10.8 金字塔表面面积

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  金字石表面面积

  

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  Janice and Julian want to make a tent that is in the shape of a square pyramid for their summer camping adventure. They want it to have a base of 6 feet wide and a slant height of 8 feet. They need to find out the of the tent so they know how much canvas to buy. How can they calculate the surface area of their future tent?
  ::Janice和Julian想造一个帐篷,这个帐篷的形状是他们的暑期露营冒险的平方金字塔形。 他们想要一个6英尺宽的底座和8英尺高的斜坡高度。 他们需要找到帐篷, 以便知道要买多少画布。 他们如何计算未来帐篷的表面面积?

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  In this concept, you will learn how to find the surface area of a pyramid. 
  ::在这个概念中,你会学会如何找到金字塔的表面区域。

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  Finding the Surface Area of a Pyramid
  ::寻找金字塔的表面区域

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  A pyramid has sides that are triangular faces and a base. The base can be any shape. Here are three examples:
  ::金字塔的侧面是三角面和基底。 基底可以是任何形状。 以下三个例子 :

  

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  The surface area is the sum of all of the areas of the faces in a . Imagine wrapping 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.
  ::表面区域是面部所有区域的总和 。 想象一下将金字塔包裹在包装纸中, 像一个礼物。 覆盖图所需的包装纸量代表其表面区域。 要找到表面区域, 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 。

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  One way to find the surface area of a three-dimensional figure is by using a . A net is a two-dimensional diagram of a three-dimensional figure. Imagine unfolding a pyramid so that it is completely flat. It would look something like this net.
  ::找到三维图的表面区域的一个方法就是使用 。 网是一个三维图的二维图。 想象一个金字塔的展开, 它会完全平坦。 它会看起来像这个网。

  

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  With a net, you can see each face of the pyramid more clearly. To find the surface area, you need to calculate the area for each face of the net, the sides and the base.
  ::使用网, 您可以更清楚地看到金字塔的每一面。 要找到表面积, 您需要计算网的每个面、 边和基座的面积 。

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  The side faces of a pyramid are always triangles, so the area formula for triangles is used to calculate their areas: $\begin{array}{r}A=\frac{1}{2}bh\end{array}$ .
  ::金字塔的侧面总是三角形,所以三角形的区域公式用来计算其区域:A=12bh。

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  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 . Use whichever area formula is appropriate for the shape. Here are some common formulas to find area.
  ::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的基座。 使用适合形状的任意区域公式。 这里有一些要找到区域的通用公式 。

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  $$\begin{array}{rl}\text{rectangle}& :A=lw\\ \text{square}& :A=s^2\\ \text{triangle}& :A=\frac{1}{2}bh\end{array}$$

  
  ::矩形: A=lwsquare: A=s2三角形: A=12bh

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  Take a look at the pyramid below. Which formula should you use to find the area of the base?
  ::看看下方的金字塔。 您应该使用哪种公式来找到基座的面积 ?

  

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  The base of the pyramid is a square with sides of 6 centimeters, so you should use the area formula for a square: $\begin{array}{r}A=s^2\end{array}$ . Now find the area of the base.
  ::金字塔的底部是一个方形,边边边为6厘米,所以您应该使用方形的区域公式:A=s2。现在找到基座的面积。

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  $$\begin{array}{rl}A& =6^2\\ A& =36sq.cm\end{array}$$

  
  ::A=62A=36平方公里

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  Next, find the area of one of the triangles. You use the formula $\begin{array}{r}A=\frac{1}{2}bh\end{array}$ . The base width is 6 cm and the slant height, the height of the side is 4 cm.
  ::下一步,找到其中三角形之一的区域。您使用公式A=12bh。基宽为6厘米,斜坡高度为4厘米。

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  $$\begin{array}{rl}A& =\frac{1}{2}(6)(4)\\ A& =\frac{1}{2}(24)\\ A& =12sq.cm\end{array}$$

