Course: 6.7 三角区面积 | The Way To Learn

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三角三角区域区域
Introduction
::导言

Trigonometry to this point has allowed us flexibility in analyzing geometric relationships that could be sketched as a combination of triangles. As we continue to explore the triangle, let's assume that an area needs to be calculated for a triangular region where only two sides and the angle between them are able to be measured. T he given sides of the triangle are 5 and 6, and the angle between the sides is $\begin{align*}\frac{\pi}{3}\end{align*}$ radians. T he height of the triangle is not given. Can the area of the triangle be calculated?

Calculating an Area Using Sine
::使用 Sine 计算区域

When we know two sides of a triangle and its included angle, we can use the sine function to calculate the area of the triangle. More specifically, the sine function allows the height of any triangle to be calculated, and that value can be used in the familiar triangle area formula.
::当我们了解三角形的两边及其包含角度时, 我们可以使用正弦函数来计算三角形的区域。更具体地说, 正弦函数允许计算任何三角形的高度, 并且该值可用于熟悉的三角形区域公式。

Using the sine function, isolate the height $\begin{align*}h\end{align*}$ :
::使用正弦函数,分离高度 h:

$\begin{aligned} \sin C & = \frac{h}{a} \\ a \sin C & = h \end{aligned}$
$\begin{align*}\sin C & = \frac{h}{a}\\ a \sin C & = h\end{align*}$
::C=h=h=h=h=h=h=h=h=h=h=h=h=h=h=h=h=h

Substituting into the area formula:
::替代区域公式:

$\begin{aligned} Area & = \frac{1}{2} b \cdot h \\ Area & = \frac{1}{2} b \cdot a \cdot \sin C \\ Area & = \frac{1}{2} \cdot a \cdot b \cdot \sin C \end{aligned}$
$\begin{align*}\text{Area} & = \frac{1}{2} b \cdot h\\ \text{Area} & = \frac{1}{2} b \cdot a \cdot \sin C\\ \text{Area} & = \frac{1}{2} \cdot a \cdot b \cdot \sin C\end{align*}$
::区域=12bhArea=12basin 区域=12aabsinc

Area Formula Given Two Sides & an Included Angle
::区域公式给定双侧和包含的角

$\begin{aligned} Area = \frac{1}{2} \cdot a \cdot b \cdot \sin C \end{aligned}$
$\begin{align*}\text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin C\end{align*}$

Calculating an Area Using Heron's Formula
::使用Heron的公式计算区域

The area formula for a triangle requires the height of the triangle to be known. However, sometimes we know only the side lengths of a triangle. When we know the three side lengths, can be used to calculate the area of the triangle.
::三角形的区域公式需要三角形的高度才能知道。然而,有时我们只知道三角形的侧长。当我们知道三角形的三个侧长时,可以用来计算三角形的区域。

Since the height of the triangle is unknown, we must do several different calculations to derive Heron's Formula.
::因为三角形的高度未知我们必须做几个不同的计算得出Heron的公式

Using the , solve for $\begin{align*}x\end{align*}$ in terms of the other two legs of the triangle $\begin{align*}ADB\end{align*}$ .
::使用三角形亚银其他两条腿的 X 解答。

$\begin{aligned} x^{2} & + h^{2} = c^{2} \\ x^{2} & = c^{2} - h^{2} \\ x & = \sqrt{c^{2} - h^{2}} \end{aligned}$
$\begin{align*}x^2&+h^2 = c^2 \\ x^2 &=c^2-h^2 \\ x &= \sqrt{c^2-h^2} \end{align*}$
::x2+h2=c2x2=c2-h2x=c2+h2=c2=h2=h2=h2=h2

Using the Pythagorean Theorem, solve for $\begin{align*}(b-x)^2\end{align*}$ in terms of the other two legs of the triangle $\begin{align*}CDB\end{align*}$ , and then expand $\begin{align*}(b-x)^2\end{align*}$ .
$\begin{aligned} (b - x)^{2} + h^{2} = a^{2} \\ (b - x)^{2} = a^{2} - h^{2} \\ b^{2} - 2 b x + x^{2} = a^{2} - h^{2} \end{aligned}$
$\begin{align*}(b-x)^2+h^2 = a^2 \\ (b-x)^2 =a^2-h^2\\ b^2-2bx+x^2 =a^2-h^2\end{align*}$
::使用 Pytagorean Theorem, 以三角 CDB 的另外两条腿为( b- x) 2 解答( b- x) 2, 然后扩展( b- x) 2, (b- x) 2+h2=a2( b- x) 2=a2 - h2b2 - 2bx+x2=a2 - h2=a2

