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 the perimeter of the rectangle with the uncertainty in each value.
Explanation:
The area of a rectangle is given by,
[tex]A=l\times w[/tex]
On substituting the known values,
[tex]\begin{gathered} A=(2.3\pm0.2)\times(1.2\pm0.2) \\ =(2.3\times1.2)\pm(\frac{0.2}{2.3}+\frac{0.2}{1.2}) \\ =2.8\pm0.3\text{ m}^2 \end{gathered}[/tex]
The perimeter of the rectangle is given by,
[tex]P=2(l+w)[/tex]
On substituting the known values,
[tex]\begin{gathered} P=2[(2.3\pm0.2)+(1.2\pm0.2)] \\ =2[(2.3+1.2)\pm(0.2+0.2)] \\ =2[3.5\pm0.4] \\ =(7.0\pm0.8)\text{ m} \end{gathered}[/tex]
Final answer:
The area of the rectangle is 2.8±0.3 m²
The perimeter of the rectangle is 7.0±0.8 m