  
  ::A=12(6)(4)A=12(24)A=12平方米

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  There are four triangles, so multiply this number by four and then add it to the area of the square.
  ::有四个三角形, 所以将这个数字乘以 4, 然后将其添加到方形的区域 。

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  $$\begin{array}{rl}SA& =4(12)+36\\ SA& =48+36\\ SA& =84sq.cm\end{array}$$

  
  ::SA=4(12)+36SA=48+36SA=84 sq.cm

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  Nets let you see each face of a pyramid so that you can calculate its area. However, you can also use a  formula to represent the surface area of a pyramid. The formula is like a short cut because you can put all of the measurements in for the appropriate variable in the formula and solve for $\begin{array}{r}SA\end{array}$ , surface area.
  ::网让您看到金字塔的每个面孔,以便您可以计算其面积。然而,您也可以使用一个公式来代表金字塔的表面面积。公式就像一个捷径,因为您可以将所有测量值都放入公式中的适当变量中,并解决 SA, 表面面积。

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  Here is the formula for finding the surface area of a pyramid.
  ::这是找到金字塔表面面积的公式

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  $$\begin{array}{r}SA=\frac{1}{2}\text{perimeter}\times \text{slant height}+B\end{array}$$

  
  ::SA=12 周边X倾斜高度+B

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  The first part of the formula, $\begin{array}{r}\frac{1}{2}\text{perimeter}\times \text{slant height}\end{array}$ , is a quick way of finding the area of all of the triangular sides of the pyramid at once. Remember, the area formula for a triangle is $\begin{array}{r}A=\frac{1}{2}bh\end{array}$ . In the formula, $\begin{array}{r}b\end{array}$ stands for base. The perimeter of the pyramid’s bottom face represents all of the bases of the triangular faces at once, because it’s their sum. The height of each triangle is always the same; this is called the slant height of the pyramid.  Therefore “ $\begin{array}{r}\frac{1}{2}\text{perimeter}\times \text{slant height}\end{array}$ ” is really the same as $\begin{array}{r}\frac{1}{2}bh\end{array}$ .
  ::公式的第一部分, 12 个周边X倾斜高度, 是同时找到金字塔所有三角边的面积的快速方法 。 记住, 三角形的面积公式是 A= 12bh。 在公式中, b 表示基数 。 金字塔底面的周边代表了三角面的所有底部, 因为它是它们的总数 。 每个三角形的高度总是一样; 这被称为金字塔的斜高度 。 因此, “ 12 周边x倾斜高度” 与 12bh 完全相同 。

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  The $\begin{array}{r}B\end{array}$ in the formula represents the area of the base. Remember, pyramids can have bases of different shapes, so the area formula is used to find $\begin{array}{r}B\end{array}$ varies. You find the area of the base first and then put it into the formula in place of $\begin{array}{r}B\end{array}$ .
  ::公式中的 B 表示基数的面积。 记住, 金字塔可以有不同形状的基数, 因此区域公式用来查找 B 不同 。 您先找到基数的面积, 然后将其放入公式, 而不是 B 。

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  Let's look at an example.
  ::让我们举个例子。

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

  

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  This is a square pyramid. The four sides of the base are all 8 inches, so the perimeter of the base is $\begin{array}{r}8\times 4=32inches\end{array}$ . You will need to use the area formula for a square to find $\begin{array}{r}B\end{array}$ .
  ::这是平方金字塔。 基底的四边都是8英寸, 所以基底的周界是8×4=32英寸。 您需要使用面积公式才能找到 B 。

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  $$\begin{array}{rl}B& =s^2\\ B& =(8{)}^2\\ B& =64in{.}^2\end{array}$$

  
  ::B=S2B=(8)2B=64,2英寸

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  Now you have all the information that you need. Put it into the formula and solve for $\begin{array}{r}SA\end{array}$ , surface area.
  ::现在您掌握了您需要的所有信息。 将其输入公式中, 并解决 SA, 表面区域 。

