1.7 毕达哥拉斯理论(毕达哥里安理论) -- -- 公式、证据、互动和实例-interactive

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• 选择节毕达哥拉斯理论(毕达哥里安理论) - 公式、 校对、 互动和实例

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  毕达哥拉斯理论(毕达哥里安理论) - 公式、 校对、 互动和实例

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  What is the Pythagoras Theorem?
  ::毕达哥拉斯神话是什么?

  The Pythagorean Theorem , also known as the Pythagoras Theorem, is one of the most fundamental theorems in mathematics and it defines the relationship between the three sides of a right-angled triangle . In a right-angled triangle, one of the angles is $\begin{align*}90^\circ,\end{align*}$ and the side that is opposite to that $\begin{align*}90^\circ\end{align*}$ (right) angle is known as the hypotenuse . The other two sides that are adjacent to the right angle are called the legs of the triangle. 

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  Let $\begin{align*}a=3\end{align*}$ , $\begin{align*}b=4\end{align*}$ be the legs and $\begin{align*}c=5\end{align*}$ be the hypotenuse of a right triangle .
  ::a=3,b=4为腿,c=5为右三角形的下限。

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  The Pythagoras Theorem states that for right triangles with legs of lengths $\begin{align*}a\end{align*}$ and $\begin{align*}b\end{align*}$ and hypotenuse of length $\begin{align*}c\end{align*}$ , $\begin{align*}a^2\end{align*}$ $\begin{align*}+\end{align*}$ $\begin{align*}b^2\end{align*}$ $\begin{align*}=\end{align*}$ $\begin{align*}c^2\end{align*}$ .
  ::《毕达哥拉斯理论论》指出,右三角形的长度为a和b,长度为c的直角三角形为a2+b2=c2。

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  The geometric interpretation of the Pythagorean theorem states that the sum of the area with side $\begin{align*}a,\end{align*}$ and the area of the square with side $\begin{align*}b,\end{align*}$ is equal to the area of the square with side $\begin{align*}c.\end{align*}$
  ::对Pythagorean定理的几何解释表明,带a侧的区域和有b侧的广场区域的总和等于有c侧的广场区域。

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  Geometric Proof of the Pythagoras Theorem (Pythagorean Theorem)
  ::毕达哥拉斯理论(毕达哥里安理论)的几何校验

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  There are many different . The following picture leads to one of those proofs.
  ::有许多不同之处。下面的景象引出其中一种证据。

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  Rearranging the triangles , we can also form another square with the same side length $\begin{align*}(a + b).\end{align*}$ Hence, this geometric proof shows that the area of the square with side $\begin{align*}c\end{align*}$ is equal to the sum of the areas of the squares with side $\begin{align*}a\end{align*}$ and side $\begin{align*}b\end{align*}$ .
  ::重新排列三角形, 我们也可以形成另一平方, 侧长相同( a+b ) 。 因此, 这个几何校验显示, 带有侧 c 的正方区域等于 侧 a 和 侧 b 的正方区域之和 。

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  Algebraic Proof of the Pythagoras Theorem (Pythagorean Theorem)
  ::毕达哥拉斯理论(毕达哥里安理论)的代数证明

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  First, you can verify that the quadrilateral in the center is a square . All sides are the same length and each angle must be $\begin{align*}90^\circ\end{align*}$ . The angles must be $\begin{align*}90^\circ\end{align*}$ because the three angles that make a straight angle at each corner of the interior quadrilateral are the same three angles that make up each of the triangles. This relationship is shown with the angle markings in the picture below.
  ::首先, 您可以验证中间的四边形是一个正方形。 方形的长度相同, 每个角度必须是 90 。 角度必须是 90 。 因为内部四边形每个角的直角的三个角度是构成每个三角形的相同三个角度。 此关系会显示在下图中的角度标记 。

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  In order to prove the Pythagorean Theorem, find the area of the interior square in two ways. First, find the area directly:
  ::为了证明毕达哥里安神话 以两种方式找到内方

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  $\begin{align*}\text{Area of square }= \end{align*}$ $\begin{align*}c^2\end{align*}$
  ::平方面积=c2

