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STEP 1: Figure Setup
first, let's name all the points, and plot this figure on a Cartesian Plane.
∵ ABCDEF is a regular hexagon, EFI is an equilateral triangle, ABHG is a square
∴ DE // HG // IF // BA
∴ HB // GA
∴ CD // IE // AF
∴ EF // CB
let's define that AB = r
∴ AB = BC = CD = DE = EF = BH = HG = GA = EI = FI = R
∴ ∠XAF = ∠ABI = ∠CBO = ∠IEF = ∠EFI = ∠FIE = ∠OBL = 60˚
Now let's define that
line LE is Line 1
line HG is Line 2
line GA is Line 3
line IF is Line 4
STEP 2 : Formular Identification
the coordinate of the point I can be written as ( R , Rsin60˚ )
the coordinate of the point E can be written as ( R + Rcos30˚ , Rsin60˚ × 2) = (1.5R , 2Rsin60˚)
∴ ∆y = 2Rsin60˚ - Rsin60˚ = Rsin60˚ = (√(3)/2)R
∴ ∆x = 1.5R - R = 0.5R
∴ m = ∆y / ∆x = ((√(3)/2)R) / (0.5R) = √(3)
∵ ∠CBO = ∠OBL
∵ ∠COB = ∠LOB = 90˚
∵ segment OB is common
∴ OC = OL = KA = EK = (√(3)/2)R = length of y-intercept of Line 1
∴ the formula for Line 1 : y = √(3)x - (√(3)/2)R
Now let's try to find the Formula of Line 2 (Line HG)
noticed that HG is a horizontal line R units away from the X-Axis
∴ the formula for Line 2 : y = R
Now let's try to find the Formula of Line 3 (Line GA)
noticed that GA is a vertical line that is R + Rcos30˚ units away from the Y-Axis
∴ the formula for Line 3 : x = R + Rcos30˚ = 1.5R
Now let's try to find the Formula of Line 4 (Line IF)
noticed that IF is a horizontal line Rsin60˚ units away from the X-Axis
∴ the formula for Line 4 : y = Rsin60˚ = (√(3)/2)R
STEP 3 : Solving the coordinates of Points I,J,G,K
let's summarize the formulas of the 4 lines we have:
Line 1 : y = √(3)x - (√(3)/2)R
Line 2 : y = R
Line 3 : x = 1.5R
Line 4 : y = (√(3)/2)R
to find the coordinate of Point J
from Line 1 and Line 2,
√(3)x - (√(3)/2)R = R
√(3)x = R + (√(3)/2)R
x = (R + (√(3)/2)R) / √(3)
y = R
to find the coordinate of Point G
from Line2 and Line3
y = R
x = 1.5R
to find the coordinate of Point K
from Line3 and Line4
x = 1.5R
y = (√(3)/2)R
to find the coordinate of Point I
from Line 4 and Line 1,
(√(3)/2)R = √(3)x - (√(3)/2)R
√(3)x = √(3)R
x = R
y = (√(3)/2)R
STEP 4 : Trapezoid Area Calculation
the requested Area IJGK is a trapezoid, so the area should be
(JG + IK) × GK / 2
So let's summarize all the points we have here:
J : ( (R + (√(3)/2)R) / √(3) , R )
G : ( 1.5R , R )
K : ( 1.5R , (√(3)/2)R )
I : ( R , (√(3)/2)R )
so the Area should be
[(1.5R - (R + (√(3)/2)R)/√(3)) + (1.5R - R)] × (R - (√(3)/2)R) / 2
in this case, R = 2 cm, I will now leave the rest of the job to you.
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