Question

mr. koomson herons area formula task#2 - 0: given a(2, 2), b(-2, 6), c(4, 8) sketch triangle abc and show that it is isosceles. task#2 - 1: use herons formula to find the area of triangle abc. task#2 - 2: use a = \frac{1}{2}bh to find the area of triangle abc. show all important steps in how you find h!

Explanation:

Step1: Calculate side - lengths using distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For side $AB$ with $A(2,2)$ and $B(-2,6)$:
\[

$$\begin{align*} AB&=\sqrt{(-2 - 2)^2+(6 - 2)^2}\\ &=\sqrt{(-4)^2+4^2}\\ &=\sqrt{16 + 16}\\ &=\sqrt{32}=4\sqrt{2} \end{align*}$$

\]
For side $BC$ with $B(-2,6)$ and $C(4,8)$:
\[

$$\begin{align*} BC&=\sqrt{(4+2)^2+(8 - 6)^2}\\ &=\sqrt{6^2+2^2}\\ &=\sqrt{36+4}\\ &=\sqrt{40}=2\sqrt{10} \end{align*}$$

\]
For side $AC$ with $A(2,2)$ and $C(4,8)$:
\[

$$\begin{align*} AC&=\sqrt{(4 - 2)^2+(8 - 2)^2}\\ &=\sqrt{2^2+6^2}\\ &=\sqrt{4 + 36}\\ &=\sqrt{40}=2\sqrt{10} \end{align*}$$

\]
Since $BC = AC=2\sqrt{10}$, triangle $ABC$ is isosceles.

Step2: Use Heron's formula for area

First, find the semi - perimeter $s=\frac{AB + BC+AC}{2}=\frac{4\sqrt{2}+2\sqrt{10}+2\sqrt{10}}{2}=2\sqrt{2}+2\sqrt{10}$.
Heron's formula is $A=\sqrt{s(s - a)(s - b)(s - c)}$. Let $a = 4\sqrt{2}$, $b = 2\sqrt{10}$, $c = 2\sqrt{10}$.
\[

$$\begin{align*} A&=\sqrt{(2\sqrt{2}+2\sqrt{10})(2\sqrt{2}+2\sqrt{10}-4\sqrt{2})(2\sqrt{2}+2\sqrt{10}-2\sqrt{10})(2\sqrt{2}+2\sqrt{10}-2\sqrt{10})}\\ &=\sqrt{(2\sqrt{2}+2\sqrt{10})(2\sqrt{10}-2\sqrt{2})(2\sqrt{2})(2\sqrt{2})}\\ &=\sqrt{(40 - 8)\times8}\\ &=\sqrt{32\times8}\\ &=\sqrt{256}=16 \end{align*}$$

\]

Step3: Use $A=\frac{1}{2}bh$ for area

Let the base $b = AB = 4\sqrt{2}$.
The mid - point of $AB$ is $M(\frac{2-2}{2},\frac{2 + 6}{2})=(0,4)$.
The slope of $AB$ is $m_{AB}=\frac{6 - 2}{-2 - 2}=\frac{4}{-4}=-1$.
The slope of the altitude from $C$ to $AB$ is $m = 1$ (since the product of slopes of perpendicular lines is - 1).
The equation of the line passing through $C(4,8)$ with slope $m = 1$ is $y - 8=1\times(x - 4)$, i.e., $y=x + 4$.
The equation of the line $AB$ is $y-2=-1\times(x - 2)$, i.e., $y=-x+4$.
Solving the system of equations

$$\begin{cases}y=x + 4\\y=-x + 4\end{cases}$$

, we get the intersection point of the altitude and $AB$. Adding the two equations gives $2y = 8\Rightarrow y = 4$, and $x = 0$.
The length of the altitude $h$ from $C$ to $AB$ is the distance between $C(4,8)$ and $(0,4)$.
\[

$$\begin{align*} h&=\sqrt{(4 - 0)^2+(8 - 4)^2}\\ &=\sqrt{16+16}\\ &=\sqrt{32}=4\sqrt{2} \end{align*}$$

\]
Then $A=\frac{1}{2}\times4\sqrt{2}\times4\sqrt{2}=\frac{1}{2}\times32 = 16$.

Answer:

The triangle $ABC$ is isosceles. The area of triangle $ABC$ using Heron's formula is 16, and using $A=\frac{1}{2}bh$ is also 16.