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  Next, find the area as the difference between the area of the large square and the area of the triangles .
  ::接下来,找到这个区域作为大广场面积和三角形面积之间的差额。

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  $\begin{align*}\text{Area of square} =\end{align*}$ $\begin{align*}(a+b)^2-4 \left(\frac{1}{2} ab \right)\end{align*}$
  ::平方面积=(a+b)2-4(12ab)

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  Since you are referring to the same square each time, those two areas must be equal.
  ::因为每次您都指同一个广场,所以这两个地区必须是平等的。

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  Converse of the Pythagorean Theorem (Converse of the Pythagoras Theorem)
  ::与毕达哥拉斯神话的对立

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  The is also true. The converse switches the "if" and "then" parts of the theorem. The converse says that if $\begin{align*}a^2+b^2=c^2\end{align*}$ , then the triangle is a right triangle.
  ::情况也是一样。 对应方转换了理论的“ 如果” 和“ 那么” 部分。 对应方表示, 如果a2+b2=c2, 那么三角形是一个右三角形 。

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  With the , you can solve many types of problems. You can:
  ::有了这些,你可以解决很多类型的问题。你可以:

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  1. Find the missing side of a right triangle when you know the other two sides.
     ::找到右三角形缺失的一面 当你认识另外两边时
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  2. Determine whether a triangle is right, acute, or obtuse.
     ::确定三角形是否正确、 急性或隐蔽 。
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  3. Find the distance between two points.
     ::查找两点之间的距离。

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  Finding the Length of the Hypotenuse using the Pythagoras Theorem
  ::使用 Pythagoras 理论查找假体长度

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  The two legs of a right triangle have lengths 3 and 4. What is the length of the hypotenuse?
  ::右三角形的两条腿有3和4的长度。 下限的长度是多少?

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  Because it is a right triangle, you can use the Pythagoras (Pythagorean) Theorem. It doesn't matter whether you assign $\begin{align*}a\end{align*}$ as 3 or $\begin{align*}b\end{align*}$ as 3.
  ::右三角形为右三角形, 您可以使用 Pytagoras (Pytagorena) 定理。 您是否指定 3 或 b 3 并不重要 。

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  Because length must be positive, the hypotenuse has a length of 5. Side lengths of 3, 4 and 5 are common in geometry. You should remember that they are the lengths of a right triangle. Triples of whole numbers that satisfy the Pythagorean Theorem are called . "3, 4, 5" is an example of a Pythagorean triple .
  ::由于长度必须是正数, 下限长度为5 。 在几何中, 侧长度为 3, 4 和 5 是常见的 。 您应该记住它们是右三角的长度 。 满足 Pytagoren Theorem 的全数的三重数字被称为 " 3, 4, 5" 。 这是 Pytagorean 三重的一个例子 。

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  Classifying Triangles using the Pythagorean Theorem
  ::使用 Pythagorean 定義定義三角形

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  A triangle has side lengths of 4, 8 and 9. What type of triangle is this?
  ::三角形的侧边长度为4、8和9,这是哪种三角形?

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  If the numbers satisfy the Pythagorean Theorem (in other words, if the lengths of the sides form a Pythagorean Triple), then it is a right triangle. If $\begin{align*}a^2+b^2 > c^2\end{align*}$ , then $\begin{align*}c\end{align*}$ is shorter than it would be in a right triangle, so the angle opposite it is smaller and it is an acute triangle . If $\begin{align*}a^2+b^2 < c^2\end{align*}$ , then $\begin{align*}c\end{align*}$ is longer than it would be in a right triangle, so the angle opposite it is larger and it is an obtuse triangle . 
  ::如果数字满足了 Pythagorean Theorem (换句话说,如果两边的长度组成了 Pytagorean Triple) , 那么它是一个右三角。 如果 a2+b2>c2, 那么 c 的长度比右三角的长度要短, 所以对角小, 它是一个急性三角。 如果 a2+b2 < c2, 那么 c 的长度比右三角的长, 所以对角的角更大, 它是一个隐形三角 。