Since $\begin{align*}a, \ b, \text{ and } c\end{align*}$ are the only known variables, eliminate the variable $\begin{align*}x\end{align*}$ using substitution.
::由于a、b和c是唯一已知变量,使用替代删除变量x。

$\begin{aligned} x & = \sqrt{c^{2} - h^{2}} \\ x^{2} & = c^{2} - h^{2} \\ b^{2} - 2 b \sqrt{c^{2} - h^{2}} & + (c^{2} - h^{2}) = a^{2} - h^{2} \\ b^{2} + c^{2} - a^{2} & = 2 b \sqrt{c^{2} - h^{2}} \end{aligned}$
$\begin{align*}x &= \sqrt{c^2-h^2} \\ x^2 &= c^2-h^2 \\ b^2-2b\sqrt{c^2-h^2}&+(c^2-h^2) =a^2-h^2 \\ b^2+c^2-a^2&= 2b\sqrt{c^2-h^2}\end{align*}$
::x=c2-h2x2=c2-h2b2-2b2-2b2-2bc2-2-h2+(c2-h2)=a2-h2b2+c2-a2=2b2-h2

Solve for the height of the triangle, $\begin{align*}h\end{align*}$ , using algebra .
::使用代数解决三角形的高度 H 。

$\begin{aligned} {(b^{2} + c^{2} - a^{2})}^{2} & = {(2 b \sqrt{c^{2} - h^{2}})}^{2} \\ {(b^{2} + c^{2} - a^{2})}^{2} & = 4 b^{2} (c^{2} - h^{2}) \\ \frac{{(b^{2} + c^{2} - a^{2})}^{2}}{4 b^{2}} & = c^{2} - h^{2} \end{aligned}$
$\begin{align*}\left( b^2+c^2-a^2 \right)^2 &= \left(2b\sqrt{c^2-h^2}\right)^2 \\ \left( b^2+c^2-a^2 \right)^2 &= 4b^2 (c^2-h^2) \\ \frac{\left( b^2+c^2-a^2 \right)^2}{4b^2} &= c^2-h^2 \end{align*}$
:b2+C2-2-a2)2=(2bc2-2-h2)、2(b2+c2-2-a2)、2=4b2(c2-2-h2(b2+c2-2-a2)、24b2=c2-2-h2)

