Question

6. Find the distance from A to B for the hexagonal nut shown below: А 1.50 in BYo I've asked tutors and they have been unable to answer, after all it's only given one side and I need some help.

Answer 1

Let

x ------> the length side of the regular polygon

we have a regular hexagon

that means

the interior angle of this polygon is

180(6-2)/6=120 degrees

A regular hexagon can be divided into 6 congruent equilateral triangles

see the attached figure to better understand the problem

in the right triangle of the figure

we have that

sin(60)=0.75/x

solve for x

x=0.75/sin(60)

Remember that

`\sin (60^o)=\frac{\sqrt[]{3}}{2}`

substitute

`\begin{gathered} x=0.75\colon\frac{\sqrt[]{3}}{2} \\ \\ x=\frac{1.50}{\sqrt[]{3}}\cdot\frac{\sqrt[]{3}}{\sqrt[]{3}}=\frac{1.50\sqrt[]{3}}{3}=\frac{\sqrt[]{3}}{2} \end{gathered}`

Part 2

Find the distance AB

Applying the Pythagorean Theorem

AB^2=1.5^2+x^2

substitute the value of x

AB^2=2.25+(3/4)

AB^2=3

`AB=\sqrt[]{3}\text{ in}`

the distance AB is the square root of 3 inches