Big Ideas Math Geometry, 2014
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Big Ideas Math Geometry, 2014 View details
7. Law of Sines and Law of Cosines
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Exercise 46 Page 515

What is the unit of the given expression?

The area of the triangle

Practice makes perfect
Let's consider a triangle with side lengths a, b, and c. Let s represent half the perimeter of the triangle. We are asked to determine what measurement of the triangle is represented by the following formula. sqrt(s(s-a)(s-b)(s-c)) Note that the unit in which s, a, b, and c are expressed is raised to the fourth power under the square root. Therefore, the above formula will be expressed in square units. This suggests that the formula represents the area of the triangle. Area? =sqrt(s(s-a)(s-b)(s-c)) To check if our assumption is correct, we will rewrite the given expression. Let's start by calculating s in terms of a, b, and c. Remember that s is half the perimeter of the triangle. \begin{gathered} s={\color{#FF0000}{\dfrac{1}{2}}}\underbrace{(a+b+c)}_\text{perimeter}=\dfrac{a+b+c}{2} \end{gathered}Let's substitute a+b+c2 for s into the given formula and simplify it as much as possible.
sqrt(s(s-a)(s-b)(s-c))
sqrt(( a+b+c/2)( a+b+c/2-a)( a+b+c/2-b)( a+b+c/2-c))
Simplify
sqrt((a+b+c/2)(a+b+c/2-2a/2)(a+b+c/2-2b/2)(a+b+c/2-2c/2))
sqrt((a+b+c/2)(b+c-a/2)(a+c-b/2)(a+b-c/2))
sqrt((ab+ac-a^2+b^2+bc-ab+bc+c^2-ac/4)(a^2+ab-ac+ac+bc-c^2-ab-b^2+bc/4))
sqrt((- a^2+b^2+2bc+c^2/4)(a^2+2bc-c^2-b^2/4))
sqrt(- a^4- 2a^2bc+ a^2c^2+ a^2b^2+ a^2b^2+2b^3c- b^2c^2-b^4+ 2a^2bc+ 4b^2c^2-2bc^3-2b^3c+ a^2c^2+2bc^3-c^4- b^2c^2/16)
sqrt(- a^4+2a^2c^2+2a^2b^2+2b^2c^2-b^4-c^4/16)
sqrt(- a^4+2a^2c^2+2a^2b^2+2b^2c^2-b^4-c^4)/4
1/4sqrt(- a^4+2a^2c^2+2a^2b^2+2b^2c^2-b^4-c^4)
Next we will rewrite the expression under the square root. To do so, we will use the formulas for the square of a sum and the square of a difference.
1/4sqrt(- a^4+2a^2c^2+2a^2b^2+2b^2c^2-b^4-c^4)
Simplify
1/4sqrt(- a^4+2a^2c^2+2a^2b^2+2b^2c^2-b^4-c^4+ 2a^2b^2- 2a^2b^2)
1/4sqrt(- a^4+2a^2c^2+4a^2b^2+2b^2c^2-b^4-c^4-2a^2b^2)
1/4sqrt(- a^4-2a^2b^2-b^4+2a^2c^2+2b^2c^2-c^4+4a^2b^2)
1/4sqrt(- (a^4+2a^2b^2+b^4)+2a^2c^2+2b^2c^2-c^4+4a^2b^2)
1/4sqrt(- (a^2+b^2)^2+2a^2c^2+2b^2c^2-c^4+4a^2b^2)
1/4sqrt(- ((a^2+b^2)^2-2a^2c^2-2b^2c^2+c^4)+4a^2b^2)
1/4sqrt(- ((a^2+b^2)^2-2(a^2c^2+b^2c^2)+c^4)+4a^2b^2)
1/4sqrt(- ((a^2+b^2)^2-2(a^2+b^2)c^2+c^4)+4a^2b^2)
1/4sqrt(- (a^2+b^2-c^2)^2+4a^2b^2)
1/4sqrt(4a^2b^2-(a^2+b^2-c^2)^2)
We will continue rewriting our expression using the Law of Cosines and the Pythagorean Trigonometric Identity. First, consider the following diagram of the triangle with side lengths a, b, and c.
Let's apply the Law of Cosines to ∠ C. c^2=a^2+b^2-2abcosC We can use the above equation to replace the expression a^2+b^2-c^2 that appears in the simplified version of our formula, 14sqrt(4a^2b^2-( a^2+b^2-c^2)^2). Let's rewrite c^2=a^2+b^2-2abcosC.
c^2=a^2+b^2-2abcosC
0=a^2+b^2-c^2-2abcosC
2abcosC=a^2+b^2-c^2
a^2+b^2-c^2=2abcosC
Now we will substitute 2abcosC for a^2+b^2-c^2 into our formula.
1/4sqrt(4a^2b^2-(a^2+b^2-c^2)^2)
1/4sqrt(4a^2b^2-( 2abcosC)^2)
Simplify
1/4sqrt(4a^2b^2-4a^2b^2cos^2C)
1/4sqrt(4a^2b^2(1-cos^2C))
1/4*sqrt(4a^2b^2)*sqrt(1-cos^2C)
1/4*2ab*sqrt(1-cos^2C)
1/2absqrt(1-cos^2C)
Finally, let's use the Pythagorean Trigonometric Identity. sin^2C+cos^2C=1 We will solve the above equation for sinC.
sin^2C+cos^2C=1
sin^2C=1-cos^2C
sinC=sqrt(1-cos^2C)
Let's substitute sinC for sqrt(1-cos^2C) into the last version of our formula. 1/2absqrt(1-cos^2C)=1/2ab sinC Recall that 12absinC is the formula for the area of a triangle. In conclusion, we have obtained the following. sqrt(s(s-a)(s-b)(s-c))=1/2absinC Therefore, the given formula represents the area of the triangle. This formula for the area of a triangle is called Heron's Formula.