$\begin{aligned} h^{2} & = c^{2} - \frac{{(b^{2} + c^{2} - a^{2})}^{2}}{4 b^{2}} \\ h^{2} & = \frac{4 b^{2} c^{2} - {(b^{2} + c^{2} - a^{2})}^{2}}{4 b^{2}} \\ h^{2} & = \frac{{(2 b c)}^{2} - {(b^{2} + c^{2} - a^{2})}^{2}}{4 b^{2}} \\ h^{2} & = \frac{[2 b c + (b^{2} + c^{2} - a^{2})] [2 b c - (b^{2} + c^{2} - a^{2})]}{4 b^{2}} \\ h^{2} & = \frac{[(b^{2} + 2 b c + c^{2}) - a^{2}] [a^{2} - (b^{2} - 2 b c + c^{2})]}{4 b^{2}} \\ h^{2} & = \frac{[{(b + c)}^{2} - a^{2}] [a^{2} - {(b - c)}^{2}]}{4 b^{2}} \\ h^{2} & = \frac{[(b + c) + a] [(b + c) - a] [a + (b - c)] [a - (b - c)]}{4 b^{2}} \\ h^{2} & = \frac{(a + b + c) (b + c - a) (a + b - c) (a - b + c)}{4 b^{2}} \\ h^{2} & = \frac{(a + b + c) (a + b + c - 2 a) (a + b + c - 2 c) (a + b + c - 2 b)}{4 b^{2}} \\ h & = \sqrt{\frac{(a + b + c) (a + b + c - 2 a) (a + b + c - 2 b) (a + b + c - 2 c)}{4 b^{2}}} \\ h & = \frac{\sqrt{(a + b + c) (a + b + c - 2 a) (a + b + c - 2 b) (a + b + c - 2 c)}}{2 b} \end{aligned}$
$\begin{align*}h^2 &= c^2 -\frac{\left( b^2+c^2-a^2 \right)^2}{4b^2} \\ h^2 &= \frac{4b^2c^2- \left(b^2+c^2-a^2 \right)^2}{4b^2} \\ h^2 &= \frac{\left(2bc\right)^2- \left(b^2+c^2-a^2 \right)^2}{4b^2} \\ h^2 &= \frac{\left[2bc + \left(b^2+c^2-a^2 \right)\right]\left[2bc - \left(b^2+c^2-a^2 \right)\right]}{4b^2} \\ h^2 &= \frac{\left[\left(b^2+2bc+c^2 \right)-a^2 \right]\left[a^2-\left(b^2-2bc+c^2 \right)\right]}{4b^2} \\ h^2 &= \frac{\left[\left(b+c\right)^2 -a^2 \right]\left[a^2-\left(b-c \right)^2\right]}{4b^2} \\ h^2 &= \frac{\left[\left(b+c\right) +a \right]\left[\left(b+c\right) -a \right]\left[a+\left(b-c\right) \right]\left[a-\left(b-c\right) \right]}{4b^2} \\ h^2 &= \frac{\left(a+b+c\right)\left(b+c-a\right)\left(a+b-c\right)\left(a-b+c\right)}{4b^2}\\ h^2 &= \frac{\left(a+b+c\right)\left(a+b+c-2a\right)\left(a+b+c-2c\right)\left(a+b+c-2b\right)}{4b^2} \\ h &= \sqrt{\frac{\left(a+b+c\right)\left(a+b+c-2a\right)\left(a+b+c-2b\right)\left(a+b+c-2c\right)}{4b^2}} \\ h &= \frac{\sqrt{\left(a+b+c\right)\left(a+b+c-2a\right)\left(a+b+c-2b\right)\left(a+b+c-2c\right)}}{2b}\end{align*}$
:b2+c2-2-2-a2),b2-2,b2,b2,b2,b2,b2,b2,2,2,2,(b)-2-(b)-2-(b2+2+c2),24,b2,2,2,(b)-(b2+2),(b2+(b)),(b2+(b)),(b)),(b+(b),(b),(b)),(b),(a),(a),(a+(b),(b),(b)),(b),(b),(b),(b),(b),(b),(b),(a),(a)-(a+(b)+(b),(b)(b),(b),(b),(b),(b),(b),(b),(b),(b),(a,(a)和(a)a)),(a,(a,(a)和(a)(a)和(a)(a)(a)(a),(a)(a),(a+)(a)(a)(a)(a)(a)(a)(a(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a))))))))))))和(a(a)(a)(a)(a)(a(a(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)和(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)

Note that the perimeter of the triangle $\begin{align*}P\end{align*}$ is equal to $\begin{align*}P=a+b+c\end{align*}$ .
::请注意三角形P的周边等于P=a+b+c。

$\begin{aligned} h & = \frac{\sqrt{(a + b + c) (a + b + c - 2 a) (a + b + c - 2 b) (a + b + c - 2 c)}}{2 b} \\ h & = \frac{\sqrt{P (P - 2 a) (P - 2 b) (P - 2 c)}}{2 b} \end{aligned}$
$\begin{align*}h &= \frac{\sqrt{\left(a+b+c\right)\left(a+b+c-2a\right)\left(a+b+c-2b\right)\left(a+b+c-2c\right)}}{2b}\\ h &= \frac{\sqrt{P\left(P-2a\right)\left(P-2b\right)\left(P-2c\right)}}{2b}\end{align*}$
::h=(a+b+c)(a+b+c-2a)(a+b+c-2b)(a+b+c-2b)(a+b+c-2c)2bh=P(P-2a)(P-2b)(P-2c)2b

The area of the triangle $\begin{align*}ABC\end{align*}$ can found be found using the formula $\begin{align*}A=\frac{1}{2}bh\end{align*}$ .
::三角形 ABC 区域, 可用公式 A= 12bh 找到。

