Q33. - Geometry

Question

a. An equilateral triangle has sides of length s. Show that its area is$\frac{s^2}{4}\sqrt{3}$

b. Find the area of an equilateral triangle with side 7

Step-by-Step Solution

Answer

a. The area of triangle is$\frac{s^2}{4}\sqrt{3}$ .

b. The area of an equilateral triangle with side 7 is.$\frac{49\sqrt{3}}{4}$ 

Step 1. Draw the equilateral triangle.

Step 2. Observe the triangle.

In $△ABC$, draw the altitude 

Now

Here two${30}^{°}-{60}^{°}-{90}^{°}$triangles are formed.

In $△ADC$

$\begin{array}{l}AC=2DC\\ AD=\sqrt{3}DC\end{array}$

Thus

$\begin{array}{l}AD=\sqrt{3}\left(\frac{AC}{2}\right)\\ AD=\sqrt{3}\left(\frac{s}{2}\right)\end{array}$

Step 3. Find the area of equilateral triangle.

Area of$△ABC$ can be found.

$\begin{array}{l}b=BC=s\\ h=AD=\frac{s}{2}\sqrt{3}\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(s\right)\left(\frac{s}{2}\sqrt{3}\right)\\ A=\frac{s^2}{4}\sqrt{3}\end{array}$

Hence, proved the area of equilateral triangle is$A=\frac{s^2}{4}\sqrt{3}$ .

Step 1. Draw the equilateral triangle.

Step 2. Observe the triangle.

In$△ABC$ , draw the altitude 

Now

Here two${30}^{°}-{60}^{°}-{90}^{°}$triangles are formed.

In $△ADC$

$\begin{array}{l}AC=2DC\\ AD=\sqrt{3}DC\end{array}$

Thus

$\begin{array}{l}AD=\sqrt{3}\left(\frac{AC}{2}\right)\\ AD=\sqrt{3}\left(\frac{s}{2}\right)\end{array}$

Step 3. Find the area of equilateral triangle.

Area of$△ABC$ can be found.

$\begin{array}{l}b=BC=s\\ h=AD=\frac{s}{2}\sqrt{3}\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(s\right)\left(\frac{s}{2}\sqrt{3}\right)\\ A=\frac{s^2}{4}\sqrt{3}\end{array}$

Step 4. Find the area of equilateral triangle with side 7 .

We get the area of triangle is $A=\frac{s^2}{4}\sqrt{3}$

Substitute$s=7$

$\begin{array}{c}A=\frac{s^2}{4}\sqrt{3}\\ =\frac{7^2}{4}\sqrt{3}\\ =\frac{49\sqrt{3}}{4}\end{array}$

Therefore, the area of equilateral triangle with side7 is$\frac{49\sqrt{3}}{4}$ .