Question

The side of each of the equilateral triangles in the figure is twice the side of the central regular hexagon. What fraction of the total area of the six triangles is the area of the hexagon?

Answer 1

Answer:

The fraction is 1/4

Step-by-step explanation:

we know that

The area of an equilateral triangle, using the law of sines is equal to

$A=\frac{1}{2}x^{2}sin(60^o)$

$A=\frac{1}{2}x^{2}(\frac{\sqrt{3}}{2})$

$A=x^{2}\frac{\sqrt{3}}{4}$

where

x is the length side of the triangle

In this problem

Let

b ----> the length side of the regular hexagon

2b ---> the length side of the equilateral triangle

step 1

Find the area of the six triangles

Multiply the area of one triangle by 6

$A=6[x^{2}\frac{\sqrt{3}}{4}]$

$A=3x^{2}\frac{\sqrt{3}}{2}$

we have

$x=2b\ units$

substitute

$A=3(2b)^{2}\frac{\sqrt{3}}{2}\\\\A=6b^{2}\sqrt{3}\ units^2$

step 2

Find the area of the regular hexagon

Remember that, a regular hexagon can be divided into 6 equilateral triangles

so

The area of the regular hexagon is the same that the area of 6 equilateral triangles

$A=3x^{2}\frac{\sqrt{3}}{2}$

we have

$x=b\ in$

substitute

$A=3(b)^{2}\frac{\sqrt{3}}{2}$

step 3

To find out what fraction of the total area of the six triangles is the area of the hexagon, divide the area of the hexagon by the total area of the six triangles

$3(b)^{2}\frac{\sqrt{3}}{2}:6b^{2}\sqrt{3}=\frac{3}{2} :6=\frac{3}{12}=\frac{1}{4}$