QUESTION IMAGE
Question
- the table shows the diameters of 4 different coins.
a. complete the table.
b. to determine how much metal is on one face of a coin,
it is more useful to use the area rather than the
circumference. explain why this is the case.
Step1: Recall the formula for the area of a circle
The area of a circle is given by the formula \(A = \pi r^{2}\), where \(r\) is the radius of the circle. Since \(r=\frac{d}{2}\) (where \(d\) is the diameter), we can also write \(A=\pi(\frac{d}{2})^{2}=\frac{\pi d^{2}}{4}\)
Step2: Calculate the area for the Nickel
For the Nickel with \(d = 2.1\) cm.
Step3: Calculate the area for the Dime
For the Dime with \(d = 1.8\) cm.
Step4: Calculate the area for the Quarter
For the Quarter with \(d = 2.4\) cm.
Step5: Explain why area is more useful than circumference
The amount of metal on one face of a coin is related to the surface - area. The circumference is a one - dimensional measure (length around the circle), while the area is a two - dimensional measure (space covered by the circular face). Since the metal covers the two - dimensional face of the coin, the area (which measures the two - dimensional space) is more relevant than the circumference (which measures a one - dimensional length around the edge)
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a.
| Coin | Diameter (cm) | Area (sq.cm) |
|---|---|---|
| Nickel | \(2.1\) | \(\frac{\pi\times(2.1)^{2}}{4}\approx 3.46\) |
| Dime | \(1.8\) | \(\frac{\pi\times(1.8)^{2}}{4}\approx 2.54\) |
| Quarter | \(2.4\) | \(\frac{\pi\times(2.4)^{2}}{4}\approx 4.52\) |
b. The amount of metal on one face of a coin is related to the two - dimensional space it covers. The circumference is a one - dimensional measure (length), while the area is a two - dimensional measure (space). Since the metal covers the face (a two - dimensional region), the area is more useful.