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  • Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20 3 question 30 sub question 2.

Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 30 sub question 2.

Answers (1)

Answer:

6.5 sq. units

Hint:

Given:

3x-2y+1=0 , 2x+3y-21=0 , x-5y+9=0

Solution:

\begin{aligned} &\\ &3 x-2 y+1=0, y_{1}=\frac{(3 x+1)}{2}\\ &2 x+3 y-21=0, y_{2}=\left(\frac{21-2 x}{3}\right)\\ &x-5 y+9=0, y_{3}=\frac{(x+9)}{5} \end{aligned}

Point off intersection of (i)  and (ii)  is A(3,5)

Point off intersection of  (ii) and (iii)  is B(6,3)

Point off intersection of  (iii) and (i)  is C(1,2)

Area of the region bonded

 

\begin{aligned} &\int_{1}^{3} y_{1} \cdot d x+\int_{3}^{6} y_{2} \cdot d x-\int_{1}^{6} y_{3} \cdot d x \\ &=\int_{3}^{1} \frac{(3 x+1)}{2} \cdot d x+\int_{3}^{6} \frac{(21-2 x)}{3} \cdot d x-\int_{1}^{6} \frac{(x+9)}{5} \cdot d x \\ &=\frac{1}{2}\left(\frac{3 x^{2}}{2}+x\right)_{1}^{3}+\frac{1}{3}\left(21 x-x^{2}\right)_{3}^{6}+\frac{1}{5}\left(\frac{x^{2}}{2}+9 x\right)_{1}^{6} \\ &=\frac{1}{2}[14]+\frac{1}{3}[36]-\frac{1}{3}\left[\frac{125}{2}\right] \\ &=7+12-12.5 \\ &=6.5 \text { sq.units } \end{aligned}

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