Conversion of Solids and Frustum of a Cone
Conversion of Solid from One Shape to Another (Volume Conservation Principle)
This concept addresses problems where a solid material, initially in one geometric shape, is transformed into one or more solid objects of different geometric shapes. This transformation can occur through physical processes such as melting a metal and recasting it into a new form, reshaping clay, or drawing a wire from a block of material.
The fundamental principle that governs such conversions is the Principle of Volume Conservation.
Statement: When a solid object (or a collection of solid objects) is melted, recast, or reshaped into one or more different solid objects, the total volume of the material remains constant. This principle holds true assuming that there is no loss or wastage of the material during the conversion process.
Therefore, to solve problems involving the conversion of solids from one shape to another:
-
Calculate the Volume of the Original Solid(s):
Determine the volume of the solid(s) in their initial shape using the appropriate volume formula(s) and given dimensions.
-
Calculate the Volume of the Final Solid(s):
Determine the volume of the solid(s) in their final shape(s). This volume calculation might involve one or more unknown dimensions (e.g., height of a new solid, radius of a new solid, or the number of new solids formed).
-
Equate the Volumes:
Apply the principle of volume conservation by setting the total volume of the original solid(s) equal to the total volume of the final solid(s).
Total Volume of Original Solid(s) = Total Volume of Final Solid(s)
-
Solve the Equation:
Solve the resulting equation for the unknown dimension(s) or quantity. Ensure all dimensions are in consistent units before performing calculations.
Example: Drawing a Wire from a Cylinder
Example 1. A solid metallic cylinder of radius $4$ cm and height $10$ cm is melted and recast into a cylindrical wire of radius $1$ mm. Find the length of the wire.
Answer:
Given:
Original solid: A solid metallic cylinder.
Radius of the original cylinder, $R = 4$ cm.
Height of the original cylinder, $H = 10$ cm.
Final solid: A cylindrical wire.
Radius of the wire, $r = 1$ mm.
To Find:
The length (which is the height) of the cylindrical wire, $h$.
Principle:
The volume of the material remains constant during the conversion.
Volume of Wire = Volume of Original Cylinder
Solution:
Step 1: Ensure Consistent Units
The dimensions of the original cylinder are in centimetres (cm), while the radius of the wire is in millimetres (mm). To perform calculations, convert all dimensions to a single unit. Let's convert everything to centimetres.
Radius of original cylinder, $R = 4$ cm
Height of original cylinder, $H = 10$ cm
Convert the radius of the wire from mm to cm. Since $1$ cm $= 10$ mm, $1$ mm $= \frac{1}{10}$ cm $= 0.1$ cm.
Radius of wire, $r = 1 \$ \text{mm} = \frac{1}{10} \$ \text{cm} = 0.1 \$ \text{cm}$
Step 2: Write down the volume formulas for the original and final shapes
Both the original shape (cylinder) and the final shape (wire, which is a very long cylinder) are cylinders. The volume formula for a cylinder is $V = \pi \times (\text{radius})^2 \times (\text{height})$.
Volume of original cylinder ($V_{\text{original}}$) $= \pi R^2 H$
Volume of wire ($V_{\text{wire}}$) $= \pi r^2 h$
Step 3: Equate the volumes based on the conservation principle
"$V_{\text{wire}} = V_{\text{original}}$"
[Principle of volume conservation]
"$\pi r^2 h = \pi R^2 H$"
[Substituting volume formulas]
We can cancel the common factor $\pi$ from both sides of the equation:
"$\cancel{\pi} r^2 h = \cancel{\pi} R^2 H$"
"$r^2 h = R^2 H$"
... (1)
Step 4: Substitute the known values (in cm) into the simplified equation
Substitute the values $R=4$ cm, $H=10$ cm, and $r=0.1$ cm into equation (1):
"$(0.1 \$ \text{cm})^2 \times h = (4 \$ \text{cm})^2 \times 10 \$ \text{cm}$"
[Substituting values]
"$0.01 \$ \text{cm}^2 \times h = 16 \$ \text{cm}^2 \times 10 \$ \text{cm}$"
[Calculating squares]
"$0.01 h \$ \text{cm}^2 = 160 \$ \text{cm}^3$"
Step 5: Solve for the unknown height (length of the wire), h
Divide both sides of the equation by $0.01 \$ \text{cm}^2$ to isolate $h$:
"$h = \frac{160 \$ \text{cm}^3}{0.01 \$ \text{cm}^2}$"
Perform the division. Dividing by 0.01 is the same as multiplying by 100:
"$h = \frac{160}{0.01} \$ \text{cm} = 160 \times 100 \$ \text{cm}$"
"$h = 16000 \$ \text{cm}$"
The length of the wire is 16000 cm. It is often more practical to express such a large length in metres or kilometres. Since $1$ m $= 100$ cm, convert centimetres to metres:
"$h = \frac{16000}{100} \$ \text{m}$"
[Converting cm to m]
"$\mathbf{h = 160 \$\$ m}$"
Therefore, the length of the wire is 160 metres (m).
Example 2: Melting Multiple Solids
Example 2. How many metallic cones, each of radius 2 cm and height 3 cm, can be made by melting a metallic sphere of radius 6 cm?
Answer:
Given:
Original solid: A metallic sphere.
