## Math 9 Chapter 4 Cones – Truncated cones – Surrounding area and volume of cones, truncated cones

## 1. Summary of theory

### 1.1. Cone

When a right angled triangle \(AOC\) is rotated around the side of a fixed right angle \(OA\), a cone is obtained.

– The side \(OC\) forms the base of the cone, which is a cone centered \(O\).

– The edge \(AC\) sweeps over the surrounding face of the cone, each of its positions is called a generation line, for example \(AD\) is a generation line.

– \(A\) is the vertex and \(AO\) is the altitude of the cone.

### 1.2. Area – volume of cone

– Surrounding area of cone: \({S_{xq}} = \pi rl\)

– Total area of the cone: \({S_{tp}} = \pi rl + \pi {r^2}\)

(\(r\) is the radius of the base circle, \( l\) is the birth line)

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

### 1.3. Conical

When the cone is cut by a plane parallel to the base, a truncated cone is obtained.

### 1.4. Surrounding area and volume of truncated cone

Given a truncated cone where \(r_1,r_2\) is the base radii, \(l\) is the length of the birth line, \(h\) is the height.

+ The area around the truncated cone is \(S_{xq}=\pi (r_1+r_2).l\)

+ The volume of truncated cone is \(V=\dfrac {1}{3}\pi h (r_1^2+r_2^2+r_1r_2)\)

## 2. Illustrated exercise

### 2.1. Basic exercises

**Question 1: **The hat has a face shape around a cone. Tell me, where is the bottom circle, where is the surrounding face, where is the cone’s birth line.

**Solution guide**

The bottom circle is the widest part of the cone’s brim

The surrounding face is the outer part of the cone, from the top of the cone to the bottom circle

The birth line is any straight line, connecting the top to the bottom circle

**Verse 2:** The figure depicts a clown’s hat made of a pyramid and two concentric circles. Know that the small circle is empty so that the clown can put on the hat.

Let \(AB=10cm; OB=6cm, OC=9cm\). Calculate the area to make that hat

**Solution guide**

We see that the cone is the total area of the cone and the area of the large circle minus the area of the small circle.

Calculating those values in turn, we have:

\(S{xq}=\pi rl=\pi .6.10=60 \pi (cm^2)\)

\(S_{(O;OC)}=\pi R^2=\pi.9^2=81 \pi (cm^2)\)

\(S_{(O;OB)}=\pi r^2=\pi.6^2=36 \pi (cm^2)\)

Area of the remainder (the bottom part has left the small circle): \(81 \pi-36\pi=45\pi (cm^2)\)

So the area to make the cone is: \(45\pi+60 \pi =105 \pi (cm^2)\)

### 2.2. Advanced exercises

**Question 1: **Given a figure formed by a cone and a cylinder, for these two shapes to have equal volumes, how many times must the height of the cone be equal to the height of the cylinder?

**Solution guide**

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

The volume of the cylinder is \(V=\pi r^2h\) so their ratio will be 3

**Verse 2: **A cone is crossed by a plane parallel to the base at the midpoint of the altitude, the cone is divided into a truncated cone and a cone. What is the volume ratio of the new cone and the truncated cone just created?

**Solution guide**

We will return to the formula for the volume of each shape to derive the proportions:

In the figure below, we have P as mid point of AO, C as mid point of AC.

It is easy to infer that in a truncated pyramid, the large base has twice the radius of the small base.

\(V_{chop}=\frac{1}{3}\pi r^2h(dvtt)\)

\(V_{chopcut}=\frac{1}{3}\pi h (r^2+4r^2+2r^2)=\frac{7}{3}\pi hr^2(dvtt)\)

So, the required problem rate is \(\frac{1}{7}\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Let \(ABC\) triangle right at \(A\), \(\widehat B = 60^\circ \) and \(BC = 2a\) (length units). Rotate the triangle around the hypotenuse \(BC\). Calculate the surrounding area and volume of the resulting shape.

**Verse 2:** Given a parallelogram \(ABCD\) with \(AB = 1, AD = x\; (x > 0)\) and \(\widehat {BAD} = 60^\circ \).

a) Calculate the total area \(S\) of the figure formed when the parallelogram \(ABCD\) is rotated exactly one circle around the side \(AB\) and the total area \(S_1\) of the formed figure when rotating around the edge \(AD\).

b) Determine the value \(x\) when \(S = S_1\) and \(S = 2S_1\).

**Question 3: **From a cone, a turner can turn a tall but “narrow” cylinder or a wide but “low” cylinder. Under what circumstances does the turner remove less material?

**Question 4: **Let \(ABC\) triangle right at \(A.\) Let \({V_1},{V_2},{V_3}\) be, in order, the volumes of the shapes generated when the triangle is rotated \(ABC) \) a loop around the edges \(BC, AB\) and \(AC.\) Prove that:

\(\displaystyle {1 \over {V_1^2}} = {1 \over {V_2^2}} + {1 \over {V_3^2}}.\)

### 3.2. Multiple choice exercises

**Question 1: **A cone has a base radius of \(5cm\) and a birth line of \(13cm\). The volume of the given cone is:

A. \(50 \pi (cm^3)\)

B. \(100 \pi (cm^3)\)

C. \(200 \pi (cm^3)\)

D. \(150 \pi (cm^3)\)

**Verse 2: **The surrounding area of a truncated cone with a large base radius of \(15cm, 8cm\) and a birth line equal to \(10cm\) is:

A. \(70\pi (cm^2)\)

B. \(230\pi (cm^2)\)

C. \(12\pi (cm^2)\)

D. \(1200\pi (cm^2)\)

**Question 3: **If the height and base radius of a cone both increase and are equal to \(\displaystyle {5 \over 4}\) from their respective initial dimensions, which of the following ratios is the ratio between the volume of the new cone and the volume of the original cone?

(A) \(\displaystyle {5 \over 4};\) (B) \(\displaystyle {{15} \over {12}};\)

(C) \(\displaystyle {{25} \over {16}};\) (D) \(\displaystyle {{125} \over {64}}.\)

**Question 4: **The figure below shows an hourglass, consisting of a large cylinder and two symmetrical cones. The ratio of the volumes of the sum of the two cones and the whole cylinder is:

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

B. \(3\)

C. \(\frac{2}{3}\)

D. \(\frac{3}{2}\)

## 4. Conclusion

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

- Know the concept of cones: base, surrounding face, birth line, altitude, cross section parallel to the base of the cone and have the concept of truncated cone.
- Formulas for calculating the perimeter and volume of a cone

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