Chapter 11.4, Problem 2CYU

To determine

To calculate: The area of the triangle of side $6\text{ m}$ .

Answer to Problem 2CYU

The area of the triangle $15.9{\text{ m}}^2$.

Explanation of Solution

Given information: The side length of the triangle is $6\text{ m}$ .

Formula Used: The area of the triangle is,

  $A=\frac{1}{2}\times \text{B}\times \text{H}$ .......(1)

• Here, B is the base of the triangle and H is the height of the triangle.

Calculation:

The length of the side of the triangle is $6\text{ m}$ .

The angle of the triangle is,

  $\begin{array}{c}\theta =\frac{{180}^{\circ }}{3}\\ ={60}^{\circ }\end{array}$

Consider the figure shown below:

  

Calculate the length of the perpendicular.

  $\begin{array}{c}\cos {30}^{\circ }=\frac{AB}{AC}\\ \frac{\sqrt{3}}{2}=\frac{AB}{6}\\ AB=\frac{6\sqrt{3}}{2}\end{array}$

Calculate the base of the triangle.

  $\text{Base}=6$

Substitute the values on equation (1).

  $\begin{array}{c}A=\frac{1}{2}\times \text{B}\times \text{H}\\ \text{=}\frac{1}{2}\times 6\times \frac{6\sqrt{3}}{2}\\ =\frac{\sqrt{3}}{4}\times {\left(6\right)}^2\\ =\frac{\sqrt{3}}{4}\times 36\\ =9\sqrt{3}\\ =15.588\\ \approx 15.9{\text{ m}}^2\end{array}$

Hence, the area of the triangle $15.9{\text{ m}}^2$.