For triangle A B C we can express its area, 𝐀, as 𝐀=(1)/(2) a b sinC . The cosine rule can be us

For triangle A B C we can express its area, 𝐀, as 𝐀=(1)/(2)...

For triangle $A B C$ we can express its area, $\mathbf{A},$ as $\mathbf{A}=\frac{1}{2} a b \sin C .$ The cosine rule can be used to write the expression $c^{2}=a^{2}+b^{2}-2 a b \cos C$ a) Using these two expressions show that $16 \mathbf{A}^{2}=4 a^{2} b^{2}\left(a^{2}+b^{2}-c^{2}\right)$ - Hint: use the Pythagorean identity $\sin ^{2} C+\cos ^{2} C=1$ b) The perimeter of the triangle is equal to $a+b+c$. Let s be the semiperimeter, that is $s=\frac{a+b+c}{2}$. Using the result from a) and that $2 s=a+b+c,$ show that $16 \mathbf{A}^{2}=2 s(2 s-2 c)(2 s-2 a)(2 s-2 b)$ c) Finally, show that the result in b) gives $A=\sqrt{s(s-c)(s-b)(s-c)} .$ This notable result expresses the area of a triangle in terms of only the length of its three sides. Although quite possibly known before his time, the formula is attributed to the ancient Greek mathematician and engineer, Heron of Alexandria (ca. $10-70 \mathrm{AD}$ ) and is thus called Heron's formula. The first written reference to the formula is Heron's proof of it in his book Metrica, written in approximately $60 \mathrm{AD}$. (FIGURE CAN'T COPY)

For triangle $A B C$ we can express its area,
$\mathbf{A},$ as $\mathbf{A}=\frac{1}{2} a b \sin C .$ The cosine rule can be
used to write the expression $c^{2}=a^{2}+b^{2}-2 a b \cos C$
a) Using these two expressions show that $16 \mathbf{A}^{2}=4 a^{2} b^{2}\left(a^{2}+b^{2}-c^{2}\right)$
- Hint: use the Pythagorean identity $\sin ^{2} C+\cos ^{2} C=1$
b) The perimeter of the triangle is equal to $a+b+c$. Let s be the semiperimeter, that is $s=\frac{a+b+c}{2}$. Using the result from a) and that $2 s=a+b+c,$ show that $16 \mathbf{A}^{2}=2 s(2 s-2 c)(2 s-2 a)(2 s-2 b)$
c) Finally, show that the result in b) gives $A=\sqrt{s(s-c)(s-b)(s-c)} .$ This notable result expresses the area of a triangle in terms of only the length of its three sides. Although quite possibly known before his time, the formula is attributed to the ancient Greek mathematician and engineer, Heron of Alexandria (ca. $10-70 \mathrm{AD}$ ) and is thus called Heron's formula. The first written reference to the formula is Heron's proof of it in his book Metrica, written in approximately $60 \mathrm{AD}$.
(FIGURE CAN'T COPY)

Key Concepts

Triangle Area Formula

The triangle area formula expresses the area of a triangle in terms of two of its sides and the sine of the included angle. It connects geometric quantity (area) with trigonometric functions, providing a way to calculate the area when direct height measurement is not readily available, which is especially useful in oblique triangles.

Cosine Rule

The cosine rule, or law of cosines, relates the lengths of the sides of a triangle to the cosine of one of its angles. This rule generalizes the Pythagorean theorem to any triangle, allowing one to compute an unknown side or angle when the other elements are known. It is fundamental in solving triangles in trigonometry.

Pythagorean Identity

The Pythagorean trigonometric identity, given by sin²? + cos²? = 1, is a cornerstone in trigonometry. It expresses the inherent relationship between the sine and cosine functions, allowing one to express one in terms of the other. This identity enables simplification and transformation of expressions involving sine and cosine.

Semiperimeter

The semiperimeter of a triangle is half the sum of its three side lengths. It is a commonly used quantity in various geometric formulas, particularly in Heron’s formula, where it serves as a key component in expressing the triangle's area solely in terms of its side lengths.

Heron's Formula

Heron's formula provides a method to compute the area of a triangle when only the lengths of its sides are known. By using the semiperimeter and the lengths of the sides, it offers an elegant solution to determine the area without needing to calculate the height or any angles. This formula has historical significance and is attributed to the ancient Greek mathematician Heron of Alexandria.

Transcript

00:01 In part a of this question, we want to show that 16 times the area squared is equal to 4a squared b squared minus a squared minus c squared all squared using the area formula and the cosine law.

00:14 So it is a show of that question.

00:16 So we'll start with the left side.

00:18 16a squared would be equal to 16 times 1 half ab sine c all squared.

00:29 So this is going to be 16.

00:31 Times 1 over 4, a squared, b squared, sine squared c.

00:36 And that'll simplify to 4, a squared, b squared, sine squared c.

00:43 Recognize that sine squared c can be written as 1 minus co -squared c.

00:50 And if we distribute, we'll get 4a squared, b squared, minus 4a squared, b squared, co -squared c and we have to do a little bit of conversion with the cosine law so we see that a squared plus b squared minus c squared is equal to to a b cosine c this is through a little bit of rearranging on the cosine law and if we square both sides we get a squared plus b squared minus c squared all squared equals 4a squared b squared, co -squared c.

01:32 And this latter expression is what we want to replace in our red equation.

01:38 So now we can rewrite this as 4a squared b squared minus a squared plus b squared minus c squared, all squared.

01:49 And that's what we wanted to show for part a.

01:54 In part b, using the result in part a, we also want to use this new.

02:03 Formula for the semi -perimeter, which is 2s equals a plus b plus c.

02:10 We want to show that 16a squared is equal to 2s times 2s minus c, times 2s minus 2a, which should be 2c, and also 2s minus 2b.

02:32 So again, starting with our result from part a, well, we know that, 16a squared is equal to 4a squared b squared minus a squared plus b squared minus c squared all squared.

02:51 And the 4a squared b squared, we can rewrite that as 2ab all squared.

02:58 And the point of this is to see that, oh, we actually have a difference of squares here.

03:06 So we can factor.

03:08 This would be written as 2ab minus a squared plus b squared minus c squared times 2ab plus a plus a squared plus b squared minus c squared...

Please give Ace some feedback