Course: 8.8 金字塔的地表面积

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金字石表面面积
Ashanti went to an alternative therapies market with her aunt and ended up buying a beautiful pyramid crystal paperweight. She’s interested in figuring out the total of her new ornament and has measured and drawn the dimensions below. How can Ashanti figure out the surface area?
::Ashanti和她姨妈去了替代疗法市场,最终买下了美丽的金字塔晶体造纸。她有兴趣了解她的新装饰品总量,并测量和绘制了下面的尺寸。 Ashanti如何找到表面积?

In this concept, you will learn to find the .
::在这一概念中,你会学会找到。

Surface Area
::地表地区

A pyramid has sides that are triangular faces and a base. The base can be any shape. Surface area is the total of the areas of each face in a . Imagine you could wrap a pyramid in wrapping paper, like a present. The amount of wrapping paper needed to cover the figure represents its surface area. To find the surface area, you must be able to calculate the area of each face and then add these areas together.
::金字塔的侧面是三角面和基底。底部可以是任何形状。表面面积是每个面部的总面积。想象一下, 您可以用包装纸将金字塔包起来, 像一个礼物一样。覆盖图所需的包装纸量代表着它的表面面积。要找到表面面积, 您必须能够计算每个面部的面积, 然后将这些区域加在一起。

One way to calculate surface area is to use a . A net is a two-dimensional diagram of a three-dimensional figure. Imagine you could unfold a pyramid so that it is completely flat. Here is what the net of a pyramid would look like.
::计算表面积的一个方法就是使用。网是一个三维图的二维图。想象一下您可以展出一个金字塔, 这样它就完全平坦。这就是金字塔的网看起来像的。

This net is of a square pyramid. You can imagine folding up the sides to create a pyramid. With a net, you can see each face of the pyramid more clearly.
::这个网是一个平方金字塔。你可以想象把两边折叠成一个金字塔。用一个网,你可以更清楚地看到金字塔的每一个面孔。

To find the surface area, you need to calculate the area for each face in the net: the sides and the base. The side faces of a pyramid are always triangles, so you use the area formula for triangles to calculate their area: $\begin{align*}A = \frac{1}{2} bh\end{align*}$ . For triangles, you will need the height , or slant height in the case of a pyramid.
::要找到表面区域, 您需要计算网中每个面的面积: 边和底。金字塔的侧面总是三角形, 所以您要使用三角形的区域公式来计算其区域 : A= 12bh。对于三角形, 您需要高, 或者金字塔的倾斜高度。

The area of the base depends on what shape it is. Remember, pyramids can have bases in the shape of a triangle , square, rectangle , or any other polygon . You use whichever area formula is appropriate for the shape.
::基点区域取决于它的形状。记住, 金字塔可以有三角形、正方形、矩形或任何其他多边形的形状。您可以使用适合形状的任意区域公式。

Here are some common area formulas:
::以下是一些共同区域公式:

$\begin{aligned} Rectangle : A = l w \end{aligned}$
$\begin{align*}\text{Rectangle}: A = lw\end{align*}$
::矩形: A=lw

$\begin{aligned} Square : A = s^{2} \end{aligned}$
$\begin{align*}\text{Square}: A = s^2\end{align*}$
::平方: A=s2

$\begin{aligned} Triangle : A = \frac{1}{2} b h \end{aligned}$
$\begin{align*}\text{Triangle}: A = \frac{1}{2} bh\end{align*}$
::三角形:A=12bh

In the net of the pyramid above, you can see that it is a square pyramid. Imagine that it has a slant height of 4 cm and a side length of 6 cm. You can use these measurements to find the area of each face of the pyramid.
::在以上金字塔的网中, 您可以看到它是一个平方金字塔。想象它有4厘米的倾斜高度和6厘米的侧长。您可以使用这些测量来找到金字塔每个面的面积。

First, find the area of the square base.
::首先,找到基地的面积

$\begin{aligned} \begin{array}{rcl} A & = & s^{2} \\ A & = & 6^{2} \\ A & = & 36 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} A &=& s^2\\ A &=& 6^2\\ A &=& 36 \end{array}\end{align*}$
::A=S2A=62A=36

Next, find the area of the triangle side.
::下一个,找到三角形边的区域。

$\begin{aligned} \begin{array}{rcl} A & = & \frac{1}{2} b h \\ A & = & \frac{1}{2} (4) (6) \\ A & = & 12 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} bh\\ A &=& \frac{1}{2} (4)(6)\\ A &=& 12 \end{array}\end{align*}$
::A=12bhA=12(4)(6)A=12

Then, add up all of the areas. Remember you have one square base and four triangle sides.
::然后,将所有的区域加起来。记住,你有一个平方基和四个三角边。

