Question

A rectangle has a length of (2.3 +0.2) m and a width of (1.2 + 0.2) m. Calculate the area and the perimeter of the rectangle, andgive the uncertainty in each value

Answer 1

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.

Explanation:

The area of the rectangle is given by,

$A=l\times w$

On substituting the known values,

$\begin{gathered} A=(2.3\pm0.2)\times(1.2\pm0.2) \\ =(2.3\times1.2)\pm(\frac{0.2}{2.3}+\frac{0.2}{1.2}) \\ =(2.8\pm0.3)\text{ m}^2 \end{gathered}$

The perimeter of the rectangle is given by,

$P=2(l+w)$

On substituting the known values,

$\begin{gathered} P=2[(2.3\pm0.2)+(1.2\pm0.2)] \\ =2[3.5\pm(0.2+0.2)] \\ =2[3.5\pm0.4] \\ =(7.0\pm0.8)\text{ m} \end{gathered}$

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.