14. (3 points) After the length and width of a rectangle are both increased by 3 cm, the area increases by 90 square cm. Then the perimeter of the original rectangle is $\qquad$ cm.
14. (3 points) After the length and width of a rectangle are both increased by 3 cm, the area increases by 90 square cm. Then the perimeter of the original rectangle is $\qquad$ cm.

【Analysis】According to the problem, the length and width are each increased by 3 cm, and the area of this rectangle increases by 90 square meters (as shown in the figure below):
The increased area includes 2 parts: the lower rectangle, with an area of 3(3+a), and the small rectangle at the upper right, with an area of $3 b$. Their sum is $3(3+a)+3 b$ equals 90, to find the sum of $a+b$, and then use $(a+b) \times 2$ to calculate.

【Solution】Solution: Let the original length of the rectangle be $a$ cm, and the width be $b$ cm. The current length is $a+3$ cm, and the width is $b+3$ cm, as shown in the figure below:
$$
\begin{array}{l}
3(3+a)+3 b=90 \\
9+3 a+3 b=90 \\
a+b=27 \\
27 \times 2=54 \text { (square cm) }
\end{array}
$$

Answer: The perimeter of the original rectangle is 54 cm.
【Comment】This problem mainly examines the calculation of the perimeter of a rectangle, the key is to find the sum of the original length and width of the rectangle, and then solve according to the formula of the rectangle.