Area of the trapezium formed in the first quadrant by the x-axis, the y-axis, and the two lines 3x + 4y = 12 and 6x + 8y = 60.

Introduction / Context:
This geometry problem asks for the area enclosed by the coordinate axes and two parallel lines in the first quadrant. Recognizing intercepts and using triangle area helps compute the area of the resulting trapezium efficiently.

Given Data / Assumptions:

• Lines: 3x + 4y = 12 and 6x + 8y = 60 (they are parallel).
• Region: bounded by x-axis (y = 0), y-axis (x = 0), and the two given lines in the first quadrant.
• Area unit: square units.

Concept / Approach:
Find intercepts with axes to determine vertices. Each line with the axes forms a right triangle with the origin. The trapezium in question is the difference between the larger triangle (from the line farther from the origin) and the smaller triangle (from the line closer to the origin).

Step-by-Step Solution:
For 3x + 4y = 12: intercepts are x = 4 (set y = 0) and y = 3 (set x = 0). Area of small triangle = 1/2 * 4 * 3 = 6.For 6x + 8y = 60, divide by 2 → 3x + 4y = 30: intercepts are x = 10 and y = 7.5. Area of large triangle = 1/2 * 10 * 7.5 = 37.5.The trapezium area = large triangle area − small triangle area = 37.5 − 6 = 31.5.

Verification / Alternative check:
Vertices on axes are (0,3), (4,0) for the first line and (0,7.5), (10,0) for the second line. The trapezium is bounded by these four axis points, reinforcing the area difference approach.

Why Other Options Are Wrong:
48 sq. unit and 36.5 sq. unit ignore the correct intercepts or the subtraction step. 37.5 sq. unit is the large triangle alone, not the difference. 24 sq. unit is an arbitrary miscalculation.

Common Pitfalls:
Not noticing that the lines are parallel; forgetting to reduce 6x + 8y = 60; mixing up intercepts; or adding instead of subtracting the triangle areas.

Final Answer:
31.5 sq. unit