Question

exercise 2
with a wooden ruler, you measure the length of a rectangular piece of sheet metal to be 12 mm. with micrometer calipers, you measure the width of the rectangle to be 5.98 mm. use the correct number of significant figures: what are (a) the area of the rectangle; (b) the ratio of the rectangle’s width to its length; (c) the perimeter of the rectangle; (d) the difference between the length and the width; and (e) the ratio of the length to the width?

Explanation:

Step1: Recall significant - figure rules

When multiplying or dividing, the result has the same number of significant figures as the number with the fewest significant figures in the values being multiplied or divided. When adding or subtracting, the result has the same number of decimal places as the number with the fewest decimal places in the values being added or subtracted. Let length $l = 12$ mm (2 significant figures) and width $w=5.98$ mm (3 significant figures).

Step2: Calculate the area $A$

The area of a rectangle is $A = l\times w$. So $A=12\times5.98 = 71.76$ mm². Since $l$ has 2 significant figures, $A = 72$ mm².

Step3: Calculate the ratio of width to length $\frac{w}{l}$

$\frac{w}{l}=\frac{5.98}{12}\approx0.4983$. Rounding to 2 significant figures, $\frac{w}{l}=0.50$.

Step4: Calculate the perimeter $P$

The perimeter of a rectangle is $P = 2(l + w)=2(12 + 5.98)=2\times17.98 = 35.96$ mm. Since $l$ has 0 decimal - places, $P = 36$ mm.

Step5: Calculate the difference between length and width $d$

$d=l - w=12-5.98 = 6.02$ mm. Since $l$ has 0 decimal - places, $d = 6.0$ mm.

Step6: Calculate the ratio of length to width $\frac{l}{w}$

$\frac{l}{w}=\frac{12}{5.98}\approx2.0067$. Rounding to 2 significant figures, $\frac{l}{w}=2.0$.

Answer:

(a) $72$ mm²
(b) $0.50$
(c) $36$ mm
(d) $6.0$ mm
(e) $2.0$