Chapter 11.2, Problem 12PSA

To determine

To find the area of triangle AOC.

Answer to Problem 12PSA

The area of triangle AOC is $44$ .

Explanation of Solution

Given:

  $\Delta AEB$ is a ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

  $\begin{array}{l}DC=7,\\ AD=10,\\ BC=14.\end{array}$

Formula Used:

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

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

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

Calculation:

  $\begin{array}{l}AB=\sqrt{{\left(11-11\right)}^2+{\left(8-0\right)}^2}\\ =\sqrt{{\left(0\right)}^2+{\left(8\right)}^2}\\ =\sqrt{64}\\ =8\\ BC=\sqrt{{\left(11-0\right)}^2+{\left(8-8\right)}^2}\\ =\sqrt{{\left(11\right)}^2+{\left(0\right)}^2}\\ =\sqrt{121}\\ =11\end{array}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ A(\Delta ABC)=\frac{1}{2}*BC*AB\\ =\frac{1}{2}*11*8\\ =44\end{array}$