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  $$\begin{array}{rl}SA& =\frac{1}{2}\text{perimeter}\times \text{slant height}+B\\ SA& =\left[\frac{1}{2}(32)\times 3\right]+64\\ SA& =(16\times 3)+64\\ SA& =48+64\\ SA& =112in{.}^2\end{array}$$

  
  ::SA=12 周边X倾斜高度+BSA=[12(32)x3]+64SA=(16x3)+64SA=48+64SA=112英寸

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  The answer is the surface area of the pyramid is 112 square inches.
  ::答案是金字塔的表面面积是112平方英寸

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were given a problem about Janice and Julian and their square pyramid tent.
  ::早些时候,你得到一个问题 珍妮丝和朱利安 和他们的平方金字塔帐篷。

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  They are planning for their tent to have a base of 6 feet wide and a slant height of 8 feet.  How much canvas will they need?
  ::他们正在计划他们的帐篷有6英尺宽的底座和8英尺高的斜坡,他们需要多少帆布?

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  First, calculate the area and perimeter of the square base:
  ::首先,计算平方基地的面积和周边:

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  $$\begin{array}{rl}B& =s^2\\ B& =6^2\\ B& =36{ft}^2\\ \\ P& =4\times 6\\ P& =24ft\end{array}$$

   
  ::B=S2B=62B=36平方英尺=4×6P=24平方英尺

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  Next, plug the values into the formula for surface area of a pyramid and multiply the values together inside the brackets:
  ::下一步,将数值插入金字塔表面积的公式中,并将括号内的数值相乘:

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  $$\begin{array}{rl}SA& =\frac{1}{2}\text{perimeter}\times \text{slant height}+B\\ SA& =\left[\frac{1}{2}(24)\times 8\right]+36\end{array}$$

  
  ::SA=12 周边x倾斜高度+BSA=[12(24)x8]+36

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  $$\begin{array}{rl}SA& =(12\times 8)+36\\ SA& =96+36\end{array}$$

  
  ::SA=(12x8)+36SA=96+36

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  Then, add the values together for the answer, making sure to include the appropriate unit of measurement:
  ::然后,将答案的数值加在一起,确保包括适当的计量单位:

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  $$\begin{array}{rl}SA& =96+36\\ SA& =132f{t}^2\end{array}$$

  
  ::SA=96+36SA=132平方英尺

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  The answer is the surface area of the tent is 132 square feet.  Janice and Julian will require 132 square feet of canvas to make their tent.
  ::答案是帐篷的面积是132平方英尺。Janice和Julian需要132平方英尺的帆布才能搭建帐篷。

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  Example 2
  ::例2

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  Find the surface area of the figure below.
  ::查找下图的表面区域。

  

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  This is a triangular pyramid because its base is a triangle. That means you need to use the area formula for triangles to find $\begin{array}{r}B\end{array}$ .  First, calculate the area of the base and the perimeter. To do this, identify the shape of the base and use the appropriate formula for  $\begin{array}{r}B\end{array}$ : 
  ::这是一个三角金字塔, 因为它的底部是一个三角形。 这意味着您需要使用三角形的面积公式来找到 B。 首先, 计算底部和周边的面积。 要做到这一点, 确定底部的形状, 并对 B 使用合适的公式 :

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  $$\begin{array}{rl}B& =\frac{1}{2}bh\\ B& =\frac{1}{2}(16)(13.86)\\ B& =8(13.86)\\ B& =110.88c{m}^2\end{array}$$

  
  ::B=12bhB=12(16)(13.86)B=8(13.86)B=110.88厘米2

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  The sides of the base are all the same length, so you can calculate the perimeter by multiplying 
  ::基底的两侧长度相同,所以您可以通过乘法计算周界

  $\begin{array}{r}16\times 3=48\end{array}$ .

