Let the width of the rectangle be \( w \) cm. Then the length \( l \) can be expressed as:
\[ l = 2w + 3 \]
The area of the rectangle is given by the formula:
\[ \text{Area} = l \times w \]
Substituting the expression for the length into the area formula, we have:
\[ (2w + 3) \times w = 90 \]
This simplifies to:
\[ 2w^2 + 3w = 90 \]
Rearranging the equation gives:
\[ 2w^2 + 3w - 90 = 0 \]
Now, we will use the quadratic formula to solve for \( w \), where \( a = 2 \), \( b = 3 \), and \( c = -90 \):
\[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Calculating the discriminant:
\[ b^2 - 4ac = 3^2 - 4 \times 2 \times (-90) = 9 + 720 = 729 \]
Now substituting the values into the quadratic formula:
\[ w = \frac{-3 \pm \sqrt{729}}{2 \times 2} = \frac{-3 \pm 27}{4} \]
Calculating the two possible values for \( w \):
- \( w = \frac{24}{4} = 6 \)
- \( w = \frac{-30}{4} = -7.5 \) (not valid since width cannot be negative)
Thus, the width is
\[ w = 6 , \text{cm} \]
Now substitute back to find the length:
\[ l = 2w + 3 = 2(6) + 3 = 12 + 3 = 15 , \text{cm} \]
Now we can calculate the perimeter of the rectangle using the formula for perimeter:
\[ \text{Perimeter} = 2(l + w) \]
Substituting the values of \( l \) and \( w \):
\[ \text{Perimeter} = 2(15 + 6) = 2(21) = 42 , \text{cm} \]
Thus, the perimeter of the rectangle is
\[ \boxed{42} , \text{cm} \]
0 answers