece length

So, second piece length = [tex]40-x[/tex]

Now, the first piece is used to form a circle of radius 'r' while the second is used for a square of side length 'a'.

Circumference of the circle is equal to the length of the first piece.

So, [tex]2\pi r=x[/tex] ------ (1)

Area of circle is given as:

[tex]A_c=\pi r^2[/tex] ------- (A)

Now, perimeter of the square is equal to the length of second piece.

So, [tex]4a=40-x[/tex]

[tex]a=\frac{40}{4}-\frac{x}{4}\\\\a=10-\frac{x}{4} ----- (2)[/tex]

Plug in 'x' value from equation (1) in equation (2). This gives,

[tex]a=10-\frac{2\pi r}{4}\\\\a=10-\frac{\pi r}{2}[/tex]

Now, area of square is given as:

[tex]A_s=a^2\\\\A_s=(10-\frac{\pi r}{2})^2----- (B)[/tex]

Now, total area is equal to the sum of the areas of circle and square. So,

Total area = Area of circle + Area of square

Total area = [tex]A_c+A_s[/tex]

Total area = [tex]\pi r^2 +(10-\frac{1}{2}\pi r)^2[/tex]

Therefore, option (E) is correct.