In regular hexagon RSTUVW with center X, ST =feet.?

Question

In regular hexagon RSTUVW with center X, ST =feet.

a. Find the area of triangle STX.
b. Find the area of hexagon RSTUVW.
c. Describe two different methods for finding the area of hexagon RSTUVW.

http://tinypic.com/view.php?pic=2qnabdu&s=2

Answer 1

The triangle STX is equilateral and its area is ST^2sqrt(3)/4
There are 6 = triangles making up the hexagon so area of hexagon = 6ST^2sqrt(3)/4 = 3ST^2sqrt(3)/2

The altitude of any one of the six congruent equilateral triangles is called the apothem. The area of the hexagon is .5*apothem*P where P is the perimeter of the hexagon.

This is true of any regular polygon.

Answer 2

a)

let ST = a ft

STX is an equilateral triangle with side a feet

drop a perpendicular XA from X to ST

SAX is a right angled triangle

now SA = AT( in eqilateral triangle perpendicular bisects the opposite side

SA = AT = a/2

XA = height , h feet

by pythagoras theorem

h ^2 = XS^2 - SA^2

= a^2 - a^2/4

= 3a^2/4

h = a sqrt(3)/2

area of triangle XAS = bh/2

= (a/2) (asqrt (3)/4 = a^2 sqrt(3)/8

b)
area of hexagon = 12* area of XAS

= 12(a^2 sqrt(3)/8

= 3a^2 sqrt(3)/2

c)
another method to find out the area of hexagon

join SW and TV to form a rectangle STVW

ST = WV = a and

SW = TV = 2XA = 2(a sqrt(3)/2)

= a sqrt(3)

area of rectangle = a(a sqrt(3)

= a^2 sqrt(3)

in triangle RSW

SW = a sqrt(3)

RS = a

and height = a/2

area of triangle RSW = bh/2 = a sqrt(3)(a/2)/2

= a^2 sqrt(3)/4

area of hexagon =

area of rectangle STVW+ 2 area of triangle RSW

= a^2 sqrt(3) + 2a^2 sqrt(3)/4

= 6a^2 sqrt(3)/4

= 3a^2 sqrt(3)/2

Answer 3

the real part of (-4+4i?3) is Cos(?/9) + ?3 Sin(?/9) which has the complicated expression of ((a million/2)(-a million+i?3))^(-a million/3) + ((a million/2)(-a million+i?3))^(a million/3) this is genuinely a root of the equation x³ -3x + a million. enable me see if i pass to discover yet in any different case of proving this. Oh, ksoileau has already crushed me to this. Kudos, ksoileau, sturdy artwork. I additionally see that (?) Dr D has fixed his answer too. Thumbs up for the two.