Question

9
for each circle, work out
i the area
ii the circumference.
give your answers correct to one decimal place (1 d.p.).
a radius = 2.7 cm
b diameter = 45 mm
10
for each semicircle, work out
i the area
ii the perimeter.
give your answers correct to one decimal place (1 d.p.).
a radius = 8.5 cm
b radius = 24 mm
c diameter = 32 cm
d diameter = 15 m

Explanation:

Problem 9 (Circles)

Part i (Area)

Step 1: Recall the formula for the area of a circle

The area \( A \) of a circle is given by \( A = \pi r^2 \), where \( r \) is the radius. If the diameter \( d \) is given, the radius \( r=\frac{d}{2} \).

Step 2: Calculate for 9a (radius = 2.7 cm)

Substitute \( r = 2.7 \) into the formula:
\( A=\pi\times(2.7)^2=\pi\times7.29\approx23.0907\approx23.1 \, \text{cm}^2 \) (to 1 d.p.)

Step 3: Calculate for 9b (diameter = 45 mm)

First, find the radius: \( r=\frac{45}{2}=22.5 \, \text{mm} \)
Then, area: \( A=\pi\times(22.5)^2=\pi\times506.25\approx1590.431\approx1590.4 \, \text{mm}^2 \) (to 1 d.p.)

Part ii (Circumference)

Step 1: Recall the formula for the circumference of a circle

The circumference \( C \) of a circle is \( C = 2\pi r \) or \( C=\pi d \) (where \( d \) is the diameter).

Step 2: Calculate for 9a (radius = 2.7 cm)

Using \( C = 2\pi r \):
\( C = 2\times\pi\times2.7 = 5.4\pi\approx16.9646\approx17.0 \, \text{cm} \) (to 1 d.p.)

Step 3: Calculate for 9b (diameter = 45 mm)

Using \( C=\pi d \):
\( C=\pi\times45\approx141.3717\approx141.4 \, \text{mm} \) (to 1 d.p.)

Problem 10 (Semicircles)

Part i (Area)

Step 1: Recall the formula for the area of a semicircle

The area of a semicircle is half the area of a full circle: \( A=\frac{1}{2}\pi r^2 \). If diameter \( d \) is given, \( r=\frac{d}{2} \).

Step 2: Calculate for 10a (radius = 8.5 cm)

Substitute \( r = 8.5 \):
\( A=\frac{1}{2}\times\pi\times(8.5)^2=\frac{1}{2}\times\pi\times72.25\approx113.490\approx113.5 \, \text{cm}^2 \) (to 1 d.p.)

Step 3: Calculate for 10c (diameter = 32 cm)

Radius \( r=\frac{32}{2}=16 \, \text{cm} \)
Area: \( A=\frac{1}{2}\times\pi\times(16)^2=\frac{1}{2}\times\pi\times256\approx402.124\approx402.1 \, \text{cm}^2 \) (to 1 d.p.)

Step 4: Calculate for 10b (radius = 24 mm)

\( A=\frac{1}{2}\times\pi\times(24)^2=\frac{1}{2}\times\pi\times576\approx904.779\approx904.8 \, \text{mm}^2 \) (to 1 d.p.)

Step 5: Calculate for 10d (diameter = 15 m)

Radius \( r=\frac{15}{2}=7.5 \, \text{m} \)
Area: \( A=\frac{1}{2}\times\pi\times(7.5)^2=\frac{1}{2}\times\pi\times56.25\approx88.357\approx88.4 \, \text{m}^2 \) (to 1 d.p.)

Part ii (Perimeter)

Step 1: Recall the formula for the perimeter of a semicircle

The perimeter of a semicircle includes the curved part (half the circumference) and the diameter: \( P=\pi r + d \) (or \( P=\pi r + 2r \) since \( d = 2r \)).

Step 2: Calculate for 10a (radius = 8.5 cm)

\( P=\pi\times8.5 + 2\times8.5\approx26.7035 + 17 = 43.7035\approx43.7 \, \text{cm} \) (to 1 d.p.)

Step 3: Calculate for 10b (radius = 24 mm)

\( P=\pi\times24 + 2\times24\approx75.398 + 48 = 123.398\approx123.4 \, \text{mm} \) (to 1 d.p.)

Step 4: Calculate for 10c (diameter = 32 cm, so radius = 16 cm)

\( P=\pi\times16 + 32\approx50.265 + 32 = 82.265\approx82.3 \, \text{cm} \) (to 1 d.p.)

Step 5: Calculate for 10d (diameter = 15 m, so radius = 7.5 m)

\( P=\pi\times7.5 + 15\approx23.5619 + 15 = 38.5619\approx38.6 \, \text{m} \) (to 1 d.p.)

Answer:

s (Summarized):

Problem 9

i. Area:

• 9a: \( \boldsymbol{23.1 \, \text{cm}^2} \)
• 9b: \( \boldsymbol{1590.4 \, \text{mm}^2} \)

ii. Circumference:

• 9a: \( \boldsymbol{17.0 \, \text{cm}} \)
• 9b: \( \boldsymbol{141.4 \, \text{mm}} \)

Problem 10

i. Area:

• 10a: \( \boldsymbol{113.5 \, \text{cm}^2} \)
• 10b: \( \boldsymbol{904.8 \, \text{mm}^2} \)
• 10c: \( \boldsymbol{402.1 \, \text{cm}^2} \)
• 10d: \( \boldsymbol{88.4 \, \text{m}^2} \)

ii. Perimeter:

• 10a: \( \boldsymbol{43.7 \, \text{cm}} \)
• 10b: \( \boldsymbol{123.4 \, \text{mm}} \)
• 10c: \( \boldsymbol{82.3 \, \text{cm}} \)
• 10d: \( \boldsymbol{38.6 \, \text{m}} \)