Question

the base of a triangle is fixed at 2.218 millimeters. determine the number of significant figures of the area of the triangle with a height of .186 mm.

Explanation:

Step1: Recall the formula for the area of a triangle

The formula for the area of a triangle is \( A=\frac{1}{2}\times b\times h \), where \( b \) is the base and \( h \) is the height. Here, \( b = 2.218\space mm \) and \( h=0.186\space mm \) (assuming the height is \( 0.186\space mm \) as the text seems to have a typo, maybe a decimal point missing before 186).

Step2: Calculate the area

Substitute the values into the formula: \( A=\frac{1}{2}\times2.218\times0.186 \). First, calculate \( 2.218\times0.186 \): \( 2.218\times0.186 = 0.412548 \). Then multiply by \( \frac{1}{2} \): \( A = 0.206274 \).

Step3: Determine significant figures

- For multiplication/division, the result should have the same number of significant figures as the least precise measurement.
- The base \( 2.218 \) has 4 significant figures, and the height (assuming \( 0.186 \)) has 3 significant figures.
- So the area should be reported with 3 significant figures. But wait, maybe the height is \( 1.86\space mm \) (if the decimal was misplaced). Let's recalculate with \( h = 1.86\space mm \). Then \( A=\frac{1}{2}\times2.218\times1.86 \). First, \( 2.218\times1.86=4.12548 \), then \( \frac{1}{2}\times4.12548 = 2.06274 \). Now, base has 4 sig figs, height has 3 sig figs. So the area should have 3 sig figs. Wait, but the question is about the number of significant figures of the area. Wait, maybe the height is \( 186\space mm \) but that would be unusual. Wait, the original problem says "height of 186 mm" but that seems too large, probably a decimal error. Assuming the height is \( 0.186\space mm \) (3 sig figs) and base is \( 2.218\space mm \) (4 sig figs). When multiplying/dividing, the result's sig figs are determined by the least number of sig figs in the inputs. So the height has 3 sig figs, base has 4. So the area will have 3 sig figs. But let's check the calculation again. Wait, maybe the height is \( 1.86\space mm \) (3 sig figs). Then \( A=\frac{1}{2}\times2.218\times1.86 \). \( 2.218\times1.86 = 4.12548 \), half of that is \( 2.06274 \). Now, the number of significant figures: the rule is that for multiplication/division, the result should have the same number of significant figures as the quantity with the least number of significant figures. The base (2.218) has 4, the height (1.86) has 3. So the area should have 3 significant figures. But wait, maybe the problem is just about the significant figures in the calculation. Wait, maybe the height is \( 186\space mm \) (but that would be 3 sig figs as well, 1, 8, 6). Wait, no, 186 has 3 significant figures. The base 2.218 has 4. So when we calculate \( \frac{1}{2}\times2.218\times186 \) (if height is 186), first \( 2.218\times186 = 412.548 \), half of that is 206.274. Now, 2.218 has 4, 186 has 3. So the result should have 3 significant figures. But the key is: when multiplying or dividing, the number of significant figures in the result is equal to the number of significant figures in the least precise measurement. So regardless of the height (assuming it's a number with 3 significant figures, and base with 4), the area will have 3 significant figures? Wait, no, maybe the height is \( 0.186\space mm \) (3 sig figs), base \( 2.218\space mm \) (4 sig figs). Then the product \( 2.218\times0.186 \) has 3 sig figs (because 0.186 has 3), then times 1/2 (which is exact, so no effect on sig figs) so the area has 3 sig figs. Wait, but maybe the problem is that the height is \( 1.86\space mm \) (3 sig figs) and base \( 2.218\space mm \) (4 sig figs). Then the area is \( \frac{1}{2}\times2.218\times1.86 = 2.06274 \), which should be rounded to 3 sig figs, so 2.06 (wait, no, 2.06 has 3 sig figs? Wait, 2.06274 rounded to 3 sig figs is 2.06? Wait, no, the first three significant figures are 2, 0, 6? Wait, no, 2.06274: the first significant figure is 2, second 0? No, wait, significant figures: all non - zero digits are significant, and zeros between non - zero digits are significant. So 2.06274 has significant figures 2, 0, 6, 2, 7, 4? No, wait, 2.06274: the number is 2.06274. The first significant figure is 2, second is 0? No, 0 is between 2 and 6, so it is significant. Wait, no, 2.06274: the significant figures are 2, 0, 6, 2, 7, 4? No, that's not right. Wait, significant figures: for a number like \( a.bcde \), all digits are significant. For \( 2.218 \), significant figures are 2, 2, 1, 8 (4 sig figs). For \( 1.86 \), significant figures are 1, 8, 6 (3 sig figs). When multiplying \( 2.218\times1.86 \), we get \( 4.12548 \). The number of significant figures in the product is determined by the least number of significant figures in the factors, which is 3 (from 1.86). Then we multiply by \( \frac{1}{2} \) (which is a exact value, so it doesn't change the number of significant figures). So \( 4.12548\times\frac{1}{2}=2.06274 \). Now, we round this to 3 significant figures. The first three significant figures are 2, 0, 6? Wait, no, 2.06274: the first significant figure is 2, the second is 0 (because it's between 2 and 6, so it's significant), the third is 6. Wait, but 2.06274 rounded to 3 significant figures is 2.06? Wait, no, 2.06274: look at the fourth significant figure, which is 2. Since 2 < 5, we round down. So 2.06 (3 significant figures). Wait, but maybe the height is \( 186\space mm \) (3 sig figs) and base \( 2.218\space mm \) (4 sig figs). Then \( A=\frac{1}{2}\times2.218\times186 \). \( 2.218\times186 = 412.548 \), half of that is 206.274. Now, 186 has 3 sig figs, 2.218 has 4. So the product \( 2.218\times186 \) should have 3 sig figs. 412.548 rounded to 3 sig figs is 413? Wait, no, 412.548: the first three significant figures are 4, 1, 2. The fourth is 5, so we round up the third digit: 413. Then half of 413 is 206.5, which is 207 when rounded to 3 sig figs? Wait, this is getting confusing. Maybe the problem is that the height is \( 1.86\space mm \) (3 sig figs) and base \( 2.218\space mm \) (4 sig figs). Then the area is \( \frac{1}{2}\times2.218\times1.86 = 2.06274 \), which has 3 significant figures (because the least number of significant figures in the inputs is 3).

Answer:

The number of significant figures of the area is 3.