Asquare is circumscribed about a circle with an 8-inch radius, as shown below. what is the area, in square inches, of the region that is inside the square but outside the circle?

choices:
16 − 64π

256−8π

64 −16 π

256 −64π

Option 4. 256 - 64π is the correct option.

Step-by-step explanation:

In the given picture a square is circumscribed about a circle with a side = 2r

where r is the radius of the circle.

Therefore area of the square = (2r)² = 4r² = 4 × 8² = 4 × 64 = 256 in²

Now area of the circle = πr² = π × 8² = 64π in²

Now area of region that is inside the square and outside the circle =

Area of square - area of circle = (256 - 64π) in².

Therefore the answer is (256 - 64π).

step-by-step explanation:

equation

you are asked to write the equation

  expression = 0

using the given expression. when you do that, you have

  2x^2-2x-12 = 0

factoring

you can see that each of the coefficients is even, so a factor of 2 can be removed immediately.

  2(x^2 -x -6) = 0

if you multiply the binomials (x+a)(x+b), you get

  (x +a)(x +b) = x^2 +(a+b)x +ab

this tells you the coefficient of x is the sum of two factors of the constant term. here, that means -1 is the sum of two factors of -6. we want to know those factors so we can factor the trinomial.

  -6 = -1×6 = -2×3 = -3×2 = -6×1

the factor pair that sums to -1 is {-3, 2}, so our factorization is

  2(x -3)(x +2) = 0

solutions

the solutions to this equation are the values of x that make the product zero. a product is zero only when only of the factors of the product is zero. here, that means

  x - 3 = 0

  x = 3 . . . add 3 to both sides of the equation

or

  x +2 = 0

  x = -2 . . . add -2 to both sides of the equation

the solutions to the equation are x=3 or x=-2. both of these values of x satisfy the equation—they make it be true.