$\begin{aligned} \begin{array}{rcl} A & = & base + 4 \times side \\ A & = & 36 + (4 \times 12) \\ A & = & 84 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} A &=& \text{base} + 4 \times \text{side}\\ A &=& 36 + (4 \times 12)\\ A &=& 84 \end{array}\end{align*}$
::A=基准+4xsideA=36+(4x12)A=84

The answer is 84.
::答案是84

The surface area of the square based pyramid is 84 cm ² .
::平方金字塔的表面面积为84厘米2。

There is also a formula is to calculate the surface area of a pyramid. Let $\begin{align*}B\end{align*}$ represent the area of the base.
::还有一个公式是计算金字塔的表面积。让 B 代表基数的区域。

$\begin{aligned} S A = \frac{1}{2} \times perimeter \times slant height + B \end{aligned}$
$\begin{align*}SA = \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B\end{align*}$
::SA=12x Periterx倾斜高度+B

Remember, pyramids can have bases of different shapes, so the area formula you use to find $\begin{align*}B\end{align*}$ varies. You will find the base’s area first and then put it into the formula in place of $\begin{align*}B\end{align*}$ .
::记住, 金字塔可以有不同形状的基座, 所以您用来找到 B 的面积公式会有所不同。您会先找到基座的面积, 然后将其放入公式中, 取代 B 。

Sometimes, you will have to find a linear measurement. This means that you will be given the surface area and one other dimension. Then you will need to work backwards to figure out the measurement for the missing dimension.
::有时候,你必须找到线性测量。这意味着你将得到表面积和另一个维度。然后你需要倒着工作,找出缺失维度的测量方法。

Let’s look at an example.
::让我们举个例子。

The base of a square pyramid has sides of 4 cm each and a surface area of 96 cm ² . What is the slant height of the pyramid?
::平方金字塔的底部每面4厘米,表面面积96厘米。金字塔的倾斜高度是多少?

First, find the area of the square base.
::首先,找到基地的面积

$\begin{aligned} \begin{array}{rcl} A & = & s^{2} \\ A & = & 4^{2} \\ A & = & 16 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} A &=& s^2\\ A &=& 4^2\\ A &=& 16 \end{array} \end{align*}$
::A=S2A=42A=16

Next, find the perimeter of the square base.
::接下来,找到广场的周边

$\begin{aligned} \begin{array}{rcl} P & = & 4 s \\ P & = & 4 (4) \\ P & = & 16 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P &=& 4s\\ P &=& 4(4)\\ P &=& 16 \end{array}\end{align*}$
::P=4sP=4(4)P=16

Then, substitute what you know in the formula for the surface area.
::然后,用公式中所知道的代替表面区域。

$\begin{aligned} \begin{array}{rcl} S A & = & \frac{1}{2} \times perimeter \times slant height + B \\ 96 & = & \frac{1}{2} \times 16 \times slant height + 16 \\ 96 & = & 8 \times slant height + 16 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ 96 &=& \frac{1}{2} \times 16 \times \text{slant height} + 16 \\ 96 &=& 8 \times \text{slant height} + 16 \end{array}\end{align*}$
::SA=12x外围计x倾斜高度+B96=12x16x倾斜高度+1696=8x倾斜高度+16

Then, solve for the slant height by subtracting 16 and dividing both sides by 8.
::然后,解决倾斜高度问题,减去16,将两边除以8。

$\begin{aligned} \begin{array}{rcl} 96 & = & 8 \times slant height + 16 \\ 96 - 16 & = & 8 \times slant height + 16 - 16 \\ 80 & = & 8 \times slant height \\ \frac{80}{8} & = & \frac{8 \times slant height}{8} \\ slant height & = & 10 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} 96 &=& 8 \times \text{slant height} + 16 \\ 96 - 16 &=& 8 \times \text{slant height} + 16 - 16 \\ 80 &=& 8 \times \text{slant height} \\ \frac{80}{8} &=& \frac{8 \times \text{slant height}} {8} \\ \text{slant height} &=& 10 \end{array}\end{align*}$
::96=8×倾斜高度+1696-16=8×倾斜高度+16-1680=8×倾斜高度808=8×倾斜高度8×倾斜高度8-8=10

The answer is 10.
::答案是10个

The slant height of the pyramid is 10 cm.
::金字塔的倾斜高度是10厘米

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about Ashanti and her crystal pyramid.
::早些时候,你被问及 Ashanti 和她晶体金字塔的问题。

Ashanti needs to use the dimensions below to calculate the surface area of her pyramid.
::Ashanti需要利用下面的维度来计算她的金字塔的表面面积。

