Circles: Additional Practice (Set 1)

Use $C = 2\pi r$ with $r = 0.5$ m. Multiply $2 \times 0.5 = 1$ first, which leaves just $\pi$.

$C = 2 \times 3.14 \times 0.5 = 3.14 \times 1 = 3.14 \text{ m}$

The diameter is $20$ cm, so the radius is $r = 10$ cm. The curved edge is $\pi r = 3.14 \times 10 = 31.4$ cm, and the straight edge is the diameter, $20$ cm.

$P = 31.4 + 20 = 51.4 \text{ cm}$

Add the square and the semicircle. The square area is $8 \times 8 = 64 \text{ cm}^2$.

The semicircle's diameter is $8$ cm, so its radius is $r = 4$ cm:

$A_{\text{semi}} = \frac{1}{2} \times 3.14 \times 4^2 = \frac{1}{2} \times 50.24 = 25.12 \text{ cm}^2$

Total:

$A = 64 + 25.12 = 89.12 \text{ cm}^2$

Use $A = \pi r^2$. Square the radius first.

$r^2 = 15^2 = 225$

Then multiply by $\pi$:

$A = 3.14 \times 225 = 706.5 \text{ cm}^2$

The outer radius is $7 + 1 = 8$ m and the inner radius is $7$ m. Subtract the inner circle from the outer circle.

$A_{\text{outer}} = 3.14 \times 8^2 = 200.96 \text{ m}^2$

$A_{\text{inner}} = 3.14 \times 7^2 = 153.86 \text{ m}^2$

$200.96 - 153.86 = 47.1 \text{ m}^2$

One turn covers one circumference. With $\pi \approx \tfrac{22}{7}$ and $d = 70$:

$C = \pi d = \frac{22}{7} \times 70 = 22 \times 10 = 220 \text{ cm}$

Divide the distance by the circumference:

$\frac{1100}{220} = 5 \text{ turns}$

The inscribed circle's diameter equals the side, $14$ cm, so the radius is $r = 7$ cm. With $\pi \approx \tfrac{22}{7}$:

$A = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154 \text{ cm}^2$

The ring follows the circumference. With the diameter given, use $C = \pi d$.

$C = 3.14 \times 6 = 18.84 \text{ m}$

The flat edge is the diameter, so the radius is $r = 5$ cm. A semicircle is half a circle.

$A = \frac{1}{2}\pi r^2 = \frac{1}{2} \times 3.14 \times 5^2 = \frac{1}{2} \times 78.5 = 39.25 \text{ cm}^2$

Replacing $r$ with $\tfrac{1}{2}r$ in $A = \pi r^2$ gives

$A_{\text{new}} = \pi \left(\tfrac{1}{2}r\right)^2 = \pi \times \tfrac{1}{4} r^2 = \tfrac{1}{4}\pi r^2.$

The area becomes one fourth as large, because area depends on the radius squared and $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$.

Use $C = 2\pi r \approx 6r$ and solve for the radius.

$30 \approx 6r \;\Rightarrow\; r \approx 5 \text{ cm}$

Check with $\pi \approx 3.14$: $C = 2 \times 3.14 \times 5 = 31.4$ cm, close to $30$ cm.

The square has the same area, $314 \text{ cm}^2$, and a square's area is side squared, so the side is the square root of the area.

$s = \sqrt{314} \approx 17.7 \text{ cm} \approx 18 \text{ cm}$

Since $17^2 = 289$ and $18^2 = 324$, the side is just under $18$ cm.