$\begin{aligned} A & = \frac{1}{2} b \frac{\sqrt{P (P - 2 a) (P - 2 b) (P - 2 c)}}{2 b} \\ A & = \frac{\sqrt{P (P - 2 a) (P - 2 b) (P - 2 c)}}{4} \\ A & = \sqrt{\frac{P (P - 2 a) (P - 2 b) (P - 2 c)}{16}} \\ A & = \sqrt{\frac{P}{2} (\frac{P - 2 a}{2}) (\frac{P - 2 b}{2}) (\frac{P - 2 c}{2})} \\ A & = \sqrt{\frac{P}{2} (\frac{P}{2} - a) (\frac{P}{2} - b) (\frac{P}{2} - c)} \end{aligned}$
$\begin{align*}A &=\frac{1}{2}b \frac{\sqrt{P\left(P-2a\right)\left(P-2b\right)\left(P-2c\right)}}{2b} \\ A &=\frac{\sqrt{P\left(P-2a\right)\left(P-2b\right)\left(P-2c\right)}}{4} \\ A &=\sqrt{\frac{P\left(P-2a\right)\left(P-2b\right)\left(P-2c\right)}{16}} \\ A &=\sqrt{\frac{P}{2}\left(\frac{P-2a}{2}\right)\left(\frac{P-2b}{2}\right)\left(\frac{P-2c}{2}\right)} \\ A &=\sqrt{\frac{P}{2}\left(\frac{P}{2}-a\right)\left(\frac{P}{2}-b\right)\left(\frac{P}{2}-c\right)} \end{align*}$
::A=12bP(P-2a)(P-2b)(P-2b)(P-2c)2bA=P(P-2a)(P-2b)(P-2c)(P-2c)4A=P(P-2a)(P-2b)(P-2c)16A=P2(P-2a)(P-2b)(P-2c)16A=P2(P-2a)(P-2b)(P-2c2)(P-2c2)A=P2(P-2a)(P-2b)(P-2b)(P-2c)

The semiperimeter of the triangle is half the perimeter: $\begin{align*}s=\frac{P}{2}=\frac{a+b+c}{2}\end{align*}$ .
::三角形的半径是周边的一半:S=P2=a+b+c2。

$\begin{aligned} A & = \sqrt{s (s - a) (s - b) (s - c)} \end{aligned}$
$\begin{align*}A &=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\end{align*}$
::A=(s-a)(s-b)(s-c)

Heron's Formula
::海隆的方程式

$\begin{aligned} Area = \sqrt{s (s - a) (s - b) (s - c)}, where a, b, and c are the side lengths \end{aligned}$
$\begin{align*}\text { Area} = \sqrt{s(s - a) (s - b) (s - c)}, \ \text{where }a, \, b, \text{ and } c \text{ are the side lengths}\end{align*}$
::区域=(s-a)(s-b)(s-c),其中a、b和c为侧长

$\begin{aligned} s = \frac{a + b + c}{2}, where s is the semiperimeter of the triangle \end{aligned}$
$\begin{align*}s=\frac{a+b+c}{2}, \ \text{where } s \text{ is the semiperimeter of the triangle}\end{align*}$
::s=a+b+c2, 其中 s 是三角形的半边距

Examples
::实例

Example 1
::例1

Given $\begin{align*}\Delta ABC\end{align*}$ with $\begin{align*}A = 22^\circ, \, b = 6, \text{ and } c =7,\end{align*}$ what is the area?
::以A=22,b=6,c=7,C=7,A=22,b=6和C=7,

Solution:
::解决方案 :

Since two sides and an angle are given, use the sine function to calculate the area.
::由于给定了两边和一个角度,使用正弦函数计算区域。

$\begin{aligned} A r e a & = \frac{1}{2} \cdot 6 \cdot 7 \cdot \sin 22^{\circ} \\ \approx 7.87 {units}^{2} \end{aligned}$
$\begin{align*}Area & = \frac{1}{2} \cdot 6 \cdot 7 \cdot \sin 22^\circ\\ &\approx 7.87 \text{ units}^2\end{align*}$
::面积=12 6 7 sin22 7.87 单位2

Example 2
::例2

Given that $\begin{align*}\Delta XYZ\end{align*}$ has an area of 28 in ² , what is the angle included between side length 8 and 9?
::鉴于XYZ的面积为28英寸2, 侧长8和9之间的角是多少?