First, find the area of the triangular base.
::首先,找到三角基点的区域。

$\begin{aligned} \begin{array}{rcl} A & = & \frac{1}{2} b h \\ A & = & \frac{1}{2} (16) (13.86) \\ A & = & 110.88 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2}bh\\ A &=& \frac{1}{2}(16)(13.86)\\ A &=& 110.88 \end{array}\end{align*}$
::A=12bhA=12(16)(13.86)A=110.88

Next, find the perimeter of the triangular base.
::接下来,找到三角基地的周边

$\begin{aligned} \begin{array}{rcl} P & = & 3 s \\ P & = & 3 (16) \\ P & = & 48 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P &=& 3s\\ P &=& 3(16)\\ P &=& 48 \end{array}\end{align*}$
::P=3sP=3(16)P=48

Then, use the surface area formula to calculate the surface area.
::然后,使用表面积公式来计算表面积。

$\begin{aligned} \begin{array}{rcl} S A & = & \frac{1}{2} \times perimeter \times slant height + B \\ S A & = & \frac{1}{2} \times 48 \times 13.86 + 110.88 \\ S A & = & 443.52 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ SA &=& \frac{1}{2} \times 48 \times 13.86 + 110.88 \\ SA &=& 443.52 \end{array}\end{align*}$
::SA=12x外围计x倾斜高度+BSA=12x48x13.86+110.88SA=443.52

The answer is 443.52.
::答案是443.52。

The surface area of the pyramid is 443.52 cm ² .
::金字塔的表面面积是443.52平方厘米。

Example 2
::例2

What is the surface area of the pyramid below?
::金字塔下面的面积是多少?

First, find the area of the square base.
::首先,找到基地的面积

$\begin{aligned} \begin{array}{rcl} A & = & s^{2} \\ A & = & 8^{2} \\ A & = & 64 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} A &=& s^2\\ A &=& 8^2\\ A &=& 64 \end{array}\end{align*}$
::A=S2A=82A=64

Next, find the perimeter of the square base.
::接下来,找到广场的周边

$\begin{aligned} \begin{array}{rcl} P & = & 4 s \\ P & = & 4 (8) \\ P & = & 32 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P &=& 4s\\ P &=& 4(8)\\ P &=& 32 \end{array}\end{align*}$
::P=4sP=4(8)P=32

Then, use the surface area formula to calculate the surface area.
::然后,使用表面积公式来计算表面积。

$\begin{aligned} \begin{array}{rcl} S A & = & \frac{1}{2} \times perimeter \times slant height + B \\ S A & = & \frac{1}{2} \times 32 \times 13.6 + 64 \\ S A & = & 281.6 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ SA &=& \frac{1}{2} \times 32 \times 13.6 + 64 \\ SA &=& 281.6\end{array}\end{align*}$
::SA=12x外围计x倾斜高度+BSA=12x32x13.6+64SA=281.6

The answer is 281.6.
::答案是281.6。

The surface area of the pyramid is 281.6 in ² .
::金字塔的表面面积是281.6英寸2。

Example 3
::例3

Find the surface area of a square pyramid with side of 8 in and slant height of 9 in.
::找到方形金字塔的表面面积,面部8英寸,倾斜高度9英寸。

First, find the area of the square base.
::首先,找到基地的面积

$\begin{aligned} \begin{array}{rcl} A & = & s^{2} \\ A & = & 8^{2} \\ A & = & 64 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} A &=& s^2\\ A &=& 8^2\\ A &=& 64 \end{array}\end{align*}$
::A=S2A=82A=64

Next, find the perimeter of the square base.
::接下来,找到广场的周边

$\begin{aligned} \begin{array}{rcl} P & = & 4 s \\ P & = & 4 (8) \\ P & = & 32 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P &=& 4s\\ P &=& 4(8)\\ P &=& 32 \end{array}\end{align*}$
::P=4sP=4(8)P=32

Then, use the surface area formula to calculate the surface area.
::然后,使用表面积公式来计算表面积。

$\begin{aligned} \begin{array}{rcl} S A & = & \frac{1}{2} \times perimeter \times slant height + B \\ S A & = & \frac{1}{2} \times 32 \times 9 + 64 \\ S A & = & 208 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ SA &=& \frac{1}{2} \times 32 \times 9 + 64 \\ SA &=& 208 \end{array}\end{align*}$
::SA=12x外围计x倾斜高度+BSA=12x32x9+64SA=208

The answer is 208.
::答案是208

The surface area of the pyramid is 208 in ² .
::金字塔的表面面积是208英寸2。

Example 4
::例4

Find the surface area of a rectangular pyramid with a length of 6 in, a width of 4 in and a slant height of 3 in.
::寻找长于 6 英寸、宽于 4 英寸、斜高为 3 英寸的矩形金字塔的表面面积。

