Question -
Answer -
Solution:
Given equation oflines x = 3, x = 5 and 2x-y-4 = 0.
Table for line 2x – y– 4 = 0
Plotting the graph, weget,
From the graph, weget,
AB = OB-OA = 5-3 = 2
AD = 2
BC = 6
Thus, quadrilateralABCD is a trapezium, then,
Area of QuadrileralABCD = ½ × (distance between parallel lines) = ½ × (AB) × (AD + BC)
= 8 sq units