A solid cylinder having radius of base as 28 cm and height as 24 cm is bisected from is height to get two identical cylinders. What will be the percentage increase in the total surface area?
53.85 percent
The problem asks us to find the percentage increase in the total surface area when a solid cylinder is bisected (cut into two equal halves) along its height.
We are given the dimensions of the original solid cylinder:
The formula for the total surface area (TSA) of a solid cylinder is:
\( \text{TSA}_{\text{original}} = 2\pi r(r+h) = 2\pi r^2 + 2\pi r h \)
Let's calculate the initial total surface area using the given values:
\( \text{TSA}_{\text{original}} = 2\pi (28 \text{ cm})(28 \text{ cm} + 24 \text{ cm}) \)
\( \text{TSA}_{\text{original}} = 2\pi (28 \text{ cm})(52 \text{ cm}) \)
\( \text{TSA}_{\text{original}} = 2\pi (1456 \text{ cm}^2) \)
\( \text{TSA}_{\text{original}} = 2912\pi \text{ cm}^2 \)
When the cylinder is bisected from its height, it is cut horizontally through the middle. This results in two identical smaller cylinders.
When the cut is made, a new circular surface is exposed on each of the two resulting pieces. The area of this new surface is \(\pi r^2\).
Each of the two new cylinders has its own total surface area. The formula for the total surface area of one of these smaller cylinders (with height h' and radius r') is the standard formula, but we must account for the surfaces present:
Each new cylinder has:
Total Surface Area of one new cylinder:
\( \text{TSA}_{\text{one new}} = \pi r^2 + \pi r^2 + 2\pi r h' = 2\pi r^2 + 2\pi r h' \)
Since there are two identical new cylinders, the total surface area of both pieces combined after cutting is:
\( \text{TSA}_{\text{after}} = 2 \times (\text{TSA}_{\text{one new}}) = 2 \times (2\pi r^2 + 2\pi r h') = 4\pi r^2 + 4\pi r h' \)
Substitute the values r = 28 cm and h' = 12 cm:
\( \text{TSA}_{\text{after}} = 4\pi (28 \text{ cm})^2 + 4\pi (28 \text{ cm})(12 \text{ cm}) \)
\( \text{TSA}_{\text{after}} = 4\pi (784 \text{ cm}^2) + 4\pi (336 \text{ cm}^2) \)
\( \text{TSA}_{\text{after}} = 3136\pi \text{ cm}^2 + 1344\pi \text{ cm}^2 \)
\( \text{TSA}_{\text{after}} = 4480\pi \text{ cm}^2 \)
The increase in total surface area is the difference between the total surface area after cutting and the original total surface area.
\( \text{Increase} = \text{TSA}_{\text{after}} - \text{TSA}_{\text{original}} \)
\( \text{Increase} = 4480\pi \text{ cm}^2 - 2912\pi \text{ cm}^2 \)
\( \text{Increase} = 1568\pi \text{ cm}^2 \)
Alternatively, the increase comes from the two new surfaces created by the cut, each with area \(\pi r^2\). The total added area is \(2 \times \pi r^2 = 2\pi (28)^2 = 2\pi (784) = 1568\pi \text{ cm}^2\). This matches our calculation.
The percentage increase is calculated using the formula:
\( \text{Percentage Increase} = \left( \frac{\text{Increase}}{\text{TSA}_{\text{original}}} \right) \times 100 \)
\( \text{Percentage Increase} = \left( \frac{1568\pi \text{ cm}^2}{2912\pi \text{ cm}^2} \right) \times 100 \)
Cancel out \(\pi\) and the units:
\( \text{Percentage Increase} = \left( \frac{1568}{2912} \right) \times 100 \)
Simplify the fraction:
\( \frac{1568}{2912} = \frac{1568 \div 16}{2912 \div 16} = \frac{98}{182} \)
\( \frac{98}{182} = \frac{98 \div 14}{182 \div 14} = \frac{7}{13} \)
So, the percentage increase is:
\( \text{Percentage Increase} = \left( \frac{7}{13} \right) \times 100 \)
\( \text{Percentage Increase} = \frac{700}{13} \)
Performing the division:
\( \frac{700}{13} \approx 53.846 \)
Rounding to two decimal places, the percentage increase is approximately 53.85 percent.
| Measurement | Original Cylinder | Each New Cylinder (after bisection) |
|---|---|---|
| Radius (r) | 28 cm | 28 cm |
| Height (h) | 24 cm | 12 cm |
| Total Surface Area Formula | \(2\pi r(r+h)\) | \(2\pi r(r+h')\) where \(h'=h/2\) (This is for one new cylinder) Total for two: \(2 \times (2\pi r^2 + 2\pi r h')\) |
| Calculated Area | \(2912\pi\) cm\(^2\) | \(2 \times (2\pi (28)^2 + 2\pi (28)(12)) = 4480\pi\) cm\(^2\) |
When a solid 3D object is cut, new surfaces are typically exposed, which increases the total surface area. The amount of increase depends on the shape of the object and the plane of the cut.
The area of the floor of a cubical room is 192 m 2. The length of the longest rod that can be kept in that room is :
A solid metallic rectangular block of dimensions 112 cm × 44 cm × 25 cm is melted and recast into a cylinder of radius 35 cm. The curved surface area (in cm 2) of the cylinder is: (Take π = 22/7)
If the volume of a cube is 175616 cm 3, what is its side?
The volume of a right circular cone is 1232 cm 3. If the height of the cone is 24 cm, then what will be the radius of its base?
A right triangle contains the right angle between the sides 5 cm and 7 cm. A cone is generated by revolving about the side 5 cm. The volume of this cone is: