## Math 9 Review chapter 4: Cylinder – Cone – Sphere

## 1. Summary of theory

### 1.1. Cylinder

**a. Area around the cylinder**

With base radius r and height h, we have:

Surrounding area: \(S_{xq}=2\pi rh\)

Total Area: \(S_{tp}=2\pi rh+2\pi r^2\)

**b. Cylindrical volume**

The volume of the cylinder is given by the formula: \(V=Sh=\pi r^2h\)

### 1.2. Cone

**a. Surrounding area of cone**

Formula: \(S_{xq}=\pi rl\)

Where: r is the radius of the base; l is the length of the birth line

So we derive the formula for total area:

\(S_{tp}=S_{xq}+S_{day}=\pi rl+\pi r^2\)

**b. Volume of cone**

Experimentally, the volume of the cone is: \(V=\frac{1}{3}\pi r^2h\)

### 1.3. Conical

Surrounding area and volume of truncated cone

We have the following formulas:

\(S_{xq}=\pi (r_1+r_2)l\)

\(V=\frac{1}{3}\pi h(r_{1}^{2}+r_{2}^{2}+r_1r_2)\)

### 1.4. Globular

**a. Area of the bridge surface**

Recalling the knowledge learned in the lower class, we have the following formula:

\(S=4\pi R^2=\pi d^2\) (where R is the radius, d is the diameter of the sphere)

**b. Volume of sphere**

The formula for calculating the volume of a sphere:

\(V=\frac{4}{3}\pi R^3\)

## 2. Illustrated exercise

### 2.1. Exercise 1

The circumference of the cylinder is \(20\pi cm\) and the height is \(4cm\). The volume of the cylinder is:

**Solution guide**

From the circumference of the circle, we deduce \(R=10 cm\); So Volume is \(V=\pi R^2h=\pi.10^2.4=400 \pi (cm^3)\)

### 2.2. Exercise 2

For drawings

Given \(OB=5cm, AB=13cm\). The volume of the upper cone is:

**Solution guide**

Using the Pythagorean theorem, we get \(OA=\sqrt{AB^2-OB^2}=12cm\)

So \(V=\frac{1}{3}.OA.\pi.OB^2=\frac{1}{3}.12.5^2.\pi=100 \pi(cm^3)\)

### 2.3. Exercise 3

The surrounding area of a truncated cone whose base radius is large and small, respectively \(14cm, 8cm\) and whose birth line is \(9cm\) is:

**Solution guide**

\(S_{xq}=\pi(R+r)l=\pi(14+8).9=198\pi (cm^2)\)

### 2.4. Exercise 4

Describe the figure below formed by a cone with a birthline of \(13cm\), a radius of \(5cm\) and a half sphere. Calculate the volume of the cube.

**Solution guide**

The height of a cone is easily calculated using the Pythagorean theorem: \(h=\sqrt{13^2-5^2}=12cm\)

So the volume of the cone is: \(V_{non}=\frac{1}{3}\pi R^2h=\frac{1}{3}\pi.5^2.12=100\pi (cm^) 3)\)

The volume of the semi-sphere is: \(V_(nuacau)=\frac{2}{3}\pi R^3=\frac{2}{3}\pi.5^3=\frac{250}{3 }\pi(cm^3)\)

So the volume of the cube is \(100\pi+\frac{250}{3}\pi=\frac{550}{3} \pi(cm^3)\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **The lengths of the sides of a triangle \(ABC\) right at \(A\), satisfy the following relations:

\(BC = AB + 2a \) (1)

\(\displaystyle AC = {1 \over 2}\left( {BC + AB} \right)\) (2)

\(a\) is a given length

a) Calculate in terms of \(a\), the lengths of the sides and the height \(AH\) of the triangle.

b) Triangle \(ABC\) is inscribed in a semicircle with center \(O.\) Calculate the area of the part of the semicircle but outside the triangle.

c) Let a triangle \(ABC\) rotate once around the hypotenuse \(BC.\) Calculate the ratio of the areas of the parts formed by the chords \(AB\) and \(AC\).

**Verse 2: **Given a cone with base circle radius \(r (cm)\) and height \(2r (cm)\) and a sphere of radius \(r (cm)\) Calculate:

a) Area of the sphere, given that the total area of the cone is \(21.06 \;\left( {c{m^2}} \right)\).

b) The volume of a cone, given the volume of the sphere is \(15.8 \;\left( {c{m^3}} \right)\)

**Question 3: **With a semi-sphere of radius \(r\) and a cylinder whose base and height are both \(h\).

a) When \(r = 12\; (cm)\) and the volumes are equal, what is the value of \(h\; (cm)\) rounded to the first decimal place?

b) When \(h = 12\, (cm)\) and the sum of the area of the half sphere and the area of the “bottom circle” is three times the total area of the cylinder, then \(r (cm)\) is equal to how many, how much?

**Question 4:** With a semi-sphere of radius r and a cylinder of radius h, base and height are both

a) When r = 12 (cm) and the volumes are equal, what is the value of h (cm) rounded to the first decimal place?

b) When h = 12 (cm) and the sum of the area of the half sphere and the area of the “bottom circle” is three times the total area of the cylinder, what is r (cm)?

### 3.2. Multiple choice exercises

**Question 1: **Let the radii of Earth and Moon be \(6371\) and \(1738\) kilometers, respectively. Which of the following numbers is the ratio of the volumes of the Earth to the Moon?

(A) \(3.67\) (C) \(15.63\)

(B) \(4.93\) (D) \(49,26\).

**Verse 2: **What is the ratio of the volumes of the inscribed cones to the cylinder and the cylinder? (know that the height of the cone is equal to \(\frac{1}{2}\) the height of the cylinder)

A. \(\frac{1}{3}\)

B. \(\frac{1}{6}\)

C. \(\frac{1}{9}\)

D. \(\frac{1}{12}\)

**Question 3: **The figure generated when orbiting the edge FI is:

A. Cylindrical

B. Rectangular shape

C. Cones and cylinders

D. Cones and truncated pyramids

**Question 4: **The volume of the above cube is: (if the base radius is 5, the height of the cone is 12)

A. \(\frac{250\pi}{3}\)

B. \(\frac{310\pi}{3}\)

C. \(\frac{125\pi}{3}\)

D. \(\frac{155\pi}{3}\)

**Question 5: **A cube is depicted as shown below:

Composed of a cylinder and two halves of the sphere on either side. Given that the length of the cylinder is \(20\), the radius of the base of the cylinder is \(4\). Calculate the total area of the cube.

A. \(56 \pi\)

B. \(64 \pi\)

C. \(120 \pi\)

D. \(184 \pi\)

## 4. Conclusion

Through this lesson, you will learn some key topics as follows:

- Remember and deepen the concepts of cylinders (base, axis, surrounding face, height line, cross section .. of the cylinder).
- Understand and know how to use the formula to calculate the surrounding area, total area, and volume of a cylinder.

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