Radius of sphere, $R = 6$ cm.
Final solids: Metallic cones.
Radius of each cone, $r = 2$ cm.
Height of each cone, $h = 3$ cm.
To Find:
The number of cones ($n$) that can be made.
Principle:
The volume of the original sphere equals the total volume of all the cones made.
Volume of Sphere = $n \times$ Volume of One Cone
Solution:
Step 1: Write down the volume formulas
Volume of Sphere ($V_{sphere}$) $= \frac{4}{3} \pi R^3$
Volume of One Cone ($V_{cone}$) $= \frac{1}{3} \pi r^2 h$
Step 2: Set up the equation based on volume conservation
"$V_{sphere} = n \times V_{cone}$"
"$\frac{4}{3} \pi R^3 = n \times \left(\frac{1}{3} \pi r^2 h\right)$"
[Substituting volume formulas]
We can cancel the common terms $\pi$ and $\frac{1}{3}$ from both sides:
"$\cancel{\frac{4}{3}}^{4} \cancel{\pi} R^3 = n \times \left(\cancel{\frac{1}{3}} \cancel{\pi} r^2 h\right)$"
"$4 R^3 = n \times r^2 h$"
... (1)
Step 3: Substitute the known values into the equation
We are given $R=6$ cm, $r=2$ cm, and $h=3$ cm. Substitute these values into equation (1):
"$4 \times (6 \$ \text{cm})^3 = n \times (2 \$ \text{cm})^2 \times 3 \$ \text{cm}$"
[Substituting values]
"$4 \times 216 \$ \text{cm}^3 = n \times 4 \$ \text{cm}^2 \times 3 \$ \text{cm}$"
[Calculating powers]
"$864 \$ \text{cm}^3 = n \times 12 \$ \text{cm}^3$"
[Calculating products]
Step 4: Solve for n
Divide both sides of the equation by $12 \$ \text{cm}^3$ to isolate $n$:
"$n = \frac{864 \$ \text{cm}^3}{12 \$ \text{cm}^3}$"
Perform the division:
"$864 \div 12 = 72$"
"$\mathbf{n = 72}$"
Therefore, 72 metallic cones can be made from the metallic sphere.
Frustum of a Cone: Definition and Formation
A frustum is a portion of a solid shape, typically a cone or a pyramid, that remains after the top part is cut off by a plane parallel to the base. While frustums of pyramids exist, the term "frustum" most commonly refers to a frustum of a right circular cone unless otherwise specified.
Definition and Formation
A frustum of a right circular cone is the part of a right circular cone that is left when a plane parallel to the base slices through the cone, and the smaller cone above the plane is removed.
Imagine a full right circular cone. Make a cut through the cone with a flat surface (a plane) that is parallel to its circular base. This cut will create a smaller circle on the cone's lateral surface. The original cone is now divided into two parts: a smaller cone at the top and a shape with two circular bases and a slanted curved surface at the bottom. The bottom part is the frustum.
The original cone is often referred to as the large cone, and the removed top part is the small cone.
Key Dimensions of a Frustum
A frustum of a right circular cone has the following important dimensions:
-
Two Circular Bases:
The frustum has two parallel circular bases. These circles are the original base of the cone and the circular cross-section created by the cutting plane.
-
Radii of the Bases:
The two bases have different radii. Let the radius of the larger circular base (the original base) be $r_1$, and the radius of the smaller circular base (the top circle created by the cut) be $r_2$. By convention, $r_1 > r_2$ for the common orientation, but the formulas work regardless of which is labelled $r_1$ as long as they correspond to the two bases.
-
Height ($h$):
The perpendicular distance between the centers of the two circular bases is the height of the frustum. This is the vertical height of the remaining portion, not the slant height.
-
Slant Height ($l$):
The shortest distance along the slanted lateral surface of the frustum between the corresponding points on the circumferences of the two bases. This is the length of the sloped edge of the frustum.
Relationship between Height, Radii, and Slant Height
In a right circular frustum, the height ($h$), the slant height ($l$), and the difference between the radii of the two bases ($r_1 - r_2$) form a right-angled triangle. This can be seen by taking a vertical cross-section through the axis of the frustum, which reveals a trapezium. By drawing a perpendicular line segment from a point on the circumference of the smaller base circle to the plane of the larger base, we form a right triangle.
In this right triangle:
- One leg is the height of the frustum, $h$.
- The other leg is the horizontal distance between the radii at the top and bottom edges, which is the difference between the radii of the two bases, $|r_1 - r_2|$ (we usually take the difference between the larger and smaller radius, $r_{max} - r_{min}$). Let's assume $r_1$ is the radius of the larger base and $r_2$ is the radius of the smaller base, so the base of the triangle is $r_1 - r_2$.
- The hypotenuse is the slant height of the frustum, $l$.
Using the Pythagorean theorem ($(\text{leg}_1)^2 + (\text{leg}_2)^2 = (\text{hypotenuse})^2$):
$\mathbf{h^2 + (r_1 - r_2)^2 = l^2}$
Therefore, the slant height ($l$) can be calculated if the height ($h$) and the radii of the bases ($r_1, r_2$) are known:
$\mathbf{l = \sqrt{h^2 + (r_1 - r_2)^2}}$
... (1)
This formula is crucial for calculating the lateral or total surface area of the frustum, as the slant height is a necessary component of those formulas.
