Question

Find the area of a regular hexagon with the given measurement. 48-inch perimeter

Answer 1

Answer:

$96\sqrt{3}$ inches

Step-by-step explanation:

Here we are going to use the formula which is

Area=$\frac{1}{2} \times P \times A$

Where P is perimeter and A is apothem

Here we are given that the Perimeter is 48 inches: Where perimeter is givenas

P=6s

Where s is the side of the hexagon  

6s=48

s=8 inches

Please refer to the image attached with this :

In a Hexagon , there are six equilateral triangle being formed by the three diagonals which meet at point O.  

Consider one of them , 0PQ  with side "s"  

As Apothem is the Altitude from point of intersection of diagonals to one of the side. Hence it divides the side in two equal parts . hence  

$PR = \frac{s}{2}$

Also OP= s

Using Pythagoras theorem ,  

$OP^2=PR^2+OR^2$

$8^2=(\frac{8}{2})^2+a^2$

$8^2=4^2+a^2$

$64-16=a^2$

$a^2=48$

$a=4\sqrt{3}$

Hence We have Apothem $a=4\sqrt{3}$

also we have the perimeter as 48

Now we put them in the main formula  

Area = $\frac{1}{2} \times 48 \times 4\sqrt{3}$

Area=$24 \times 4\sqrt{3}$

Area=$96\sqrt{3}$