Solution:
::解决方案 :

Since the included angle needs to be determined for two given side lengths , use the sine function .
::由于包含的角度需要为两个给定的侧边长度确定,使用正弦函数。

$\begin{aligned} Area & = \frac{1}{2} \cdot a \cdot b \cdot \sin C \\ 28 & = \frac{1}{2} \cdot 8 \cdot 9 \cdot \sin C \\ \sin C & = \frac{28 \cdot 2}{8 \cdot 9} \\ \sin^{- 1} (\sin C) & = \sin^{- 1} (\frac{28 \cdot 2}{8 \cdot 9}) \\ C & = \sin^{- 1} (\frac{28 \cdot 2}{8 \cdot 9}) \\ \approx {51.06}^{\circ} or {128.94}^{\circ} \end{aligned}$
$\begin{align*}\text{Area} & = \frac{1}{2} \cdot a \cdot b \cdot \sin C\\ 28 & = \frac{1}{2} \cdot 8 \cdot 9 \cdot \sin C\\ \sin C & = \frac{28 \cdot 2}{8 \cdot 9}\\ \sin^{-1} \left(\sin C \right) & = \sin^{-1} \left(\frac{28 \cdot 2}{8 \cdot 9} \right) \\ C & = \sin^{-1} \left( \frac{28 \cdot 2}{8 \cdot 9} \right)\\ &\approx 51.06 ^\circ \text{ or } 128.94 ^\circ\end{align*}$
::区域=12absinC28=1289CsinC=282891(sinC)=sin-11(282899)C=sin-1(2828289) 51.06或128.94

Example 3
::例3

What is the area of $\begin{align*}\Delta ABC\end{align*}$ with $\begin{align*}A = 31^\circ, \, b = 12, \text{ and }c = 14\end{align*}$ ?
::使用 A=31,b=12,c=14的 ABC 区域是什么?

Solution:
::解决方案 :

Since t wo sides and the included angle are given, use the area formula.
::由于给出了两边和包括角度,请使用区域公式。

$\begin{aligned} A r e a = \frac{1}{2} \cdot 12 \cdot 14 \cdot \sin 31^{\circ} \approx 43.26 {units}^{2} \end{aligned}$
$\begin{align*}Area = \frac{1}{2} \cdot 12 \cdot 14 \cdot \sin 31^\circ \approx 43.26 \text { units}^2\end{align*}$
::面积=12=12=12=14=16=31=43.26单位2

Example 4
::例4

In the problem in the Introduction , the sides of the given triangle are 5 and 6, and the angle between the sides is $\begin{align*}\theta = \frac{\pi}{3}\end{align*}$ . Calculate the area.
::在导言中的问题,三角形的两边是5和6,两边之间的角是3。计算面积。

Solution:
::解决方案 :

Since two sides and an angle are given, use the sine function to calculate the area.
::由于给定了两边和一个角度,使用正弦函数计算区域。

$\begin{aligned} Area = \frac{1}{2} \cdot 5 \cdot 6 \cdot \sin \frac{π}{3} \approx 12.99 {units}^{2} \end{aligned}$
$\begin{align*}\text {Area} = \frac{1}{2} \cdot 5 \cdot 6 \cdot \sin \frac{\pi}{3} \approx 12.99 \text { units}^2\end{align*}$
::面积=12 5 6 sin 3 12.99 单位2

Example 5
::例5

Calculate the area of $\begin{align*}\Delta ABC\end{align*}$ with side lengths $\begin{align*}a = 11, \, b = 12, \text{ and } z = 13.\end{align*}$
::a=11,b=12,z=13。

Solution:
::解决方案 :

Since three sides are given, use Heron's Formula to calculate the area.
::既然有三面,请使用Heron的公式来计算面积。

$\begin{aligned} s & = \frac{11 + 12 + 13}{2} = 18 \\ Area & = \sqrt{18 \cdot (18 - 11) \cdot (18 - 12) \cdot (18 - 13)} \\ Area & = \sqrt{18 \cdot (7) \cdot (6) \cdot (5)} \\ Area & = \sqrt{3, 780} \\ Area & \approx 61.48 {units}^{2} \end{aligned}$
$\begin{align*}s&=\frac{11+12+13}{2}=18\\ \text { Area} &= \sqrt{18\cdot \left(18-11 \right) \cdot \left(18-12 \right) \cdot \left(18-13 \right)}\\ \text { Area} &= \sqrt{18\cdot \left(7 \right) \cdot \left(6 \right) \cdot \left(5 \right)}\\ \text { Area} &= \sqrt{3,\!780}\\ \text { Area} &\approx 61.48 \text { units}^2\end{align*}$
::=11+12+132=18区域=18(18-11)_(18-12)_(18-13)_(18-13)区域=18(7)(7)(6)(5)区域=3,780区域=61.48单位2

