There is a triangular park \(ABC\) in the society. The park is divided into two sections as shown in the figure.
In the region OAC, children are allowed to play games like cricket, football,
while in the region AOB, activities which involve running are not allowed. The vertices of the triangular park that is ABC are \(A(0, 4), B(-2, 0)\) and \(C(3, 0)\).
Based on the above information, answer the following questions :
Question: 1
Write the equation of the boundary line \(AB\) of the park.
Show Hint
If one point is on the y-axis, like \(A(0,4)\), the y-intercept \(c\) is immediately known, and the equation is simply \(y = mx + c\).
• Use the intercept form of a line: \(\frac{x}{a} + \frac{y}{b} = 1\), where \(a\) is the x-intercept and \(b\) is the y-intercept.
Step 1: Identify the intercepts
For line \(AC\), it passes through \(A(0, 4)\) and \(C(3, 0)\).
x-intercept \(a = 3\).
y-intercept \(b = 4\). Step 2: Formulate the equation
\[ \frac{x}{3} + \frac{y}{4} = 1 \]
Multiplying by \(12\):
\[ 4x + 3y = 12 \]
Alternatively, in terms of \(y\): \(y = -\frac{4}{3}x + 4\).
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Question: 3
Using integration, find the area of region OAC, in which
children are allowed to play cricket, football.
Show Hint
For simple shapes like a triangle, you can check your answer using \(\frac{1}{2} \times \text{base} \times \text{height}\).
For \(OAC\), \(\frac{1}{2} \times 3 \times 4 = 6\), which matches our integration result.
Concept: • The area under a curve \(y = f(x)\) from \(x = a\) to \(x = b\) is given by \(\int_a^b f(x) \, dx\).
• Here, the region is bounded by the y-axis (\(x=0\)), the x-axis (\(y=0\)), and line \(AC\).
Step 1: Identify the limits and the function
The region \(OAC\) starts at \(x = 0\) (origin \(O\)) and ends at \(x = 3\) (point \(C\)).
The upper boundary is line \(AC\): \(y = -\frac{4}{3}x + 4\).
Step 2: Set up the integral
\[ \text{Area}(OAC) = \int_0^3 \left( -\frac{4}{3}x + 4 \right) \, dx \]
Step 3: Evaluate the integration
\[ \text{Area}(OAC) = \left[ -\frac{4}{3} \cdot \frac{x^2}{2} + 4x \right]_0^3 \]
\[ \text{Area}(OAC) = \left[ -\frac{2}{3}x^2 + 4x \right]_0^3 \]
\[ \text{Area}(OAC) = \left( -\frac{2}{3}(9) + 4(3) \right) - 0 \]
\[ \text{Area}(OAC) = -6 + 12 = 6 \text{ sq. units} \]
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Question: 4
Using integration, find the area of region AOB.
Show Hint
Area must always be positive. If your definite integral comes out negative (due to a line below the x-axis or incorrect direction), take the absolute value.
In this case, the line \(y = 2x + 4\) is above the x-axis for the chosen limits, so the result is naturally positive.
Concept: • Area in the second quadrant can be found by integrating with absolute values or by correctly defining the boundary.
• Area = \(\int_{x_1}^{x_2} y \, dx\).
Step 1: Identify limits and function
The region \(AOB\) is bounded between \(x = -2\) (point \(B\)) and \(x = 0\) (point \(O\)).
The upper boundary is line \(AB\): \(y = 2x + 4\).
Step 2: Set up the integral
\[ \text{Area}(AOB) = \int_{-2}^0 (2x + 4) \, dx \]
Step 3: Evaluate the integration
\[ \text{Area}(AOB) = [x^2 + 4x]_{-2}^0 \]
\[ \text{Area}(AOB) = (0 + 0) - ((-2)^2 + 4(-2)) \]
\[ \text{Area}(AOB) = 0 - (4 - 8) \]
\[ \text{Area}(AOB) = -(-4) = 4 \text{ sq. units} \]
