Question

The area of the region enclosed by the lines \( 2x + y = 10 \), \( y = 1 \), \( y = 5 \) and the y-axis is

• A. 28 sq units
• B. 9.5 sq units
• C. 14 sq units
• D. 37.5 sq units

Hint:

For finding the area of a trapezoid, use the formula \( \text{Area} = \frac{1}{2} \times (b_1 + b_2) \times h \), where \( b_1 \) and \( b_2 \) are the lengths of the parallel sides, and \( h \) is the height.

Answer 1

The Correct Option is C

Step 1: Find the points of intersection.
We are given the lines:
- \( 2x + y = 10 \) (Line 1)
- \( y = 1 \) (Line 2)
- \( y = 5 \) (Line 3)
Let's first find the points where Line 1 intersects with \( y = 1 \) and \( y = 5 \).

Step 2: Intersection of Line 1 with \( y = 1 \).
Substitute \( y = 1 \) in \( 2x + y = 10 \):
\[ 2x + 1 = 10 \Rightarrow 2x = 9 \Rightarrow x = \frac{9}{2} \] So, the point of intersection is \( \left( \frac{9}{2}, 1 \right) \).

Step 3: Intersection of Line 1 with \( y = 5 \).
Substitute \( y = 5 \) in \( 2x + y = 10 \):
\[ 2x + 5 = 10 \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2} \] So, the point of intersection is \( \left( \frac{5}{2}, 5 \right) \).

Step 4: Find the area of the trapezoid.
The region is a trapezoid with the vertices \( \left( 0, 1 \right) \), \( \left( \frac{9}{2}, 1 \right) \), \( \left( \frac{5}{2}, 5 \right) \), and \( \left( 0, 5 \right) \). The area of the trapezoid is given by the formula:
\[ \text{Area} = \frac{1}{2} \times (b_1 + b_2) \times h \] where \( b_1 = \frac{9}{2} \), \( b_2 = \frac{5}{2} \), and the height \( h = 4 \) (the difference in \( y \)-coordinates).
Thus, the area is:
\[ \text{Area} = \frac{1}{2} \times \left( \frac{9}{2} + \frac{5}{2} \right) \times 4 = \frac{1}{2} \times \left( \frac{14}{2} \right) \times 4 = 14 \, \text{sq units} \]

Step 5: Final Answer.
Thus, the area of the region enclosed by the lines and the y-axis is \( 14 \) sq units.