Example 6
::例6

The area of a triangle is 3 units ² . Two sides of the triangle are 4 units and 5 units. What is the measure of their included angle?
::三角形区域为 3 个单位。三角形的两边为 4 个单位和 5 个单位。三角形所包括角度的度量是多少 ?

Solution:
::解决方案 :

Since the included angle needs to be determined for two given side lengths, use the sine function .
::由于包含的角度需要为两个给定的侧边长度确定,使用正弦函数。

$\begin{aligned} 3 & = \frac{1}{2} \cdot 4 \cdot 5 \cdot \sin θ \\ θ & = \sin^{- 1} (\frac{3 \cdot 2}{4 \cdot 5}) \\ \approx {17.46}^{\circ} and {162.54}^{\circ} \end{aligned}$
$\begin{align*}3 &= \frac{1}{2} \cdot 4 \cdot 5 \cdot \sin \theta \\ \theta &= \sin^{-1} \left( \frac{3 \cdot 2}{4 \cdot 5} \right)\\ &\approx 17.46 ^\circ \text{ and } 162.54 ^\circ\end{align*}$
::3=124556111(3245517.46和162.542244444444444444222222222222244444444444444454444444444444444444444444444444444444444444444444444444444444444444

Example 7
::例7

Given $\begin{align*}\Delta ABC\end{align*}$ with $\begin{align*}A = 12^\circ, \, b = 4,\end{align*}$ and $\begin{align*}\text{Area} = 1.7 \text{ units}^2,\end{align*}$ what is the length of side $\begin{align*}c\end{align*}$ ?
::根据A=12,b=4和A=1.7单位2的 ABC 和A=12,b=4和Area=1.7单位2计算,C侧的长度是多少?

Solution:
::解决方案 :

Since a side length, an angle, and an area are given , use the sine function.
::如果给定一个侧长、一个角度和一个区域,则使用正弦函数。

$\begin{aligned} A r e a & = \frac{1}{2} \cdot c \cdot b \cdot \sin A \\ 1.7 & = \frac{1}{2} \cdot c \cdot 4 \cdot \sin 12^{\circ} \\ c & = \frac{1.7 \cdot 2}{4 \cdot \sin 12^{\circ}} \approx 4.09 units \end{aligned}$
$\begin{align*}Area & = \frac{1}{2} \cdot c \cdot b \cdot \sin A\\ 1.7 & = \frac{1}{2} \cdot c \cdot 4 \cdot \sin 12^\circ\\ c & = \frac{1.7 \cdot 2}{4 \cdot \sin 12^\circ} \approx 4.09 \text{ units}\end{align*}$
::面积=12=12cbsin=A1.7=12c4sin=12c=1.724sin=12c4.09单位

Summary
::摘要

If two sides of a triangle and an included angle are known, the sine function can be used to calculate the area: $\begin{align*}A=\frac{1}{2}ab{}{}\sin{C}\end{align*}$ .
::如果知道三角形的两边和包含角度,可使用正弦函数计算区域:A=12absinC。

If three sides of a triangle are known, Heron's Formula can be used to calculate the area: $\begin{align*}\text { Area} = \sqrt{s(s - a) (s - b) (s - c)}\end{align*}$ where $\begin{align*}s\end{align*}$ is the semiperimeter $\begin{align*}s = \frac{a + b + c}{2}\end{align*}$ .
::如果已知三角形的三边,可使用Heron的公式计算区域:面积=(s-a)(s-b)(s-c),其中的半径=a+b+c2。

Review
::回顾

For 1-11, calculate the area of each triangle.
::1-11,计算每个三角形的区域。

$\begin{align*}\Delta ABC\end{align*}$ if $\begin{align*}a=13, b=15\end{align*}$ , and $\begin{align*}\angle C = 70^\circ\end{align*}$ .
::a=13,b=15,和a=70,ABC=70。

