Chapter 11.5, Problem 3PSA

a

To determine

To calculate : The area of the equilateral triangle with apothem equal to $6$

Answer to Problem 3PSA

The area is $A=12\sqrt{3}$

Explanation of Solution

Given information : The apothem of the equilateral triangle $(h)=6$

Formula used : We use the following formula to find the area of an equilateral triangle

  $A=\frac{h^2}{\sqrt{3}}$

Calculation :

Substituting the value of $h=6$ in the equation, we get

  $\begin{array}{l}A=\frac{{(6)}^2}{\sqrt{3}}\\ A=12\sqrt{3}\end{array}$

Thus, we get

  $A=12\sqrt{3}$

b

To determine

To calculate : The area of the equilateral triangle with apothem $4$

Answer to Problem 3PSA

The area is $A=\frac{16}{\sqrt{3}}$

Explanation of Solution

Given information : The apothem of the equilateral triangle is, $h=4$

Formula used : We use the following formula to find the area of an equilateral triangle

  $A=\frac{h^2}{\sqrt{3}}$

Calculation : Substituting the value of $h=4$ in the equation, we get

  $\begin{array}{l}A=\frac{{(4)}^2}{\sqrt{3}}\\ A=\frac{16}{\sqrt{3}}\end{array}$

Thus, we get

  $A=\frac{16}{\sqrt{3}}$

c

To determine

To calculate : The area of the equilateral triangle with apothem $3$

Answer to Problem 3PSA

The area is $A=3\sqrt{3}$

Explanation of Solution

Given information :

The apothem of the equilateral triangle is, $h=3$

Formula used : We use the following formula to find the area of an equilateral triangle

  $A=\frac{h^2}{\sqrt{3}}$

Calculation : Substituting the value of $h=3$ in the equation, we get

  $\begin{array}{l}A=\frac{{(3)}^2}{\sqrt{3}}\\ A=3\sqrt{3}\end{array}$

Thus, we get

  $A=3\sqrt{3}$

d

To determine

To calculate : The area of the equilateral triangle with apothem $2\sqrt{3}$

Answer to Problem 3PSA

The area is $A=4\sqrt{3}$

Explanation of Solution

Given information :

The apothem of the equilateral triangle is, $h=2\sqrt{3}$

Formula used : We use the following formula to find the area of an equilateral triangle

  $A=\frac{h^2}{\sqrt{3}}$

Calculation : Substituting the value of $h=2\sqrt{3}$ in the equation, we get

  $\begin{array}{l}A=\frac{{\left(2\sqrt{3}\right)}^2}{\sqrt{3}}\\ A=4\sqrt{3}\end{array}$

Thus, we get

  $A=4\sqrt{3}$