Chapter 11.2, Problem 27PSC

a.

To determine

To find the length of side for an equilateral triangleABC.

Answer to Problem 27PSC

The length of side for an equilateral triangleABCis $6$ .

Explanation of Solution

Given:

  $Areaofequilateraltriangle=9\sqrt{3}$

Formula Used:

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:

  

Let 2x = length of side.

In right angled $\Delta ADC$ , we get

  ${\left(AC\right)}^2={\left(AD\right)}^2+{\left(DC\right)}^2$

  $\begin{array}{l}{\left(2x\right)}^2={\left(AD\right)}^2+{\left(x\right)}^2\\ {\left(AD\right)}^2=4x^2-x^2\\ {\left(AD\right)}^2=3x^2\\ AD=\sqrt{3}x\end{array}$

  $\begin{array}{l}A(△ABC)=\frac{1}{2}\times BC\times AD\\ 9\sqrt{3}=\frac{1}{2}\times 2x\times x\sqrt{3}\\ 9\sqrt{3}=x^2\sqrt{3}\\ 9=x^2\\ x=\sqrt{9}=3\end{array}$

Since each side is 2x , length is 6.

b.

To determine

To compute the formula for equilateral triangleABC with sides s unit long.

Answer to Problem 27PSC

The formula for equilateral triangle ABC with side s unit long is $A=\frac{s^2}{4}\sqrt{3}$ .

Explanation of Solution

Given:

  $\begin{array}{l}AB=BC=CA=s,\\ \angle A=\angle B=\angle C={60}^o.\end{array}$

Formula Used:

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

Calculation:

  

In right angled $\Delta ADC$ , we get

  ${\left(AC\right)}^2={\left(AD\right)}^2+{\left(DC\right)}^2$

  $\begin{array}{l}{\left(s\right)}^2={\left(AD\right)}^2+{\left(\frac{s}{2}\right)}^2\\ {\left(AD\right)}^2=s^2-\frac{s^2}{4}\\ {\left(AD\right)}^2=\frac{3s^2}{4}\\ AD=\frac{\sqrt{3}s}{2}\end{array}$

  $\begin{array}{l}A(△ABC)=\frac{1}{2}*BC*AD\\ =\frac{1}{2}*s*\frac{s}{2}\sqrt{3}\\ A=\frac{s^2}{4}\sqrt{3}\end{array}$