Circles: Additional Practice (Set 2)

The perimeter of a semicircle is the curved half of the circumference plus the straight diameter.

The curved part is $\pi r = 3.14 \times 10 = 31.4$ cm. The straight diameter is $d = 2r = 20$ cm. Add them:

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

From $C = \pi d$, divide the circumference by $\pi$.

$d = \frac{C}{\pi} = \frac{18.84}{3.14} = 6 \text{ cm}$

Start from $C = 2\pi r$ and solve for $r$ by dividing by $2\pi$.

First $2\pi \approx 2 \times 3.14 = 6.28$, then

$r = \frac{C}{2\pi} = \frac{43.96}{6.28} = 7 \text{ m}$

From $A = \pi r^2$, divide the area by $\pi$ to get $r^2$.

$r^2 = \frac{78.5}{3.14} = 25$

Then take the square root:

$r = \sqrt{25} = 5 \text{ cm}$

Large area: $A = 3.14 \times 8^2 = 3.14 \times 64 = 200.96 \text{ cm}^2$. Small area: $A = 3.14 \times 4^2 = 3.14 \times 16 = 50.24 \text{ cm}^2$.

$\frac{200.96}{50.24} = 4$

The radius doubled, and area depends on the radius squared, so the area grew by $2^2 = 4$ times.

Subtract the fountain's area from the whole garden's area.

$A_{\text{garden}} = 3.14 \times 10^2 = 314 \text{ m}^2$

$A_{\text{fountain}} = 3.14 \times 2^2 = 12.56 \text{ m}^2$

$314 - 12.56 = 301.44 \text{ m}^2$

Start with the rectangle, then subtract the semicircle that has been cut out.

Rectangle area:

$A_{\text{rect}} = 12 \times 6 = 72 \text{ cm}^2$

The removed semicircle has diameter $6$ cm, so radius $r = 3$ cm:

$A_{\text{semi}} = \frac{1}{2} \times 3.14 \times 3^2 = \frac{1}{2} \times 28.26 = 14.13 \text{ cm}^2$

Subtract:

$72 - 14.13 = 57.87 \text{ cm}^2$

Halve the diameter for the radius: $r = 7$ in. The whole pizza's area with $\pi \approx \tfrac{22}{7}$ is

$A = \frac{22}{7} \times 7^2 = 22 \times 7 = 154 \text{ in}^2.$

One of four equal slices is

$\frac{154}{4} = 38.5 \text{ in}^2$

First find the distance around, the circumference, using $C = \pi d$.

$C = 3.14 \times 100 = 314 \text{ m}$

Then divide the distance by the walking speed:

$\frac{314 \text{ m}}{50 \text{ m per minute}} = 6.28 \text{ minutes}$

A semicircle's boundary is the curved half of the circumference, $\tfrac{1}{2}(2\pi r) = \pi r$, plus the straight diameter, $2r$.

$P = \pi r + 2r$

Using only $\pi r$ forgets the straight edge, and $\pi r^2$ is an area, not a length.

The grazing area is a circle whose radius is the rope length. Doubling the radius from $6$ m to $12$ m multiplies the area by $2^2 = 4$.

Check: $3.14 \times 6^2 = 113.04 \text{ m}^2$ and $3.14 \times 12^2 = 452.16 \text{ m}^2$, and

$\frac{452.16}{113.04} = 4.$

The square has area $144 \text{ cm}^2$, so its side is $\sqrt{144} = 12$ cm. The inscribed circle touches all four sides, so its diameter equals the side, $12$ cm.

Use $C = \pi d$:

$C = 3.14 \times 12 = 37.68 \text{ cm}$