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  Next, plug the values into the formula for surface area of a pyramid and multiply the values together inside the brackets:
  ::下一步,将数值插入金字塔表面积的公式中,并将括号内的数值相乘:

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  $$\begin{array}{rl}SA& =\frac{1}{2}\text{perimeter}\times \text{slant height}+B\\ SA& =\left[\frac{1}{2}(48)\times 6\right]+110.88\end{array}$$

  
  ::SA=12 周边X倾斜高度+BSA=[12(48)x6]+110.88

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  $$\begin{array}{rl}SA& =(24\times 6)+110.88\\ SA& =144+110.88\end{array}$$

  
  ::SA=(24x6)+110.88SA=144+110.88

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  Then, add the values together, making sure to include the appropriate unit of measurement:
  ::然后,将数值加在一起,确保包括适当的计量单位:

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  $$\begin{array}{rl}SA& =144+110.88\\ SA& =254.88c{m}^2\end{array}$$

  
  ::SA=144+110.88SA=254.88 cm2

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  The answer is the surface area of this triangular pyramid is 254.88 square centimeters.
  ::答案是这个三角金字塔的表面面积是254.88平方厘米。

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  Example 3
  ::例3

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  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。

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  First, calculate the area and perimeter of the square base: 
  ::首先,计算平方基地的面积和周边:

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  $$\begin{array}{rl}B& =s^2\\ B& =6^2\\ B& =36{cm}^2\\ \\ P& =4\times 6\\ P& =24cm\end{array}$$

   
  ::B=S2B=62B=36B=36cm2P=4x6P=24厘米

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  Next, plug the values into the formula for surface area of a pyramid and multiply the values together inside the brackets:

  $$\begin{array}{rl}SA& =\frac{1}{2}\text{perimeter}\times \text{slant height}+B\\ SA& =\left[\frac{1}{2}(24)\times 5\right]+36\end{array}$$

   
  

  $$\begin{array}{rl}SA& =(12\times 5)+36\\ SA& =60+36\end{array}$$

  Then, add the values together for the answer, remembering the appropriate unit of measurement.

  $$\begin{array}{rl}SA& =60+36\\ SA& =96c{m}^2\end{array}$$

  
  ::下一步,将数值插入金字塔表面面积的公式中,并将括号内值相乘:SA=12 周边x倾斜高度+BSA=[12(24)xx5]+36 SA=(12x5)+36SA=60+36 然后,将答案的值相加,记住适当的计量单位。SA=60+36SA=96cm2

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  The answer is the surface area of the square pyramid is   $\begin{array}{r}96c{m}^2\end{array}$ .
  ::答案是方形金字塔的表面面积是96厘米2。

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  Example 4
  ::例4

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  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。

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  First, calculate the area and perimeter of the rectangular base:
  ::首先,计算矩形基的面积和周界:

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  $$\begin{array}{rl}B& =lw\\ B& =6\times 4\\ B& =24{in}^2\\ \\ P& =2l+2w\\ P& =2(6)+2(4)\\ P& =12+8\\ P& =20in\end{array}$$

  
  ::B=lwB=6x4B=24 in2P=2l+2wP=2(6)+2(4)P=12+8P=20

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  Next, plug the values into the formula for surface area of a pyramid and multiply the values together inside the brackets:
  ::下一步,将数值插入金字塔表面积的公式中,并将括号内的数值相乘:

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  $$\begin{array}{rl}SA& =\frac{1}{2}\text{perimeter}\times \text{slant height}+B\\ SA& =\left[\frac{1}{2}(20)\times 3\right]+24\end{array}$$

   
  

  $$\begin{array}{rl}SA& =(10\times 3)+24\\ SA& =30+24\end{array}$$

  Then, add the values together for the answer.  Remember the appropriate unit of measurement.
  ::SA=12 周界x倾斜高度+BSA=[12(20)x3)+24 SA=(10x3)+24 SA=(10x3)+24SA=30+24 然后,将答案的值加在一起。记住适当的计量单位。

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  $$\begin{array}{rl}SA& =30+24\\ SA& =54i{n}^2\end{array}$$

  
  ::SA=30+24SA=54,2英寸

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  The answer is  $\begin{array}{r}54i{n}^2\end{array}$ .
  ::答案是54英寸2

