A circle is inscribed in a square as shown - Leaving Cert Mathematics - Question 1 - 2010

The area of the square can be calculated using the formula:

$$ext{Area} = ext{side length}^2 = (18)^2 = 324 ext{ cm}^2.$$

With the field dimensions as follows: 7.5 m, 6 m, 8.5 m, 7 m, 4.5 m, 9.5 m over intervals of 5 m, we apply Simpson's rule:

ext{Area} ext{ (estimate)} = rac{h}{3} [f_0 + 4f_1 + 2f_2 + 4f_3 + 2f_4 + f_5]

Where:

• $h = 5$ m (the distance between intervals)
• $f_0 = 7.5, f_1 = 6, f_2 = 8.5, f_3 = 7, f_4 = 4.5, f_5 = 9.5$.

Plugging in these values:

ext{Area} = rac{5}{3} [7.5 + 4(6) + 2(8.5) + 4(7) + 2(4.5) + 9.5] = rac{5}{3} [7.5 + 24 + 17 + 28 + 9 + 9.5] = rac{5}{3} [95] = 158.33 ext{ m}^2.

To find the percentage error, use the formula:

ext{Percentage Error} = rac{| ext{Estimated Area} - ext{Actual Area}|}{ ext{Actual Area}} imes 100 ext{.}

Substituting the values:

ext{Percentage Error} = rac{| 158.33 - 200 |}{200} imes 100 = rac{41.67}{200} imes 100 = 20.835 ext{ \\%}.

The volume of the sphere is calculated using the formula:

V = rac{4}{3} imes ext{π} imes r^3

Where the radius $r$ is 4.5 cm (since the diameter is 9 cm). Thus,

V = rac{4}{3} imes ext{π} imes (4.5)^3 = rac{4}{3} imes ext{π} imes 91.125 = 121.5 ext{ } ext{cm}^3.

To find the radius $r$ of the cone and cylinder combined, we first equate the volume of the sphere to the combined volume of the cone and cylinder:

$$V_{ ext{cylinder}} + V_{ ext{cone}} = ext{Volume of sphere}.$$

The cylinder has a height of 8 cm and a radius $r$, while the cone has a height of 8 cm and a radius $r$. Thus:

$$V_{ ext{cylinder}} = ext{π}r^2 imes 8,$$ V_{ ext{cone}} = rac{1}{3} imes ext{π}r^2 imes 8.

Setting up the equation:

ext{π}r^2 (8 + rac{8}{3}) = 121.5 ext{ } ext{cm}^3.

Solving for $r$ gives us:

ext{π}r^2 imes rac{32}{3} = 121.5 ightarrow r^2 = rac{121.5 imes 3}{32 ext{π}} ightarrow r ≈ 3.4 ext{ cm}.