First, find the area of the rectangular base.
::首先,找到矩形基地的区域。

$\begin{aligned} \begin{array}{rcl} A & = & l w \\ A & = & (6) (4) \\ A & = & 24 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} A &=& lw\\ A &=& (6)(4)\\ A &=& 24 \end{array}\end{align*}$
::A=lwA=(6)(4)A=24

Next, find the perimeter of the rectangular base.
::接下来,找到长方形基地的周边

$\begin{aligned} \begin{array}{rcl} P & = & 2 l + 2 w \\ P & = & 2 (6) + 2 (4) \\ P & = & 20 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P &=& 2l+2w\\ P &=& 2(6)+2(4)\\ P &=& 20 \end{array}\end{align*}$
::P=2l+2wP=2(6)+2(4)P=20

Then, use the surface area formula to calculate the surface area.
::然后,使用表面积公式来计算表面积。

$\begin{aligned} \begin{array}{rcl} S A & = & \frac{1}{2} \times perimeter \times slant height + B \\ S A & = & \frac{1}{2} \times 20 \times 3 + 24 \\ S A & = & 54 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ SA &=& \frac{1}{2} \times 20 \times 3 + 24 \\ SA &=& 54 \end{array}\end{align*}$
::SA=12x外围计x倾斜高度+BSA=12x20x3+24SA=54

The answer is 54.
::答案是54。

The surface area of the pyramid is 54 in ² .
::金字塔的表面面积是54英寸2。

Example 5
::例5

Find the surface area of a square pyramid with side of 6 cm, slant height of 5 cm.
::找到方形金字塔的表面面积,面部6厘米,斜高5厘米。

First, find the area of the square base.
::首先,找到基地的面积

$\begin{aligned} \begin{array}{rcl} A & = & s^{2} \\ A & = & 6^{2} \\ A & = & 36 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} A &=& s^2\\ A &=& 6^2\\ A &=& 36 \end{array}\end{align*}$
::A=S2A=62A=36

Next, find the perimeter of the square base.
::接下来,找到广场的周边

$\begin{aligned} \begin{array}{rcl} P & = & 4 s \\ P & = & 4 (6) \\ P & = & 24 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P &=& 4s\\ P &=& 4(6)\\ P &=& 24 \end{array}\end{align*}$
::P=4sP=4(6)P=24

Then, use the surface area formula to calculate the surface area.
::然后,使用表面积公式来计算表面积。

$\begin{aligned} \begin{array}{rcl} S A & = & \frac{1}{2} \times perimeter \times slant height + B \\ S A & = & \frac{1}{2} \times 24 \times 5 + 36 \\ S A & = & 96 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} SA &=& \frac{1}{2} \times \text{perimeter} \times \text{slant height} + B \\ SA &=& \frac{1}{2} \times 24 \times 5 + 36 \\ SA &=& 96 \end{array}\end{align*}$
::SA=12x外围计x倾斜高度+BSA=12x24x5+36SA=96

The answer is 96.
::答案是96。

The surface area of the pyramid is 96 cm ² .
::金字塔的表面面积为96厘米2。

Review
::回顾

Answer the questions below each of the diagrams.
::回答每个图表下面的问题。

1. What is the name of the figure represented by this net?
::1. 这一净额所代表数字的名称是什么?

2. What is the length of each of the sides of the base?
::2. 基地两侧的长度是多少?

3. What is the surface area of the figure?
::3. 数字的表面积是多少?

4. What is the name of this figure?
::4. 这个数字叫什么名字?

5. What is the shape of the base?
::5. 基地的形状是什么?

6. How many faces does this figure have?
::6. 这个数字有多少面孔?

7. What is the surface area of this figure?
::7. 这一数字的表面积是多少?

8. What is the name of the figure represented by this net?
::8. 这一净额所代表数字的名称是什么?

9. What is the length of each of the sides of the base?
::9. 基地两侧的长度是多少?

10. What is the surface area of the figure?
::10. 数字的表面积是多少?

11. What is the name of this figure? (Hint the base is an equilateral triangle)
::11. 这个数字叫什么名字? (基数是一个等边三角形)

12. What is the length of each of the sides of the base?
::12. 基地两侧的长度是多少?

13. What is the surface area of the figure? Round your answer to the nearest hundredths place.
::13. 图的表面面积是多少?

14. True or false. The B in the formula for surface area of a pyramid stands for bottom.
::14. 真实的或假的,金字塔表面积公式中的B代表底部。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Resources
::资源

Section outline

General

金字石表面面积

Surface Area ::地表地区

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Resources ::资源

Surface Area
::地表地区

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)

Resources
::资源