Surface Area of a Frustum of a Cone
The surface area of a frustum of a right circular cone consists of the areas of its two circular bases and the area of its curved lateral surface. Calculating these areas and summing them gives the total surface area.
Let the dimensions of the frustum of a right circular cone be:
- Radius of the larger circular base = $r_1$
- Radius of the smaller circular base = $r_2$
- Height of the frustum (perpendicular distance between bases) = $h$
- Slant height of the frustum = $l$
As derived in the previous section (I2), the slant height $l$ is related to the height $h$ and the radii $r_1$ and $r_2$ by the formula:
$\mathbf{l = \sqrt{h^2 + (r_1 - r_2)^2}}$
... (1)
Curved Surface Area (CSA) of Frustum
The Curved Surface Area (CSA) of a frustum of a cone is the area of its slanted lateral surface, which connects the circumferences of the two bases. This area can be derived by considering the frustum as the difference between the original larger cone and the smaller cone that was removed from the top.
Derivation of the Formula:
Let the original larger cone have height $H$ and slant height $L$. The smaller cone removed from the top has radius $r_2$, let its height be $H'$ and its slant height be $L'$. The height of the frustum is $h = H - H'$, and the slant height of the frustum is $l = L - L'$.
The CSA of the frustum is the difference between the CSA of the large cone and the CSA of the small cone:
$\text{CSA}_{\text{frustum}} = \text{CSA}_{\text{large cone}} - \text{CSA}_{\text{small cone}}$
... (2)
The formula for the CSA of a cone is $\pi \times \text{radius} \times \text{slant height}$.
$\text{CSA}_{\text{frustum}} = \pi r_1 L - \pi r_2 L'$
[Substituting CSA formulas into (2)]
From similar triangles formed by the axis and slant heights of the large and small cones, we have the proportionality $\frac{r_1}{L} = \frac{r_2}{L'}$. This can be rearranged as $r_1 L' = r_2 L$.
Also, $L = l + L'$. Substitute $L = l+L'$ into $r_1 L' = r_2 L$:
"$r_1 L' = r_2 (l + L')$"
"$r_1 L' = r_2 l + r_2 L'$"
Group terms involving $L'$:
"$r_1 L' - r_2 L' = r_2 l$"
"$L'(r_1 - r_2) = r_2 l$"
$\mathbf{L' = \frac{r_2 l}{r_1 - r_2}}$
... (3)
Now, substitute $L' = L - l$ (since $L = l + L'$ implies $L-L'=l$, so $L=l+L'$) back into the proportionality $\frac{r_1}{L} = \frac{r_2}{L'}$: $\frac{r_1}{L} = \frac{r_2}{L-l}$. This is getting complicated. Let's use the relationship $r_1 L' = r_2 L$ directly in the CSA formula.
From $r_1 L' = r_2 L$, we have $L' = \frac{r_2}{r_1} L$. Also $L = l + L'$. Substitute $L'$ into the CSA formula $ \pi r_1 L - \pi r_2 L'$:
$\text{CSA}_{\text{frustum}} = \pi r_1 L - \pi r_2 \left(\frac{r_2}{r_1} L\right) = \pi L \left(r_1 - \frac{r_2^2}{r_1}\right) = \pi L \left(\frac{r_1^2 - r_2^2}{r_1}\right)$
$= \pi L \frac{(r_1 - r_2)(r_1 + r_2)}{r_1}$
We also have $\frac{r_1}{L} = \frac{r_1 - r_2}{l}$, which means $l = \frac{L(r_1 - r_2)}{r_1}$, or $L = \frac{l r_1}{r_1 - r_2}$. Substitute this expression for $L$ into the formula:
$\text{CSA}_{\text{frustum}} = \pi \left(\frac{l r_1}{r_1 - r_2}\right) \frac{(r_1 - r_2)(r_1 + r_2)}{r_1}$
Cancel common terms $(r_1 - r_2)$ and $r_1$:
$= \pi \left(\frac{l \cancel{r_1}}{\cancel{r_1 - r_2}}\right) \frac{\cancel{(r_1 - r_2)}(r_1 + r_2)}{\cancel{r_1}}$
$\mathbf{CSA_{\text{frustum}} = \pi (r_1 + r_2) l}$
... (4)
Formula:
The formula for the Curved Surface Area (CSA) of a frustum of a right circular cone is:
$\textbf{CSA of Frustum} = \mathbf{\pi \times (Sum \$\$ of \$\$ Radii \$\$ of \$\$ Bases) \times Slant \$\$ Height}$
Or simply, $\mathbf{CSA = \pi (r_1 + r_2) l}$, where $r_1$ and $r_2$ are the radii of the two bases and $l$ is the slant height of the frustum.
Total Surface Area (TSA) of Frustum
The Total Surface Area (TSA) of a solid frustum of a cone is the sum of the area of its curved surface and the areas of its two circular bases (top and bottom). Since the bases are circles, their areas are calculated using the standard formula for the area of a circle, $\pi r^2$.
Derivation of the Formula:
$\text{TSA}_{\text{frustum}} = \text{CSA}_{\text{frustum}} + \text{Area of Larger Base} + \text{Area of Smaller Base}$
... (5)
The area of the larger base (radius $r_1$) is $\pi r_1^2$.
