Chapter 9.6, Problem 32E

To determine

Tofind:Area of equilateral triangle

Answer to Problem 32E

  $A=16\sqrt{3}$

Explanation of Solution

Given information:

Concept Used:

In a ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ right angled triangle, length of longer leg is product of shorter length and square root of 3 and length of hypotenuse is given by twice the length of shorter leg.

Calculation:

Area of triangle is-

  $A=\frac{1}{2}bh$

Where $b$ is the base of triangle and $h$ is its height.

Now, height of given equilateral triangle is longer leg of ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle with hypotenuse 8

Let length of shorter leg of ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle be $x$

Length of hypotenuse is twice the length of shorter leg-

Therefore

  $\begin{array}{l}8=2x\\ \Rightarrow x=4\end{array}$

Letlength of longer legof ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ be $b$

Length of longer leg is product of shorter length and square root of 3

Therefore

  $\begin{array}{l}y=x\sqrt{3}\\ \Rightarrow y=4\sqrt{3}\end{array}$

Therefore, height of equilateral triangle is $4\sqrt{3}$

Base length of equilateral triangle will be sum of shorter leg length of both ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle

Length of shorter leg of ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle with hypotenuse 8 is

  $x=4$

Let length of shorter leg of second ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle be $z$

Its longer length is height of equilateral triangle i.e. $y=4\sqrt{3}$

Length of longer leg is product of shorter length and square root of 3

Therefore

  $\begin{array}{l}y=z\sqrt{3}\\ \Rightarrow 4\sqrt{3}=z\sqrt{3}\\ \Rightarrow z=4\end{array}$

Base of equilateral triangle will be-

  $\begin{array}{l}b=x+z=4+4\\ b=8\end{array}$

Now,

  $\begin{array}{l}A=\frac{1}{2}·8·4\sqrt{3}\\ \Rightarrow A=16\sqrt{3}\end{array}$