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  In this case, $\begin{align*}4^2+8^2=80\end{align*}$ and $\begin{align*}9^2 = 81\end{align*}$ . So $\begin{align*}a^2+b^2 < c^2\end{align*}$ , hence the triangle is obtuse.
  ::在此情况下, 42+82=80和 92=81。 所以 a2+b2 <c2, 所以三角形是模糊的 。

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  CK-12 Interactive - The Pythagorean Theorem
  ::CK-12 互动 - 毕达哥里安定理

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  Derive Distance Formula using the Pythagoras Theorem
  ::使用 Pytagoras 定理的带宽距离公式

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  Use the Pythagorean Theorem to derive the : $\begin{align*}d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\end{align*}$ .
  ::使用 Pythagorean 定理来产生 : d( x2- x1) 2+ (y2-y1) 2。

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  It can help to draw in a right triangle, where  $\begin{align*}d\end{align*}$ is the hypotenuse. Then, find the lengths of the sides in terms of $\begin{align*}x_1\end{align*}$ , $\begin{align*}y_1\end{align*}$ , $\begin{align*}x_2\end{align*}$ , $\begin{align*}y_2\end{align*}$ . Finally, use the Pythagorean Theorem to find $\begin{align*}d\end{align*}$ .
  ::它可以帮助绘制右三角形, d 是下方。 然后, 找到以 x1 y1 y1, x2, y2 y2 值为边的长度 。 最后, 使用 Pytagorean 理论来找到 d 。

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  CK-12 PLIX Interactive - The Pythagorean Theorem
  ::CK-12 PLIX 互动 - 毕达哥里安理论

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  Examples - The Pythagoras Theorem
  ::实例 - 毕达哥拉斯理论

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  Example 1
  ::例1

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  The lengths of the three sides of the triangle are 5, 6 and 10. Is this a right triangle?
  ::三角形三边的长度是5、6和10 这是右三角形吗?

  

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  The lengths of the three sides of the triangle are 5, 6, and 10.
  ::三角形三边的长度是5、6和10

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  $\begin{align*}5^2 + 6^2 & = 61 \\ 10^2 & = 100 \\ 61 & < 100 \\\end{align*}$
  $\begin{align*}a^2+b^2 < c^2\end{align*}$ so this is not a right triangle, it is an obtuse triangle.
  ::52+62=61102=10061<100 a2+b2<c2,所以这不是右三角,而是隐形三角。

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  Example 2
  ::例2

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  Will a multiple of a Pythagorean triple always also be a Pythagorean triple? For example, "6, 8, 10" is a multiple of "3, 4, 5". Is "6, 8, 10" a Pythagorean triple? Is any multiple of "3, 4, 5" (or any other Pythagorean triple) also a Pythagorean triple?
  ::毕达哥里安三重力的倍数是否也总是毕达哥里安三重力?例如,“6,8,10”是“3,4,5”的倍数。“6,8,10”是毕达哥里安三重力的倍数吗?“3、4,5”的倍数是否也是毕达哥里安三重力?

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  Yes. Assume " $\begin{align*}a\end{align*}$ , $\begin{align*}b\end{align*}$ , $\begin{align*}c\end{align*}$ " is a Pythagorean triple, so $\begin{align*}a^2+b^2=c^2\end{align*}$ . " $\begin{align*}ka\end{align*}$ , $\begin{align*}kb\end{align*}$ , $\begin{align*}kc\end{align*}$ " where $\begin{align*}k\end{align*}$ is a whole number is a multiple of this Pythagorean triple.
  ::是的, 假设“ a, b, c” 是毕达哥林的三倍, 所以a2+b2=c2. “ k, kb, kc” K是整数是比达哥林的三倍的倍数 。

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  $\begin{align*}(ka)^2 + (kb)^2 & = k^2a^2 + k^2b^2 \\ & = k^2(a^2 + b^2) \\ & = k^2c^2 \\ & = (kc)^2\\\end{align*}$
  :k)2+(kb)2=k2a2+k2b2=k2(a2+b2)2=k2c2=(kc)2

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  Since $\begin{align*}(ka)^2+(kb)^2=(kc)^2\end{align*}$ , " $\begin{align*}ka\end{align*}$ , $\begin{align*}kb\end{align*}$ , $\begin{align*}kc\end{align*}$ " is also a Pythagorean triple.
  ::由于 (ka) 2+( kb) 2=( kc) 2, “ ka, kb, kc” 也是毕达哥林的三联。

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  Example 3
  ::例3

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  Find the distance between $\begin{align*}(3, -4)\end{align*}$ and $\begin{align*}(-1, 5)\end{align*}$ .
  ::查找(3,-4)和(-1,5)之间的距离。

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  Use the Pythagorean Theorem or the .
  ::使用毕达哥里定理或...

