We can start by finding the height of trapezium ABCE by using the formula for the area of a trapezium:
Area = (1/2) x sum of parallel sides x height
Substituting the given values, we get:
60 = (1/2) x (AB + CE) x h
We can also find the height of trapezium ABCD using the same formula:
48 = (1/2) x (AB + CD) x h
Since AD is parallel to BC, we know that AB + CD = AE.
Adding the above equations, we get:
108 = (AB + CE + AE) x h
We also know that DE = 4 cm, and AD is parallel to BC, so triangles ADE and BCD are similar.
Therefore:
DE/BC = AD/AB
4/BC = AD/AB
AD = 2AB
Substituting this into the above equation, we get:
108 = (AB + CE + 2AB) x h
108 = (3AB + CE) x h
h = 108 / (3AB + CE)
Next, we can use the Pythagorean theorem to find the length of CE:
CE^2 = DE^2 + CD^2
CE^2 = 4^2 + AB^2
CE^2 = 16 + AB^2
We can also find the length of AE by using the fact that AD is parallel to BC:
AE = AB + CD
AE = AB + (CE - DE)
AE = AB + CE - 4
AE = AB + √(16 + AB^2) - 4
Now we can substitute our expressions for h, CE, and AE into the formula for trapezium area:
Area of ABCE = (1/2) x (AB + CE) x h
60 = (1/2) x (AB + √(16 + AB^2)) x (108 / (3AB + √(16 + AB^2)))
Simplifying this equation:
120 = (AB + √(16 + AB^2)) x (108 / (3AB + √(16 + AB^2)))
120(3AB + √(16 + AB^2)) = (AB + √(16 + AB^2)) x 108
Squaring both sides, we get:
(120(3AB + √(16 + AB^2)))^2 = (AB + √(16 + AB^2)) x 108)^2
Simplifying the left-hand side:
(120(3AB + √(16 + AB^2)))^2 = 1296AB^2 + 6912AB + 20736
Expanding the right-hand side and simplifying:
AB^2 + 32AB - 96 = 0
Solving for AB using the quadratic formula:
AB = (-32 ± √(32^2 - 4(-96)) / 2
AB = (-32 ± √1760) / 2
AB ≈ 9.8 cm or AB ≈ -3.12 cm (extraneous)
Since we know that AB + CE = AE, we can solve for CE:
CE = AE - AB
CE = (√(16 + AB^2) + AB) - AB
CE = √(16 + AB^2)
Now we can substitute AB and CE into the given expressions for f and g:
f = (AB - CE) / 2
f = (9.8 - √(16 + 9.8^2)) / 2
f ≈ -0.96 cm (since f represents a length, we take the positive value)
g = (AD + CD) / 2 - DE
g = (2AB + CE) / 2 - 4
g = AB + √(16 + AB^2) / 2 - 4
g ≈ 6.3 cm