Mensuration: Area and Perimeter - Interpret pi as irrational and use approximations for accurate computation

A circular wire has a radius of $14\text{ cm}$. Find its circumference using $\pi = \frac{22}{7}$.

$C = 2\pi r$ $C = 2 \times \frac{22}{7} \times 14$ $C = 2 \times 22 \times 2$ $C = 88\text{ cm}$

Since the radius $14$ is a multiple of $7$, using the approximation $\frac{22}{7}$ allows us to cancel the denominator, resulting in an accurate and quick computation.

Find the area of a circle whose radius is $10\text{ cm}$. (Use $\pi = 3.14$)

$A = \pi r^2$ $A = 3.14 \times (10)^2$ $A = 3.14 \times 100$ $A = 314\text{ cm}^2$

When the radius is a power of $10$, using the decimal approximation $3.14$ is more convenient for calculation than using the fraction $\frac{22}{7}$.

The circumference of a circular sheet is $154\text{ m}$. Find its radius and area. (Use $\pi = \frac{22}{7}$)

First, find the radius $r$: $2\pi r = 154$ $2 \times \frac{22}{7} \times r = 154$ $r = \frac{154 \times 7}{2 \times 22}$ $r = \frac{7 \times 7}{2} = 24.5\text{ m}$ Now, find the area $A$: $A = \pi r^2$ $A = \frac{22}{7} \times 24.5 \times 24.5$ $A = \frac{22}{7} \times \frac{49}{2} \times \frac{49}{2}$ $A = 11 \times 7 \times \frac{49}{4} = \frac{3773}{2} = 1886.5\text{ m}^2$

We first use the circumference formula to isolate $r$. Then, we substitute the value of $r$ into the area formula to find the total surface region.

1. Inner radius ($r$) = $21\text{ m}$
2. Outer radius ($R$) = $21 + 7 = 28\text{ m}$
3. Area of path = Outer Area - Inner Area
4. $A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$
5. $A = \frac{22}{7} \times (28^2 - 21^2)$
6. $A = \frac{22}{7} \times (28 - 21)(28 + 21)$
7. $A = \frac{22}{7} \times 7 \times 49 = 22 \times 49 = 1078\text{ m}^2$

To find the area of a ring (annulus), we subtract the area of the smaller circle from the larger circle. Using the identity $a^2 - b^2 = (a-b)(a+b)$ simplifies the calculation when using $\pi = \frac{22}{7}$.

1. Diameter ($d$) = $14\text{ cm}$
2. Radius ($r$) = $\frac{14}{2} = 7\text{ cm}$
3. Perimeter = Curved Arc + Diameter
4. $P = \pi r + d$
5. $P = (\frac{22}{7} \times 7) + 14$
6. $P = 22 + 14 = 36\text{ cm}$

The perimeter of a semi-circular object like a protractor consists of the semi-circular arc length ($\frac{1}{2} \times 2\pi r$) plus the straight base which is the diameter.