Question

1. a solid is made up of a conical frustum and a hemispherical top as shown in the figure below. the dimensions are as indicated in the figure.

(a) find the area of:
(i) the circular base
(ii) the curved surface of the frustum
(iii) the hemispherical surface.
(b) a similar solid has a total surface area of 81.51 cm². determine the radius of its base.
2000 pp1 no. 20

Explanation:

Part (a)

(i) Area of the circular base

The formula for the area of a circle is \( A = \pi r^2 \), where \( r = 3.5 \, \text{cm} \).

Step 1: Substitute \( r = 3.5 \) into the formula

\( A = \pi \times (3.5)^2 \)

Step 2: Calculate \( (3.5)^2 = 12.25 \)

\( A = \pi \times 12.25 \approx 3.1416 \times 12.25 \approx 38.48 \, \text{cm}^2 \)

(ii) Curved surface area of the frustum

The formula for the curved surface area (CSA) of a frustum of a cone is \( \text{CSA} = \pi (R + r) l \), where \( R = 4.2 \, \text{cm} \), \( r = 3.5 \, \text{cm} \), and \( l = 8 \, \text{cm} \).

Step 1: Substitute the values

\( \text{CSA} = \pi \times (4.2 + 3.5) \times 8 \)

Step 2: Simplify \( 4.2 + 3.5 = 7.7 \)

\( \text{CSA} = \pi \times 7.7 \times 8 \)

Step 3: Calculate \( 7.7 \times 8 = 61.6 \)

\( \text{CSA} = \pi \times 61.6 \approx 3.1416 \times 61.6 \approx 193.52 \, \text{cm}^2 \)

(iii) Hemispherical surface area

The formula for the curved surface area of a hemisphere is \( \text{CSA} = 2\pi R^2 \), where \( R = 4.2 \, \text{cm} \).

Step 1: Substitute \( R = 4.2 \)

\( \text{CSA} = 2\pi \times (4.2)^2 \)

Step 2: Calculate \( (4.2)^2 = 17.64 \)

\( \text{CSA} = 2\pi \times 17.64 \approx 2 \times 3.1416 \times 17.64 \approx 110.88 \, \text{cm}^2 \)

Part (b)

Let the scale factor between the similar solids be \( k \). The total surface area (TSA) of the original solid is the sum of the frustum CSA, hemisphere CSA, and circular base area:
\( \text{TSA}_{\text{original}} = 193.52 + 110.88 + 38.48 \approx 342.88 \, \text{cm}^2 \)

For similar solids, the ratio of surface areas is \( k^2 \). Let \( \text{TSA}_{\text{new}} = 81.51 \, \text{cm}^2 \).

Step 1: Set up the ratio

\( \frac{\text{TSA}_{\text{new}}}{\text{TSA}_{\text{original}}} = k^2 \)

Step 2: Substitute values

\( \frac{81.51}{342.88} \approx 0.2377 = k^2 \)

Step 3: Solve for \( k \)

\( k = \sqrt{0.2377} \approx 0.4876 \)

Step 4: Find the new radius (base of the frustum)

The original base radius is \( r = 3.5 \, \text{cm} \). The new radius \( r' = k \times r \).

\( r' = 0.4876 \times 3.5 \approx 1.707 \, \text{cm} \) (or use exact proportionality for surface area ratio \( k^2 \), so radius scales by \( k = \sqrt{\frac{81.51}{342.88}} \approx 0.4876 \), leading to \( r' \approx 1.71 \, \text{cm} \))

Final Answers

(a)(i) \( \boldsymbol{\approx 38.48 \, \text{cm}^2} \)
(a)(ii) \( \boldsymbol{\approx 193.52 \, \text{cm}^2} \)
(a)(iii) \( \boldsymbol{\approx 110.88 \, \text{cm}^2} \)
(b) \( \boldsymbol{\approx 1.71 \, \text{cm}} \) (or more precise calculation based on exact \( \pi \))

Answer:

Part (a)

(i) Area of the circular base

The formula for the area of a circle is \( A = \pi r^2 \), where \( r = 3.5 \, \text{cm} \).

Step 1: Substitute \( r = 3.5 \) into the formula

\( A = \pi \times (3.5)^2 \)

Step 2: Calculate \( (3.5)^2 = 12.25 \)

\( A = \pi \times 12.25 \approx 3.1416 \times 12.25 \approx 38.48 \, \text{cm}^2 \)

(ii) Curved surface area of the frustum

The formula for the curved surface area (CSA) of a frustum of a cone is \( \text{CSA} = \pi (R + r) l \), where \( R = 4.2 \, \text{cm} \), \( r = 3.5 \, \text{cm} \), and \( l = 8 \, \text{cm} \).

Step 1: Substitute the values

\( \text{CSA} = \pi \times (4.2 + 3.5) \times 8 \)

Step 2: Simplify \( 4.2 + 3.5 = 7.7 \)

\( \text{CSA} = \pi \times 7.7 \times 8 \)

Step 3: Calculate \( 7.7 \times 8 = 61.6 \)

\( \text{CSA} = \pi \times 61.6 \approx 3.1416 \times 61.6 \approx 193.52 \, \text{cm}^2 \)

(iii) Hemispherical surface area

The formula for the curved surface area of a hemisphere is \( \text{CSA} = 2\pi R^2 \), where \( R = 4.2 \, \text{cm} \).

Step 1: Substitute \( R = 4.2 \)

\( \text{CSA} = 2\pi \times (4.2)^2 \)

Step 2: Calculate \( (4.2)^2 = 17.64 \)

\( \text{CSA} = 2\pi \times 17.64 \approx 2 \times 3.1416 \times 17.64 \approx 110.88 \, \text{cm}^2 \)

Part (b)

Let the scale factor between the similar solids be \( k \). The total surface area (TSA) of the original solid is the sum of the frustum CSA, hemisphere CSA, and circular base area:
\( \text{TSA}_{\text{original}} = 193.52 + 110.88 + 38.48 \approx 342.88 \, \text{cm}^2 \)

For similar solids, the ratio of surface areas is \( k^2 \). Let \( \text{TSA}_{\text{new}} = 81.51 \, \text{cm}^2 \).

Step 1: Set up the ratio

\( \frac{\text{TSA}_{\text{new}}}{\text{TSA}_{\text{original}}} = k^2 \)

Step 2: Substitute values

\( \frac{81.51}{342.88} \approx 0.2377 = k^2 \)

Step 3: Solve for \( k \)

\( k = \sqrt{0.2377} \approx 0.4876 \)

Step 4: Find the new radius (base of the frustum)

The original base radius is \( r = 3.5 \, \text{cm} \). The new radius \( r' = k \times r \).

\( r' = 0.4876 \times 3.5 \approx 1.707 \, \text{cm} \) (or use exact proportionality for surface area ratio \( k^2 \), so radius scales by \( k = \sqrt{\frac{81.51}{342.88}} \approx 0.4876 \), leading to \( r' \approx 1.71 \, \text{cm} \))

Final Answers

(a)(i) \( \boldsymbol{\approx 38.48 \, \text{cm}^2} \)
(a)(ii) \( \boldsymbol{\approx 193.52 \, \text{cm}^2} \)
(a)(iii) \( \boldsymbol{\approx 110.88 \, \text{cm}^2} \)
(b) \( \boldsymbol{\approx 1.71 \, \text{cm}} \) (or more precise calculation based on exact \( \pi \))