Chapter 11.2, Problem 9PSA

a.

To determine

To find the area of a triangle ABCif triangle ABC is a ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$ .

Answer to Problem 9PSA

The area of a triangle ABCis $18\sqrt{3}$ .

Explanation of Solution

Given:

  $\Delta ABC$ isa ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$ .

Formula Used:

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

Calculation:

  

  $\Delta ABC$ isa ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$ .

  $\begin{array}{l}AB=\frac{1}{2}(AC)\\ =\frac{1}{2}(12)\\ AB=6\\ CB=(AB)\sqrt{3}\\ CB=6\sqrt{3}\end{array}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ =\frac{1}{2}(6\sqrt{3})*6\\ =\frac{1}{2}(36\sqrt{3})\\ =18\sqrt{3}\end{array}$

b.

To determine

To calculate the area of a triangle ABC with side BC as $12\sqrt{2}$ .

Answer to Problem 9PSA

The area of a triangle ABC is $72$ .

Explanation of Solution

Given:

  $\Delta ABC$ isa ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

Formula Used:

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

Calculation:

  

  $\Delta ABC$ isa ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

So

  $\begin{array}{l}BC=(AB)\sqrt{2}\\ 12\sqrt{2}=(AB)\sqrt{2}\\ AB=12\end{array}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ =\frac{1}{2}(6\sqrt{3})*6\\ =\frac{1}{2}(12*12)\\ =\frac{1}{2}(144)\\ =72\end{array}$

c.

To determine

To compute the area of a triangle ABC with side AC as $12$ .

Answer to Problem 9PSA

The area of a triangle ABCis $36\sqrt{3}$ .

Explanation of Solution

Given:

  $\Delta ABC$ is equilateral and equiangular.

Formula Used:

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

Calculation:

  

  $\Delta ABC$ isequilateral and equiangular.

Each angle is ${60}^{\circ }$ .

Draw altitude $\overline{AD}$ .

In $\Delta ADB$ , shorter leg is 6 and altitude is opposite ${60}^{\circ }$ angle.

So, altitude= $6\sqrt{3}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ =\frac{1}{2}*12*(6\sqrt{3})\\ =\frac{1}{2}(72\sqrt{3})\\ =36\sqrt{3}\end{array}$