/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 The base of a hemisphere has an ... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 14

The base of a hemisphere has an area of \(256 \pi \mathrm{cm}^{2}\) Find its volume.

Short Answer

Expert verified
2730.66\(\pi\) \(\text{cm}^3\)

Step by step solution

01

- Understand the given data

The problem states that the base area of a hemisphere is given as 256\(\pi\) \(\text{cm}^2\). Since the base of a hemisphere is a circle, we need to relate this area to the radius of the circle.
02

- Calculate the radius

The area of a circle is given by the formula \(\text{Area} = \pi r^2\). Thus, we equate \(\text{Area} = 256 \pi\) and solve for the radius (r).\[256\pi = \pi r^2\implies r^2 = 256\implies r = \sqrt{256}\implies r = 16 \ \text{cm}\]
03

- Use the formula for the volume of a hemisphere

The volume of a hemisphere is given by \[\text{Volume} = \frac{2}{3} \pi r^3\]. We already found that the radius (r) is 16 cm. Plug this value into the formula:\[\text{Volume} = \frac{2}{3} \pi (16)^3\]
04

- Calculate the volume

Now compute the volume using the formula from Step 3:\[\text{Volume} = \frac{2}{3} \pi (16 \times 16 \times 16)\]\[\text{Volume} = \frac{2}{3} \times 4096 \pi\]\[\text{Volume} = 2730.66 \pi \ \text{cm}^3\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hemisphere Volume
To find the volume of a hemisphere, we use the formula \[\text{Volume} = \frac{2}{3} \pi r^3\]. This formula is derived from the volume of a full sphere.
A sphere's volume formula is \[\text{Volume}_{sphere} = \frac{4}{3} \pi r^3\]. Since a hemisphere is exactly half of a sphere, we take half of this volume. Then we get our equation.
This gives us the formula to use for any hemisphere.
Radius Calculation
The radius of a hemisphere can be found if we know the area of the base. The base of a hemisphere is a circle.
The area of a circle is given by the formula \[\text{Area} = \pi r^2\]. Rearrange this equation to solve for the radius (r):\[\text{r} = \sqrt{\frac{\text{Area}}{\pi}}\].
In our problem, we started with a base area of 256\pi \text{cm}^{2}. By substituting this value in and solving, we found that the radius is 16 cm.
Circle Area Formula
Understanding the circle area formula \[\text{Area} = \pi r^2\] is key to solving problems involving both circles and hemispheres.
This formula connects the radius of a circle with its area, allowing us to easily find one if we know the other. It's essential for calculating other properties, such as the volume of a hemisphere.
With a base area given, you can substitute that into the area formula to find the radius, just like we did in the problem.
Mathematical Problem-Solving
Problem-solving in mathematics involves understanding given data, using relevant formulas, and performing computations step by step.
In this problem, we first identified the given base area and related it to the radius using the circle area formula.
After finding the radius, we used another formula specific to hemispheres to find the volume.
Breaking down the problem like this makes it easier to manage and solve accurately.
Geometry Formulas
Geometry formulas are essential tools in problem-solving.
They offer robust methods for calculating properties of shapes, such as area, volume, and more.
Understanding and memorizing key geometry formulas will allow you to tackle a wide range of problems more confidently and efficiently.
For example, the circle area formula and the volume formula for spheres and hemispheres are crucial to solve many geometry problems involving circular shapes.

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Most popular questions from this chapter

Although the Exxon Valdez oil spill (11 million gallons of oil) is one of the most notorious oil spills, it was small compared to the 250 million gallons of crude oil that were spilled during the 1991 Persian Gulf War. A gallon occupies 0.13368 cubic foot. How many rectangular swimming pools, each 20 feet by 30 feet by 5 feet, could be filled with 250 million gallons of crude oil?

Match each real object with a geometry term. You may use a geometry term more than once or not at all. A. Cylinder B. Cone C. Square prism D. Square pyramid E. Sphere F. Triangular pyramid G. Octagonal prism H. Triangular prism I. Trapezoidal prism J. Rectangular prism K. Heptagonal pyramid L. Hexagonal prism M. Hemisphere Can of tuna fish

How many cubes measuring \(1 \mathrm{cm}\) on each edge will fit into the container? (figure not copy) A box measuring \(3 \mathrm{cm}\) by \(4 \mathrm{cm}\) by \(5 \mathrm{cm}\) on the inside edges

Refer to the table. $$\begin{array}{|l|l|l|l|} \hline \text { Metal } & \text { Density } & \text { Metal } & \text { Density } \\\\\hline \text { Aluminum } & 2.81 \mathrm{g} / \mathrm{cm}^{3} & \text { Nickel } & 8.89 \mathrm{g} / \mathrm{cm}^{3} \\\\\hline \text { Copper } & 8.97 \mathrm{g} / \mathrm{cm}^{3} & \text { Platinum } & 21.40 \mathrm{g} / \mathrm{cm}^{3} \\\\\hline \text { Gold } & 19.30 \mathrm{g} / \mathrm{cm}^{3} & \text { Potassium } & 0.86 \mathrm{g} / \mathrm{cm}^{3} \\\\\hline \text { Lead } & 11.30 \mathrm{g} / \mathrm{cm}^{3} & \text { Silver } & 10.50 \mathrm{g} / \mathrm{cm}^{3} \\\\\hline \text { Lithium } & 0.54 \mathrm{g} / \mathrm{cm}^{3} & \text { Sodium } & 0.97 \mathrm{g} / \mathrm{cm}^{3} \\\\\hline\end{array}$$ A square-prism container with a base \(5 \mathrm{cm}\) by \(5 \mathrm{cm}\) is partially filled with water. You drop a clump of metal with mass \(525 \mathrm{g}\) into the container, and the water level rises \(2 \mathrm{cm} .\) What is the density of the metal? Assuming the metal is pure, what is the metal?

Find the surface area of a sphere whose volume is \(288 \pi \mathrm{cm}^{3}\) (IMAGE CAN'T COPY)
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