To find the dimensions of the rectangular poster, we can use the given perimeter and area measurements.
First, let's recall the formulas for the perimeter and the area of a rectangle:
The perimeter (P) of a rectangle is calculated by adding together the lengths of all four sides. Since opposite sides of a rectangle are equal, the perimeter can be expressed as P = 2 * (length + width). Given that the perimeter is 76 inches, we can set up an equation as follows:
\[ 2 \cdot (length + width) = 76 \]
Dividing both sides by 2 gives us the sum of the length and width:
\[ length + width = 38 \] (1)
The area (A) of a rectangle is calculated as the product of its length and its width, so A = length * width. Given that the area is 352 square inches, we set up another equation:
\[ length \cdot width = 352 \] (2)
Now, we have a system of two equations with two unknowns (length and width). We can use equation (1) to express one variable in terms of the other and then substitute it into equation (2). Let's solve for length in terms of width using equation (1):
\[ length = 38 - width \]
Now we can substitute this expression into equation (2) to find the width:
\[ (38 - width) \cdot width = 352 \]
Expanding the left side, we get a quadratic equation:
\[ 38 \cdot width - width^2 = 352 \]
Rearranging terms, we bring them all to one side to set the equation to zero:
\[ width^2 - 38 \cdot width + 352 = 0 \]
To solve this quadratic equation, we can factor it, or we can use the quadratic formula. Since it's not obvious how to factor this, let's use the quadratic formula:
\[ width = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, a = 1, b = -38, and c = 352. Now let's calculate the discriminant (b² - 4ac) to see if the equation has real solutions:
\[ discriminant = (-38)^2 - 4 \cdot 1 \cdot 352 = 1444 - 1408 = 36 \]
Since the discriminant is positive, we have two real solutions. We calculate the two potential widths:
\[ width = \frac{-(-38) \pm \sqrt{36}}{2 \cdot 1} \]
\[ width = \frac{38 \pm 6}{2} \]
\[ width = \frac{38 + 6}{2} \; or \; \frac{38 - 6}{2} \]
\[ width = \frac{44}{2} \; or \; \frac{32}{2} \]
\[ width = 22 \; or \; 16 \]
Now that we have two possible widths, we can find the corresponding lengths using the relation length = 38 - width:
For width = 22 inches:
\[ length = 38 - 22 = 16 \; inches \]
For width = 16 inches:
\[ length = 38 - 16 = 22 \; inches \]
Therefore, the dimensions of the poster are either 16 inches by 22 inches or 22 inches by 16 inches. These are essentially the same because a rectangle's length and width can be interchanged. Option (d) 16 in. by 22 in. is the correct choice.
