Question

3) Smurfette is glad you're getting on with your adventure, but she wants to warn you that this last

Answer 1

Answer:

take x=2.515834 and r=14.31386

Step-by-step explanation:

Let us call the the lenght of the side of the square as x. The total lenght of the wire is 100. Note that if we have a square of side x, it has a perimeter of 4x. That means, that we need a wire of length 4x to build a square of side x. Note that if we take a wire of length 4x, we have that the remainder is 100-4x. We will use this to form a circle. As for the square, since we are using the wire of length 100-4x to build a circle, it must happen that the perimeter of the circle is 100-4x.

Recall that the perimeter of a circle of radius r is $2\pi r$, so in this case , we have that $2\pi r = 100-4x$ which implies that $r =\frac{100-4x}{2\pi}$.

Now, recall that the area of a square of side x is $x^2$. Also, recall that the area of a circle of radius r is $\pi r^2$. In our case, the area of the circle is given by

$\pi r^2 = \pi (\frac{100-4x}{2\pi})^2$

Also, we are given that the sum of the areas of both figures is 650. That is

$x^2 + \pi (\frac{100-4x}{2\pi})^2 = x^2+\frac{(100-4x)^2}{4\pi}=650$

By substracting 650 on both sides and multiplying by $4\pi$ this equation is equivalent to

$4\pi x^2 + (100-4x)^2-650\cdot 4\pi =0$

if we expand $(100-4x)^2$ and then group each term of x, x^2 we get :

$(4\pi+16)x^2-800x+100^2-650\cdot 4\pi =0$

Recall that given a cuadractic equation of the form $ax^2+bx+c=0$ the solution is given by

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In this case, the existence of a real solution is given by the expression $b^2-4ac$. For it to have a real solution, it must happen that $b^2-4ac\geq 0$

In this case, b=-800, c = 100^2-650\cdot 4\pi, a = 4\pi+16[/tex]

For this values, we get that $b^2-4ac = 430681\geq 0$, so the is a solution for our problem.

In this case,

$x = \frac{-b - \sqrt[]{b^2-4ac}}{2a}= 2.515834$

and $x = \frac{-b + \sqrt[]{b^2-4ac}}{2a}= 25.4891$ are the two possible solutions for this problem. We can check that this values of x fulfill our restrictions. Note that if x=25.4891, by replacing in the value of r, we get that r=-0.3113853. Since r is the radius of the circle, it must be positive. Hence, x=25.4891 is discarted.For x=2.515834 we get a value of r=14.31386, so this is the answer