Question

a square is constructed on each side of an equilateral triangle, and segments are drawn between adjacent external vertices of the three squares to form hexagon ABCDEF as shown. If the equilateral triangle has sides of length √3 - 1 units, what is the perimeter of the hexagon ABCDEF?

Answer 1

Answer:

$P=6\ units$

Step-by-step explanation:

we know that

The perimeter of the hexagon is equal to

$P=AB+BC+CD+DE+EF+FA$

Remember that

$AB+CD=EF=(\sqrt{3}-1)\ units$

$BC=DE=FA$  ----> is the length base of an isosceles triangle

so

The perimeter is equal to

$P=3(\sqrt{3}-1)+3BC$

Find the length side BC

Let

x ----> the measure of the vertex angle in the isosceles triangle

The measure of the vertex angle in the isosceles triangle is equal to

$90^o+60^o+90^o+x=360^o$

$x=120^o$

Applying trigonometric property

In the right triangle  of the attached figure

$cos(30^o)=\frac{BC/2}{\sqrt{3}-1}$ ----> by CAH (adjacent side divided buy the hypotenuse

Remember that

$cos(30^o)=\frac{\sqrt{3}}{2}$

substitute

$\frac{\sqrt{3}}{2}=\frac{BC/2}{\sqrt{3}-1}$

$BC=(3-\sqrt{3})\ units$

Find the perimeter

$P=3(\sqrt{3}-1)+3(3-\sqrt{3})$

$P=3\sqrt{3}-3+9-3\sqrt{3})$

$P=6\ units$