The area of the smaller base (radius $r_2$) is $\pi r_2^2$.
Substitute the formula for CSA (4) and the areas of the two bases into equation (5):
$\text{TSA}_{\text{frustum}} = \pi (r_1 + r_2) l + \pi r_1^2 + \pi r_2^2$
[Substituting areas into (5)]
$\mathbf{TSA_{\text{frustum}} = \pi l (r_1 + r_2) + \pi r_1^2 + \pi r_2^2}$
... (6)
Formula:
The formula for the Total Surface Area (TSA) of a solid frustum of a right circular cone is:
$\textbf{TSA of Frustum} = \mathbf{\pi \times Slant \$\$ Height \times (Sum \$\$ of \$\$ Radii) + \pi \times (Radius \$\$ of \$\$ Large \$\$ Base)^2 + \pi \times (Radius \$\$ of \$\$ Small \$\$ Base)^2}$
Or simply, $\mathbf{TSA = \pi l (r_1 + r_2) + \pi r_1^2 + \pi r_2^2}$, where $r_1$ and $r_2$ are the radii of the two bases and $l$ is the slant height.
Remember that if the height $h$ is given instead of the slant height $l$, you must first calculate $l$ using formula (1): $l = \sqrt{h^2 + (r_1 - r_2)^2}$.
Example
Example 1. The radii of the top and bottom bases of a frustum of a cone are $10$ cm and $4$ cm respectively. Its height is $8$ cm. Find its curved surface area and total surface area. (Use $\pi = 3.14$)
Answer:
Given:
Frustum of a cone.
Radius of larger base, $r_1 = 10$ cm.
Radius of smaller base, $r_2 = 4$ cm.
Height of frustum, $h = 8$ cm.
Value of $\pi = 3.14$.
To Find:
1. Curved Surface Area (CSA) of the frustum.
2. Total Surface Area (TSA) of the frustum.
Solution:
First, we need to calculate the slant height ($l$) of the frustum using the given height and radii. Using formula (1):
"$l = \sqrt{h^2 + (r_1 - r_2)^2}$"
Substitute the given values:
"$l = \sqrt{(8 \$ \text{cm})^2 + (10 \$ \text{cm} - 4 \$ \text{cm})^2}$"
[Substituting values]
"$l = \sqrt{8^2 + (6)^2} \$ \text{cm} = \sqrt{64 + 36} \$ \text{cm}$"
"$l = \sqrt{100} \$ \text{cm} = 10 \$ \text{cm}$"
The slant height of the frustum is $10$ cm.
Part 1: Calculate the Curved Surface Area (CSA)
Using the formula for CSA of a frustum, $CSA = \pi (r_1 + r_2) l$. Using formula (4) derived above:
"$CSA = \pi (r_1 + r_2) l$"
Substitute the given values for $\pi, r_1, r_2$, and $l$:
"$CSA = 3.14 \times (10 \$ \text{cm} + 4 \$ \text{cm}) \times 10 \$ \text{cm}$"
[Substituting values]
"$CSA = 3.14 \times (14 \$ \text{cm}) \times 10 \$ \text{cm}$"
[Calculating $10+4$]
"$CSA = 3.14 \times 140 \$ \text{cm}^2$"
[Calculating $14 \times 10$]
"$\mathbf{CSA = 439.60 \$\$ cm^2}$"
[Calculating $3.14 \times 140$]
The curved surface area of the frustum is 439.60 square centimetres ($\text{cm}^2$).
Part 2: Calculate the Total Surface Area (TSA)
Using the formula for TSA of a frustum, $TSA = \pi (r_1 + r_2) l + \pi r_1^2 + \pi r_2^2$. Using formula (6) derived above:
"$TSA = \pi (r_1 + r_2) l + \pi r_1^2 + \pi r_2^2$"
Substitute the known values for $\pi, r_1, r_2$, and $l$:
"$TSA = 3.14 \times (10 \$ \text{cm} + 4 \$ \text{cm}) \times 10 \$ \text{cm} + 3.14 \times (10 \$ \text{cm})^2 + 3.14 \times (4 \$ \text{cm})^2$"
[Substituting values]
We already calculated the first term (CSA). Calculate the areas of the bases:
Area of Larger Base $= \pi r_1^2 = 3.14 \times (10 \$ \text{cm})^2 = 3.14 \times 100 \$ \text{cm}^2 = 314 \$ \text{cm}^2$"
Area of Smaller Base $= \pi r_2^2 = 3.14 \times (4 \$ \text{cm})^2 = 3.14 \times 16 \$ \text{cm}^2 = 50.24 \$ \text{cm}^2$"
Sum the CSA and the areas of the two bases:
"$TSA = 439.60 \$ \text{cm}^2 + 314 \$ \text{cm}^2 + 50.24 \$ \text{cm}^2$"
[Substituting CSA and base areas]
"$TSA = (439.60 + 314 + 50.24) \$ \text{cm}^2$"
"$TSA = (753.60 + 50.24) \$ \text{cm}^2$"
"$\mathbf{TSA = 803.84 \$\$ cm^2}$"
The total surface area of the frustum is 803.84 square centimetres ($\text{cm}^2$).
