Chapter 11.2, Problem 23PSB

To determine

To find the area of triangle ABC.

Answer to Problem 23PSB

The area of triangle ABC is $12$ .

Explanation of Solution

Given:

  $\Delta ABC$ with vertices $A=(1,3),B=(7,3),$ and $C=(4,-1)$ .

Formula Used:

The distance between two points by using the distance formula, which is an application of the Pythagorean theorem.

D $=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ b=Base,\\ h=Height\end{array}$

Calculation:

  

  $\begin{array}{l}AB=\sqrt{{\left(7-1\right)}^2+{\left(3-3\right)}^2}\\ =\sqrt{{\left(6\right)}^2+{\left(0\right)}^2}\\ =\sqrt{36}\\ =6\\ BC=\sqrt{{\left(4-7\right)}^2+{\left(-1-3\right)}^2}\\ =\sqrt{{\left(-3\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\end{array}$

In right angled $\Delta CDA$ , we get

  ${\left(CA\right)}^2={\left(CD\right)}^2+{\left(DA\right)}^2$

  $\begin{array}{l}{\left(5\right)}^2={\left(CD\right)}^2+{\left(3\right)}^2\\ {\left(CD\right)}^2=25-9\\ CD=\sqrt{16}=4\end{array}$

  $\begin{array}{l}A(\Delta ABC)=\frac{1}{2}\times AB\times CD\\ =\frac{1}{2}(6)(4)\\ =12\end{array}$