(a) The area of a trapezium is given by the formula:
Area = (1/2) × sum of parallel sides × distance between them
Substituting the given values, we get:
10 = (1/2) × (AB + DC) × h
where h is the perpendicular distance between AB and DC.
We can express h in terms of AB and DC using the Pythagorean theorem, since AD is the height of the right triangle ACD:
AD² + h² = DC²
(x - 1)² + h² = (x + 3)²
x² - 2x + 1 + h² = x² + 6x + 9
h² = 8x + 8
Substituting this value of h² in the equation for the area, we get:
10 = (1/2) × (AB + DC) × (sqrt(8x + 8) / sqrt(1))
10 = (1/2) × (AB + DC) × sqrt(8x + 8)
Substituting the given values of AB and DC in terms of x, we get:
10 = (1/2) × ((3x + 2) + (x + 3)) × sqrt(8x + 8)
10 = (2x + 5) × sqrt(8x + 8)
(2x + 5)² × (8x + 8) = 100
(2x + 5)² × 2(x + 1) = 25
4x² + 4x + 25 = 25
4x² + x - 25 = 0
(b) We can solve the quadratic equation 4x² + x - 25 = 0 using the quadratic formula:
x = [-b ± sqrt(b² - 4ac)] / 2a
where a = 4, b = 1, and c = -25.
Substituting the values, we get:
x = [-1 ± sqrt(1² - 4(4)(-25))] / 2(4)
x = [-1 ± sqrt(401)] / 8
x = (-1 ± 20.025) / 8
x = -3.128 or x = 1.628
Since the length of a side cannot be negative, we reject the negative value of x and take x = 1.628.
Substituting this value of x in the expression for AB, we get:
AB = 3x + 2 = 3(1.628) + 2 = 7.884 ≈ 7.88 cm
Therefore, the length of AB is approximately 7.88 cm, correct to 2 significant figures.