Volume of a Frustum of a Cone
The volume of a frustum of a right circular cone is the measure of the space it occupies. Similar to the surface area, the volume of a frustum can be derived by considering it as the difference between the volume of the original larger cone and the volume of the smaller cone that was removed from the top.
Let the dimensions of the frustum be:
- Radius of the larger circular base = $r_1$
- Radius of the smaller circular base = $r_2$
- Height of the frustum = $h$
Let the original large cone (from which the frustum is formed) have height $H$ and radius $r_1$. Let the smaller cone that is removed from the top have height $H'$ and radius $r_2$. The height of the frustum is the difference between the heights of the two cones: $h = H - H'$.
Formula Derivation:
The volume of the frustum is the difference between the volume of the large cone and the volume of the small cone:
$\text{Volume}_{\text{frustum}} = \text{Volume}_{\text{large cone}} - \text{Volume}_{\text{small cone}}$
... (1)
The formula for the volume of a cone is $\frac{1}{3} \pi \times (\text{radius})^2 \times (\text{height})$.
$\text{Volume}_{\text{frustum}} = \frac{1}{3} \pi r_1^2 H - \frac{1}{3} \pi r_2^2 H'$
[Substituting volume formulas into (1)]
To express this formula solely in terms of $h, r_1,$ and $r_2$, we need to relate $H$ and $H'$ to these dimensions. From the similar triangles formed by the axis and radii of the large and small cones:
$\frac{r_1}{H} = \frac{r_2}{H'}$
[From similar triangles]
From this proportionality, we can write $H' = \frac{r_2}{r_1} H$. We also know $h = H - H' = H - \frac{r_2}{r_1} H = H \left(1 - \frac{r_2}{r_1}\right) = H \left(\frac{r_1 - r_2}{r_1}\right)$.
Solving for $H$: $\mathbf{H = \frac{h r_1}{r_1 - r_2}}$.
Now we can find $H'$: $\mathbf{H' = \frac{r_2}{r_1} H = \frac{r_2}{r_1} \left(\frac{h r_1}{r_1 - r_2}\right) = \frac{h r_2}{r_1 - r_2}}$.
Substitute these expressions for $H$ and $H'$ back into the volume formula (1):
$\text{Volume}_{\text{frustum}} = \frac{1}{3} \pi r_1^2 \left(\frac{h r_1}{r_1 - r_2}\right) - \frac{1}{3} \pi r_2^2 \left(\frac{h r_2}{r_1 - r_2}\right)$
Factor out the common terms $\frac{1}{3} \pi h$ and the denominator $(r_1 - r_2)$:
$= \frac{1}{3} \pi h \left( \frac{r_1^3}{r_1 - r_2} - \frac{r_2^3}{r_1 - r_2} \right)$
$= \frac{1}{3} \pi h \left( \frac{r_1^3 - r_2^3}{r_1 - r_2} \right)$
Use the algebraic identity for the difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Here $a=r_1$ and $b=r_2$.
$= \frac{1}{3} \pi h \left( \frac{\cancel{(r_1 - r_2)}(r_1^2 + r_1 r_2 + r_2^2)}{\cancel{r_1 - r_2}} \right)$
Cancel the term $(r_1 - r_2)$ from the numerator and denominator:
$\mathbf{Volume_{\text{frustum}} = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)}$
... (2)
Formula:
The formula for the Volume of a frustum of a right circular cone is:
$\textbf{Volume of Frustum} = \mathbf{\frac{1}{3} \pi \times Height \times ((Radius \$\$ of \$\$ Large \$\$ Base)^2 + (Radius \$\$ of \$\$ Small \$\$ Base)^2 + (Product \$\$ of \$\$ Radii))}$
Or simply, $\mathbf{V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)}$, where $r_1$ and $r_2$ are the radii of the two bases and $h$ is the perpendicular height of the frustum. The unit is cubic units (e.g., $\text{cm}^3$, $\text{m}^3$).
Example
Example 1. A bucket is in the shape of a frustum of a cone. Its height is $24$ cm and the radii of its top and bottom bases are $20$ cm and $8$ cm respectively. Find the volume of the bucket. (Use $\pi = \frac{22}{7}$)
Answer:
Given:
Bucket in the shape of a frustum of a cone.
Height of the frustum, $h = 24$ cm.
Radius of the top base (larger radius), $r_1 = 20$ cm.
Radius of the bottom base (smaller radius), $r_2 = 8$ cm.
Value of $\pi = \frac{22}{7}$.
To Find:
Volume of the bucket (frustum), $V$.