$\begin{align*}\Delta ABC\end{align*}$ if $\begin{align*}b=8, c=4\end{align*}$ , and $\begin{align*}\angle A = 58^\circ\end{align*}$ .
::b=8,c=4,和A=58,ABC=8,c=4。

$\begin{align*}\Delta ABC\end{align*}$ if $\begin{align*}b=34, c=29\end{align*}$ , and $\begin{align*}\angle A = 125^\circ\end{align*}$ .
::b=34,c=29,和A=125

$\begin{align*}\Delta ABC\end{align*}$ if $\begin{align*}a=3, b=7\end{align*}$ , and $\begin{align*}\angle C = 81^\circ\end{align*}$ .
::ABC=3,b=7,和QC=81。

$\begin{align*}\Delta ABC\end{align*}$ if $\begin{align*}a=4.8, c=3.7\end{align*}$ , and $\begin{align*}\angle B = 54^\circ\end{align*}$ .
::a=4.8,c=3.7,和B=54。

$\begin{align*}\Delta ABC\end{align*}$ if $\begin{align*}a=12, b=5\end{align*}$ , and $\begin{align*}\angle C = 22^\circ\end{align*}$ .
::a=12,b=5,和a=22,ABC=22。

$\begin{align*}\Delta ABC\end{align*}$ if $\begin{align*}a=3, b=10\end{align*}$ , and $\begin{align*}\angle C = 65^\circ\end{align*}$ .
::ABC 如果a=3,b=10,和C=65。

$\begin{align*}\Delta ABC\end{align*}$ if $\begin{align*}a=5, b=9\end{align*}$ , and $\begin{align*}\angle C = 11^\circ\end{align*}$ .
::ABC如果a=5,b=9,和c=11。

$\begin{align*}\Delta ABC\end{align*}$ if $\begin{align*}a=5, b=7\end{align*}$ , and $\begin{align*}c=8\end{align*}$ .
::a=5,b=7,c=8的ABC。

$\begin{align*}\Delta ABC\end{align*}$ if $\begin{align*}a=7, b=8\end{align*}$ , and $\begin{align*}c=14\end{align*}$ .
::a=7,b=8,c=14的ABC。

$\begin{align*}\Delta ABC\end{align*}$ if $\begin{align*}a=12, b=14\end{align*}$ , and $\begin{align*}c=13\end{align*}$ .
::a=12,b=14,c=13的ABC。

The area of a triangle is 12 square units. Two sides of the triangle are 8 units and 4 units. What is the measure of their included angle?
::三角形区域为 12 平方单位。三角形的两边为 8 个单位和 4 个单位。包括角度的测量值是多少 ?

The area of a triangle is 23 square units. Two sides of the triangle are 14 units and 5 units. What is the measure of their included angle?
::三角形区域为23平方单位,三角形两侧为14平方单位和5平方单位。三角形角的度量是多少?

Given $\begin{align*}\Delta DEF\end{align*}$ has area 32 square inches, what is the angle included between side length 9 and 10?
::鉴于ZQDEF的面积为32平方英寸,侧长9至10之间的角是多少?

Given $\begin{align*}\Delta GHI\end{align*}$ has area 15 square inches, what is the angle included between side length 7 and 11?
::鉴于GHI的面积为15平方英寸,侧长7至11之间的角是多少?

Review ( Answers)
::回顾(答复)

Please see the Appendix.
::请参看附录。

Section outline

General

三角三角区域区域

Introduction ::导言

Calculating an Area Using Sine ::使用 Sine 计算区域

Area Formula Given Two Sides & an Included Angle ::区域公式给定双侧和包含的角

Calculating an Area Using Heron's Formula ::使用Heron的公式计算区域

Heron's Formula ::海隆的方程式

Examples ::实例

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

Summary ::摘要

Review ::回顾

Review ( Answers) ::回顾(答复)

Introduction
::导言

Calculating an Area Using Sine
::使用 Sine 计算区域

Area Formula Given Two Sides & an Included Angle
::区域公式给定双侧和包含的角

Calculating an Area Using Heron's Formula
::使用Heron的公式计算区域

Heron's Formula
::海隆的方程式

Examples
::实例

Example 4
::例4

Example 5
::例5

Example 6
::例6

Example 7
::例7

Summary
::摘要

Review
::回顾

Review ( Answers)
::回顾(答复)