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  Example 4
  ::例4

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  The length of one leg of a triangle is 5 and the length of the hypotenuse is 8. What is the length of the other leg?
  ::三角形一条腿的长度是5, 下限的长度是8。 另一条腿的长度是多少?

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  Use the Pythagorean Theorem.
  ::使用毕达哥里神话

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  $\begin{align*}a^2 + 5^2 & = 8^2 \\ a^2 + 25 & = 64 \\ a^2 & = 64 - 25 = 39 \\ a & = \sqrt{39} \approx 6.24\end{align*}$
  ::a2+52=82a2+25=64a2=64-25=39a396.24

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  CK-12 PLIX Interactive - Pythagorean Theorem and Pythagorean Triples
  ::CK-12 PLIX 互动 - 毕达哥里安理论和毕达哥里安三重奏

  Summary

  播放段落 The Pythagorean Theorem defines the relationship between the three sides of a right-angled triangle, stating that the square of the hypotenuse c is equal to the sum of the squares of the other two sides a and b: $\begin{align*}c^2 = a^2 + b^2.\end{align*}$   ::Pythagorean Theorem 定义了右三角形三侧之间的关系,指出下方c的平方等于其他两面a和b:c2=a2+b2的平方之和。 播放段落 The converse of the Pythagorean Theorem states that if $\begin{align*}c^2 = a^2 + b^2,\end{align*}$  then the triangle is a right triangle. ::Pythagorean 理论的反义词表示,如果 c2=a2+b2, 三角形是一个右三角形。 播放段落 Pythagorean triples are sets of three whole numbers that satisfy the Pythagorean Theorem, such as (3, 4, 5). Multiples of Pythagorean triples are also Pythagorean triples. ::毕达哥伦三联赛是满足毕达哥伦定理理论的三组三整数字,例如(3,4,5),比达哥伦三联赛的数数也多为毕达哥伦三联赛。 播放段落 If $\begin{align*}a^2 + b^2 > c^2,\end{align*}$  the triangle is acute. ::如果 a2+b2>c2,则三角形为急性。 播放段落 If $\begin{align*}a^2 + b^2 < c^2,\end{align*}$  the triangle is obtuse. ::如果 a2+b2 <c2, 三角形是模糊的 。

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  Review Questions on the Pythagoras Theorem (The Pythagorean Theorem)
  ::审查关于毕达哥拉斯理论(毕达哥里安理论)的问题

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  Use the Pythagorean Theorem to solve for $\begin{align*}x\end{align*}$ in each right triangle below. Round your answer to the nearest tenths place.
  ::使用 Pythagorena 理论解析下方每个右三角形的 x。 将您的答案转至最近的十分位 。

  1.

  

  2.

  

  3.

  

  4.

  

  5.

  

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  Three side lengths for triangles are given. Determine whether or not each triangle is right, acute, or obtuse.
  ::给三角形给定三个侧边长度。 确定每个三角形是否正确、 急性或模糊 。

  6. 2, 5, 6

  7. 4, 7, 8

  8. 6, 8, 10

  9. 6, 9, 10

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  Find the distance between each pair of points.
  ::查找每对点之间的距离 。

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  10. $\begin{align*}(2, 5)\end{align*}$ and $\begin{align*}(1, -3)\end{align*}$
  ::10.(2,5)和1,3)

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  11. $\begin{align*}(-4.5, 2)\end{align*}$ and $\begin{align*}(1.6, 5)\end{align*}$
  ::11. (-4.5,2)和(1.6,5)

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  12. $\begin{align*}(-3.7, 2.1)\end{align*}$ and $\begin{align*}(-3.2, -1.5)\end{align*}$
  ::12. (-3.7.2.1)和(-3.2,-1.5)