Solution:
Using the formula for the volume of a frustum of a cone, $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$. Using formula (2) derived above:
"$V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$"
Substitute the given values for $\pi, h, r_1$, and $r_2$:
"$V = \frac{1}{3} \times \frac{22}{7} \times 24 \$ \text{cm} \times ((20 \$ \text{cm})^2 + (8 \$ \text{cm})^2 + (20 \$ \text{cm} \times 8 \$ \text{cm}))$"
[Substituting values]
Simplify the first part and calculate the squares and product inside the parenthesis:
"$V = \frac{1}{\cancel{3}_1} \times \frac{22}{7} \times \cancel{24}^8 \$ \text{cm} \times (400 \$ \text{cm}^2 + 64 \$ \text{cm}^2 + 160 \$ \text{cm}^2)$"
[Simplify $\frac{1}{3} \times 24$, Calculate squares and product]
"$V = \frac{22}{7} \times 8 \$ \text{cm} \times (400 + 64 + 160) \$ \text{cm}^2$"
Calculate the sum inside the parenthesis:
"$400 + 64 + 160 = 464 + 160 = 624$"
"$V = \frac{22}{7} \times 8 \times 624 \$ \text{cm}^3$"
"$V = \frac{176 \times 624}{7} \$ \text{cm}^3$"
Perform the multiplication in the numerator: $176 \times 624$
$\begin{array}{cc}& & & 6 & 2 & 4 \\ \times & & & 1 & 7 & 6 \\ \hline && 37 & 4 & 4 \\ & 436 & 8 & \times \\ 624 & \times & \times \\ \hline 1098 & 2 & 4 \\ \hline \end{array}$"$V = \frac{109824}{7} \$ \text{cm}^3$"
Perform the division. $109824 \div 7$
$\begin{array}{r} 15689.14... \\ 7{\overline{\smash{\big)}\,109824.00...}} \\ \underline{-~\phantom{(}7\phantom{00000.}} \\ 39\phantom{00000.} \\ \underline{-~\phantom{(}35\phantom{00000.}} \\ 48\phantom{0000.} \\ \underline{-~\phantom{()42\phantom{000.}}} \\ 62\phantom{00.} \\ \underline{-~\phantom{()56\phantom{00.}}} \\ 64\phantom{0.} \\ \underline{-~\phantom{()63\phantom{0.}}} \\ 10\phantom{.} \\ \underline{-~\phantom{()7\phantom{.}}} \\ 30\phantom{} \\ \underline{-~\phantom{()28\phantom{}}}\\ 2\phantom{}\end{array}$"$V \approx 15689.14 \$ \text{cm}^3$"
The question doesn't specify rounding, so the exact value is $\frac{109824}{7}$ cm³, or approximately 15689.14 cm³.
Therefore, the volume of the bucket is approximately 15689.14 cm³.
Problems involving Conversion and Frustum
This section combines the concept of converting solids from one shape to another (conservation of volume) with the geometric properties and formulas for a frustum of a cone. These problems often involve calculating the volume of a frustum and then equating it to the volume of another shape formed by the same material, or vice versa.
Key Problem Types
- Calculating the volume or surface area of a frustum given its dimensions.
- Calculating the volume or surface area of a combined solid that includes a frustum.
- Problems where a frustum is melted and recast into another shape (equating volumes).
- Problems where another shape is melted and recast into a frustum (equating volumes).
- Problems where a solid is cut to form a frustum, and properties of the original solid or the cut-off part are used.
General Approach
Follow the problem-solving steps outlined in the previous sections:
- Understand the problem and draw a diagram.
- Identify the shapes involved (original, final, or parts of a frustum).
- List all given dimensions and identify unknowns. Calculate any required dimensions (like slant height for surface area).
- Apply the principle of volume conservation if it's a conversion problem.
- Use the correct formulas for volumes or surface areas of cones, frustums, or other relevant shapes.
- Set up the equation(s) and solve for the unknown(s).
- Check units and present the final answer clearly.
Example 1: Capacity and Surface Area of a Bucket (Frustum)
This is a direct application of the formulas for the volume and surface area of a frustum.
Example 1. A drinking glass is in the shape of a frustum of a cone of height $14$ cm. The diameters of its two circular ends are $4$ cm and $2$ cm. Find the capacity of the glass and the area of sheet required to make it (assuming it's open at the top). (Use $\pi = \frac{22}{7}$)
Answer:
Given:
Shape: Drinking glass in the shape of a frustum of a cone.
Height of the frustum, $h = 14$ cm.
Diameter of the larger circular end (bottom base), $D_1 = 4$ cm.
Diameter of the smaller circular end (top opening), $D_2 = 2$ cm.
The glass is open at the top (the smaller circular end).
Value of $\pi = \frac{22}{7}$.
To Find:
1. Capacity (Volume) of the glass.
2. Area of sheet required to make the glass (Surface Area, excluding the open top).
Solution:
First, find the radii from the given diameters:
Radius of the larger base, $r_1 = \frac{D_1}{2} = \frac{4}{2} \$ \text{cm} = 2 \$ \text{cm}$
Radius of the smaller base, $r_2 = \frac{D_2}{2} = \frac{2}{2} \$ \text{cm} = 1 \$ \text{cm}$
Height of the frustum, $h = 14$ cm.
