Given:
The length of the rectangle, l=2.3±0.2 m
The width of the rectangle, w=1.2±0.2 m
To find:
The area and perimeter of the rectangle and the uncertainty in each value.
Step-by-step explanation:
The area of the rectangle is given by,

On substituting the known values,

The perimeter of the rectangle is given by,

On substituting the known values,
![\begin{gathered} P=2[(2.3\pm0.2)+(1.2\pm0.2)] \\ =2[3.5\pm(0.2+0.2)] \\ =2[3.5\pm0.4] \\ =(7.0\pm0.8)\text{ m} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/zibbgu4csh3vwv9m5rnvhf2s0v1o9v58dl.png)
Final answer:
Thus the area of the rectangle is (2.8±0.3) m².
The perimeter of the rectangle is (7.0±0.8) m.