## Math 8 Chapter 4 Lesson 8: Area around a regular pyramid

## 1. Theoretical Summary

We usually use the following formulas:

– The perimeter of a regular pyramid is equal to the product of the semicircle of the base and the midpoint.

\({S_{xq}} = pd\) (p is the half circumference of the base; d is the midpoint of the regular pyramid).

– The total area of the chop is equal to the sum of the surrounding area and the area of the base.

– With pyramids, to calculate the area around we calculate the sum of the areas of the side faces.

– To calculate the surrounding area of a regular truncated pyramid, we calculate the area of one side and multiply it by the number of sides, or subtract the surrounding area of a large pyramid from the surrounding area of a small pyramid.

## 2. Illustrated exercise

### 2.1. Exercise 1

If a regular quadrilateral pyramid has a base length of 6 cm and a height of 4 cm, what is the surrounding area?

**Solution guide**

Assume there is a regular apex \(S.ABCD\), \(O\) is the center of the bottom, \(I\) is the midpoint of \(CD\).

Apply Pythagorean theorem to right triangle \(SOI\), We have:

\(S{I^2} = S{O^2} + O{I^2}\)

\(\Rightarrow SI = \sqrt {S{O^2} + O{I^2}}\)\(\, = \sqrt {{4^2} + {3^2}} = 5\,\ left( {cm} \right)\)

The perimeter of the pyramid is:

\(\displaystyle {S_{xq}} = {1 \over 2}.4.6.5 = 60\;(c{m^2})\)

### 2.2. Exercise 2

A regular quadrilateral pyramid has side length \(25cm\), base is a square \(ABCD\) side \(30cm\).

Calculate the total area of the pyramid.

**Solution guide**

Let \(H\) be the midpoint of \(BC\).

Then we have: \(BH = HC = \dfrac{1}{2}BC =\dfrac{1}{2}. 30=15 \,cm \)

Since the triangle \(SBC\) is isosceles at \(S\), \( SH\perp BC \).

We have: \(d = SH = \sqrt{SB^2- BH^2}\) \( = \sqrt{25^2 -15^2} = \sqrt{400}=20(cm)\)

The circumference of the base is: \(4. 30 = 120 (cm)\)

Surrounding area of pyramid:

\(S_{xq} = pd =\dfrac{1}{2} .120.20 = 1200 (cm^2) \)

The area of the bottom is:

\( S_{đ} = 30.30 = 900 (cm^2)\)

The total area of the pyramid is:

\( S_{tp} =S_{xq}+ S_{đ} = 1200 + 900 \) \(= 2100 (cm^2) \)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Calculate the total area of the equilateral triangular pyramid according to the dimensions in the figure.

**Verse 2: **If a regular quadrilateral pyramid has base length \(6cm\) and height \(4cm\), what is the surrounding area?

**Question 3:** Regular pyramid \(S.ABC\) has base side \(a = 12cm\), height \(h = 8cm.\) Calculate the surrounding area of that pyramid.

**Question 4: **The figure is a tent at a summer camp with the dimensions on the ABC being an isosceles right triangle.

a) Calculate the volume of the tent.

b) How much canvas is needed to set up the tent?

### 3.2. Multiple choice exercises

**Question 1:** An equilateral quadrilateral has an area around \(30cm^{2}\), the midpoint of 5cm. The base length is:

A. 6cm

B. 12cm

C. 3cm

D. 1.5cm

**Verse 2:** Let the equilateral triangular pyramid S.ABC have total area \(25\sqrt{3}\) . SB=x. We have:

A. x=5

B. x=25

C. x=2.5

D. x=\(5\sqrt{3}\)

**Question 3: **Let the pyramid of a regular quadrilateral have all sides equal to a. Calculate the total area of the pyramid:

A. \(a^{2}\sqrt{3}\)

B. \(2a^{2}\)

C. \(a^{2}(\sqrt{3}+1)\)

D. \(2a^{2}\sqrt{3}\)

**Question 4:** A regular tetrahedron has side 2, and what is its total area?

A. 8

B. 16

C. \(4\sqrt{6}\)

D. \(4\sqrt{3}\)

**Question 5:** The four vertices of a cube of the vertex of an equilateral triangular pyramid. Find the ratio of the total area of the cube to the total area of the pyramid of an equilateral triangle.

A. \(\sqrt{2}\)

B. \(\sqrt{3}\)

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

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

## 4. Conclusion

Through this lesson, you will learn some of the main topics as follows:

- Know how to calculate the area around a regular pyramid
- Knowing how to apply calculation formulas to specific shapes, gradually perfecting the skills of folding shapes, consolidating learned geometric concepts

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