The height of the regular hexagon shown is 24 in. calculate the area to the nearest tenth.

The complete question in the attached figure N 1
see the attached figure N 2 to better understand the problem 

we know that
In a regular hexagon, all sides equals the same length and all interior angles have the same measure
the regular hexagon can be divided into 6 equilateral triangles

angle ∡BXC=60°
therefore
∡BXY=30°
AD=24 in
XY=24/2-----> 12 in

in the triangle BXY
cos 30°=XY/XB-----------> XB=XY/cos 30°----> XB=12/(√3/2)---> XB=8√3 in
BC=8√3 in

[area of one equilateral triangle]=BC*XY/2-----> 8√3*12/2---> 48√3 in²
[area of six equilateral triangles]=6*48√3----> 288√3 in²----> 498.83----> 498.8 in²

the answer is
The area of the regular hexagon is 498.8 in²

Abcd stupid son of a bitxh