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  Example 5
  ::例5

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  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。

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  First, calculate the area and perimeter of the square base.  

  $$\begin{array}{rl}B& =s^2\\ B& =8^2\\ B& =64{in}^2\\ \\ P& =4\times 8\\ P& =32in\end{array}$$

  Next, plug the values into the formula for surface area of a pyramid and multiply the values together inside the brackets:

  $$\begin{array}{rl}SA& =\frac{1}{2}\text{perimeter}\times \text{slant height}+B\\ SA& =\left[\frac{1}{2}(32)\times 9\right]+64\end{array}$$

   
  

  $$\begin{array}{rl}SA& =(16\times 9)+64\\ SA& =144+64\end{array}$$

  
  ::首先,计算平方基的面积和周边。 B=s2B=82B=64in2P=4x8P=32 下一步,将数值插入金字塔表面积的公式中,并将括号内值相乘:SA=12 周边x倾斜高度+BSA=[12(32)xx9]+64 SA=(16x9)+64SA=144+64

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  Then, add the values together for the answer, remembering the appropriate unit of measurement:
  ::然后,将答案的数值加在一起,记住适当的计量单位:

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  $$\begin{array}{rl}SA& =144+64\\ SA& =208i{n}^2\end{array}$$

  
  ::SA=144+64SA=208 英寸2

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  The answer is the surface area of the square pyramid is  $\begin{array}{r}208i{n}^2\end{array}$ .
  ::答案是平方金字塔的表面面积是208英寸2。

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  Review
  ::回顾

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  Find the surface area of each square pyramid. Remember that $\begin{array}{r}b\end{array}$ means base and $\begin{array}{r}sh\end{array}$ means slant height.
  ::找到每个方形金字塔的表面积。 记住 b 意指底部和 sh 意指倾斜高度 。

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  1. $\begin{array}{r}b=4in,sh=5in\end{array}$
     ::b=4英寸,sh=5英寸
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  2. $\begin{array}{r}b=4in,sh=6in\end{array}$
     ::b=4英寸,sh=6英寸
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  3. $\begin{array}{r}b=6in,sh=8in\end{array}$
     ::b=6英寸,sh=8英寸
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  4. $\begin{array}{r}b=5in,sh=7in\end{array}$
     ::b=5英寸,sh=7英寸
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  5. $\begin{array}{r}b=7m,sh=9m\end{array}$
     ::b=7米,sh=9米
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  6. $\begin{array}{r}b=8m,sh=10m\end{array}$
     ::b=8米,sh=10米
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  7. $\begin{array}{r}b=9cm,sh=11cm\end{array}$
     ::b=9厘米,sh=11厘米
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  8. $\begin{array}{r}b=11m,sh=13m\end{array}$
     ::b=11米,sh=13米
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  9. $\begin{array}{r}b=6in,sh=9in\end{array}$
     ::b=6英寸,sh=9英寸
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  10. $\begin{array}{r}b=10cm,sh=12cm\end{array}$
      ::b=10厘米,sh=12厘米

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  Find the surface area of each rectangular pyramid.
  ::找到每个矩形金字塔的表面区域。

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  11. $\begin{array}{r}l=10in,w=8in,sh=6in\end{array}$
      ::l=10英寸,w=8英寸,sh=6英寸
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  12. $\begin{array}{r}l=12ft,w=8ft,sh=10ft\end{array}$
      ::l=12英尺,w=8英尺,sh=10英尺
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  13. $\begin{array}{r}l=4in,w=2in,sh=3in\end{array}$
      ::l=4 英寸,w=2 英寸,sh=3 英寸
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  14. $\begin{array}{r}l=16m,w=12m,sh=11m\end{array}$
      ::l=16米,w=12米,sh=11米
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  15. $\begin{array}{r}l=15in,w=10in,sh=12in\end{array}$
      ::l=15,w=10,sh=12,in

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  Review (Answers)
  ::回顾(答复)

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