Part 1: Calculate the Capacity (Volume) of the glass
The capacity of the glass is its volume. Using the formula for the volume of a frustum of a cone, $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$. Using formula (2) from section I4:
"$V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$"
Substitute the given values for $\pi, h, r_1$, and $r_2$:
"$V = \frac{1}{3} \times \frac{22}{7} \times 14 \$ \text{cm} \times ((2 \$ \text{cm})^2 + (1 \$ \text{cm})^2 + (2 \$ \text{cm} \times 1 \$ \text{cm}))$"
[Substituting values]
Simplify the terms:
"$V = \frac{1}{3} \times \frac{22}{\cancel{7}_1} \times \cancel{14}^2 \$ \text{cm} \times (4 \$ \text{cm}^2 + 1 \$ \text{cm}^2 + 2 \$ \text{cm}^2)$"
[Simplify $\frac{14}{7}$, Calculate squares and product]
"$V = \frac{1}{3} \times 22 \times 2 \$ \text{cm} \times (4 + 1 + 2) \$ \text{cm}^2$"
Calculate the sum inside the parenthesis:
"$V = \frac{44}{3} \$ \text{cm} \times 7 \$ \text{cm}^2$"
[Calculate $22 \times 2$ and $4+1+2$]
"$V = \frac{44 \times 7}{3} \$ \text{cm}^3 = \frac{308}{3} \$ \text{cm}^3$"
The volume (capacity) of the glass is $\frac{308}{3}$ cubic centimetres ($\text{cm}^3$). This can also be expressed as a mixed number or decimal:
"$\mathbf{V = 102 \frac{2}{3} \$\$ cm^3 \approx 102.67 \$\$ cm^3}$"
The capacity of the glass is $\frac{308}{3}$ cm³ or approximately 102.67 cm³.
Part 2: Calculate the Area of Sheet Required
The glass is open at the top (the smaller circular end, radius $r_2$). The sheet required to make the glass covers the bottom base (larger circle) and the curved surface area of the frustum.
Area of Sheet = Area of Bottom Base + CSA of Frustum
Area of bottom base $= \pi r_1^2$.
CSA of frustum $= \pi (r_1 + r_2) l$.
We need to find the slant height, $l$, using the formula $l = \sqrt{h^2 + (r_1 - r_2)^2}$. Using formula (1) from section I2:
"$l = \sqrt{h^2 + (r_1 - r_2)^2}$"
Substitute the values $h=14$ cm, $r_1=2$ cm, $r_2=1$ cm:
"$l = \sqrt{(14 \$ \text{cm})^2 + (2 \$ \text{cm} - 1 \$ \text{cm})^2}$"
[Substituting values]
"$l = \sqrt{196 + (1)^2} \$ \text{cm} = \sqrt{196 + 1} \$ \text{cm} = \sqrt{197} \$ \text{cm}$"
The slant height is $\sqrt{197}$ cm. Now, substitute this into the formula for the area of the sheet:
Area of Sheet $= \pi r_1^2 + \pi (r_1 + r_2) l$"
"$= \frac{22}{7} \times (2 \$ \text{cm})^2 + \frac{22}{7} \times (2 \$ \text{cm} + 1 \$ \text{cm}) \times \sqrt{197} \$ \text{cm}$"
[Substituting values]
"$= \frac{22}{7} \times 4 \$ \text{cm}^2 + \frac{22}{7} \times 3 \$ \text{cm} \times \sqrt{197} \$ \text{cm}$"
"$= \frac{88}{7} \$ \text{cm}^2 + \frac{66\sqrt{197}}{7} \$ \text{cm}^2$"
Factor out $\frac{22}{7}$ or $\frac{1}{7}$:
"$= \frac{1}{7} (88 + 66\sqrt{197}) \$ \text{cm}^2$"
"$\mathbf{= \frac{22}{7} (4 + 3\sqrt{197}) \$\$ cm^2}$"
Since $\sqrt{197}$ is irrational, the exact area is $\frac{22}{7} (4 + 3\sqrt{197})$ cm². If an approximate numerical answer is needed, we would use an approximation for $\sqrt{197}$. For instance, $\sqrt{197} \approx 14.0357$.
Area $\approx \frac{22}{7} (4 + 3 \times 14.0357) \approx \frac{22}{7} (4 + 42.1071) \approx \frac{22}{7} \times 46.1071 \approx 3.1428 \times 46.1071 \approx 144.95$ cm².
The capacity of the glass is $\frac{308}{3}$ cm³ and the area of sheet required is $\frac{22}{7} (4 + 3\sqrt{197})$ cm².
Example 2: Conversion involving Frustum
This example combines the concept of volume conservation with the volume formula for a frustum.
Example 2. A metallic right circular cone $20$ cm high and whose vertical angle is $60^\circ$ is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter $\frac{1}{16}$ cm, find the length of the wire.
Answer:
Given:
Original solid part: A frustum obtained by cutting a right circular cone.
Original cone dimensions: Height $H = 20$ cm, Vertical angle $= 60^\circ$.
Cut: By a plane parallel to the base, at the middle of the cone's height.
Conversion: The frustum is melted and drawn into a cylindrical wire.
Diameter of the wire $= \frac{1}{16}$ cm.
To Find:
The length of the wire, $L_{wire}$.
Principle:
When the frustum is drawn into a wire, the volume of the material is conserved.
Volume of Wire = Volume of Frustum
Solution:
First, we need to determine the dimensions of the frustum (height and radii of bases).
The original cone has height $H = 20$ cm and semi-vertical angle $\alpha = \frac{60^\circ}{2} = 30^\circ$.
Let the radius of the base of the original cone be $r_1$. In the right triangle formed by the height, radius, and slant height of the large cone, $\tan(\alpha) = \frac{\text{radius}}{\text{height}}$.
"$\tan(30^\circ) = \frac{r_1}{H} = \frac{r_1}{20}$"
We know $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.