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  13. $\begin{align*}(-3, -5)\end{align*}$ and $\begin{align*}(5, 6)\end{align*}$
  ::13. (-3,-5)和(-5,6)

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  14. Find two more that are not multiples of "3, 4, 5".
  ::14. 再找两个不是"3,4,5"的倍数

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  15. Pick any two whole numbers $\begin{align*}m\end{align*}$ and $\begin{align*}n\end{align*}$ with $\begin{align*}n > m\end{align*}$ . Then $\begin{align*}n^2-m^2\end{align*}$ , $\begin{align*}2mn\end{align*}$ , and $\begin{align*}n^2+m^2\end{align*}$ will be a Pythagorean triple. Test this with a few values of $\begin{align*}n\end{align*}$ and $\begin{align*}m\end{align*}$ and then show why this process works using algebra.
  ::15. 选取任何两个整数 m 和 n, 加上 n>m。 然后 n2- m2, 2mn, 和 n2+m2 将是一个比达哥里安三重数值, 测试这个数值, 数值为 n 和 m , 然后用代数来显示这个过程为什么起作用 。

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  16. There are many ways to prove the Pythagorean Theorem. In the diagram below, $\begin{align*}\triangle BCE\end{align*}$ is given as shown. How can this triangle be rotated and translated to create $\begin{align*}\triangle DEF\end{align*}$ ? Is $\begin{align*}DFCB\end{align*}$ a trapezoid? How do you know ? Is $\begin{align*}\triangle DEB\end{align*}$ a right triangle? How do you know ? Find the area of the trapezoid two different ways. Set the results equal to each other. Then perform algebra to establish the Pythagorean Theorem. Do you prefer this proof or the one in the notes above ? Why?
  ::16. 有许多方法可以证明毕达哥里安理论。在下图中给出了 BCE 。这个三角形如何旋转和翻译来创建 DEF ? DFCB 是一个隐形体吗? 你怎么知道? DFCB 是右三角体吗? 你如何知道 ? 找到隐形体区域有两种不同的方式。 将结果设置为对等。 然后使用代数来建立 Pythagorean理论。 您更喜欢这个证据还是上面的注解? 为什么?

  

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  17. The diagram below is one visual representation of the squares of the sides of a right triangle. Explore the relationship between the area of the squares using the interactive in the discussion above. Is it true that the area of a square of side length 50 has the same area as the sum of two squares having side lengths 30 and 40? How about the area of a circle whose radius is 13, compared with two circles whose radii are 5 and 12? Use this relationship to find the radii of three different circles such that areas of the first two sum to that of the third .
  ::17. 下面的图表是右三角形两侧方形的直观表示图。 使用上文讨论中的互动讨论来探讨方形区域之间的关系; 侧长50的方形区域是否与侧长30和40的两个方形之和具有相同的区域? 半径为13的圆圈区域与半径为5和12的两个圆圈之间的区域如何? 利用这一关系找到三个不同圆圈的半径,即前两个半径至第三个圆的圆圈。

  

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  18. Find the dimensions of three cubes such that the surface areas of the first two add to the of the third.
  ::18. 找出三个立方体的尺寸,使前两个立方体的表面积增加第三个立方体的面积。

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  19. Will the relationship in the last problem hold true for the if the side lengths of the cubes are integers? Why or why not?
  ::19. 如果立方体的侧长度是整数,最后一个问题中的关系是否会维持下去?为什么或为什么没有?

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  20. In the diagram below there is a line with two triangles. Find the dimensions of the large one. Use these to find the dimensions of the smaller one. Round the value for the hypotenuse to the hundredths place. Now imagine creating more triangles, increasing x by 1 at a time. For every increase of x by 1, how much does the hypotenuse increase by? How long would the hypotenuse be for an x of 29?
  ::20. 在下图中有一个带有两个三角的直线。 查找大三角的维度。 使用这些直线查找小三角的维度。 将下限值四舍五入到百位值。 现在想象一下创建更多三角, 一次增加x1乘1。 每增加x1乘1, 下限值增加多少? 29x的下限需要多久?

  

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  Review (Answers)
  ::审查(答复)

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  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 。