"$\frac{1}{\sqrt{3}} = \frac{r_1}{20}$"
"$r_1 = \frac{20}{\sqrt{3}} \$ \text{cm}$"
The cone is cut at the middle of its height by a plane parallel to the base. This means the height of the smaller cone removed from the top is half the height of the original cone.
Height of small cone, $H' = \frac{H}{2} = \frac{20 \$ \text{cm}}{2} = 10 \$ \text{cm}$
The radius of the base of the small cone ($r_2$, which is the radius of the top base of the frustum) can be found using the same semi-vertical angle:
"$\tan(30^\circ) = \frac{r_2}{H'} = \frac{r_2}{10}$"
"$\frac{1}{\sqrt{3}} = \frac{r_2}{10}$"
"$r_2 = \frac{10}{\sqrt{3}} \$ \text{cm}$"
The height of the frustum ($h_{frustum}$) is the difference between the heights of the large and small cones:
"$h_{frustum} = H - H' = 20 \$ \text{cm} - 10 \$ \text{cm} = 10 \$ \text{cm}$"
Step 1: Find dimensions of the Frustum
The frustum has:
- Height, $h_{frustum} = 10$ cm.
- Radius of larger base, $r_1 = \frac{20}{\sqrt{3}}$ cm.
- Radius of smaller base, $r_2 = \frac{10}{\sqrt{3}}$ cm.
Step 2: Calculate Volume of the Frustum
Using the formula for the volume of a frustum of a cone, $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$. Using formula (2) from section I4:
"$V_{frustum} = \frac{1}{3} \pi h_{frustum} (r_1^2 + r_2^2 + r_1 r_2)$"
Substitute the dimensions of the frustum:
"$V_{frustum} = \frac{1}{3} \pi (10 \$ \text{cm}) \left[ \left(\frac{20}{\sqrt{3}} \$ \text{cm}\right)^2 + \left(\frac{10}{\sqrt{3}} \$ \text{cm}\right)^2 + \left(\frac{20}{\sqrt{3}} \$ \text{cm}\right)\left(\frac{10}{\sqrt{3}} \$ \text{cm}\right) \right]$"
[Substituting values]
"$V_{frustum} = \frac{10 \pi}{3} \$ \text{cm} \left[ \frac{400}{3} \$ \text{cm}^2 + \frac{100}{3} \$ \text{cm}^2 + \frac{200}{3} \$ \text{cm}^2 \right]$"
[Calculating squares and product]
"$V_{frustum} = \frac{10 \pi}{3} \left[ \frac{400 + 100 + 200}{3} \right] \$ \text{cm}^3$"
[Adding fractions with common denominator]
"$V_{frustum} = \frac{10 \pi}{3} \left[ \frac{700}{3} \right] \$ \text{cm}^3$"
[Calculating sum]
"$\mathbf{V_{frustum} = \frac{7000 \pi}{9} \$\$ cm^3}$"
[Multiplying fractions]
Step 3: Find dimensions and Volume of the Wire
The wire is a cylinder. The diameter of the wire is given as $\frac{1}{16}$ cm.
Radius of wire, $r_{wire} = \frac{1}{2} \times \text{Diameter} = \frac{1}{2} \times \frac{1}{16} \$ \text{cm} = \frac{1}{32} \$ \text{cm}$
Let the length of the wire be $L_{wire}$. This is the height of the cylindrical wire.
Using the formula for the volume of a cylinder, $V = \pi \times (\text{radius})^2 \times (\text{height})$:
"$V_{wire} = \pi (r_{wire})^2 L_{wire} = \pi \left(\frac{1}{32} \$ \text{cm}\right)^2 L_{wire}$"
"$V_{wire} = \pi \frac{1}{32 \times 32} L_{wire} = \pi \frac{1}{1024} L_{wire} \$ \text{cm}^2$"
Step 4: Equate Volumes and Solve for $L_{wire}$
By the principle of volume conservation, the volume of the frustum is equal to the volume of the wire.
"$V_{wire} = V_{frustum}$"
"$\pi \frac{1}{1024} L_{wire} = \frac{7000 \pi}{9}$"
[Equating the volumes]
Cancel $\pi$ from both sides of the equation:
"$\cancel{\pi} \frac{1}{1024} L_{wire} = \frac{7000 \cancel{\pi}}{9}$"
"$\frac{L_{wire}}{1024} = \frac{7000}{9}$"
Multiply both sides by 1024 to solve for $L_{wire}$:
"$L_{wire} = \frac{7000}{9} \times 1024 \$ \text{cm}$"
"$L_{wire} = \frac{7000 \times 1024}{9} \$ \text{cm} = \frac{7168000}{9} \$ \text{cm}$"
The length of the wire is $\frac{7168000}{9}$ cm. Convert this to metres for a more practical unit. Since 1 m = 100 cm, divide by 100:
"$L_{wire} = \frac{7168000}{9 \times 100} \$ \text{m} = \frac{7168000}{900} \$ \text{m}$"
Simplify the fraction by cancelling out 100 from numerator and denominator:
"$L_{wire} = \frac{71680}{9} \$ \text{m}$"
Performing the division: $71680 \div 9 \approx 7964.44$.
"$\mathbf{L_{wire} = \frac{71680}{9} \$\$ m \approx 7964.44 \$\$ m}$"
Therefore, the length of the wire is $\frac{71680}{9}$ metres or approximately 7964.44 metres.