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  • Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 23.

Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 23.

Answers (1)

Answer:

4 sq. units

Hint: use concept.

Given:

The vertices of triangle are (-1,2), (1,5) and (3,4).

Solution:

Equation of,

AB is:y=\frac{1}{2}(3x+7)

BC is: y=\frac{1}{2}(11-x)

AC is: y=\frac{1}{2}(x+5)

Required area

\begin{aligned} &=\frac{1}{2} \int_{-1}^{1}(3 x+7) d x+\frac{1}{2} \int_{1}^{3}(11-x) d x-\frac{1}{2} \int_{-1}^{3}(3 x+5) d x \\ &=\left[\frac{1}{12}(3 x+7)^{2}\right]_{-1}^{1}-\frac{1}{4}\left[(11-x)^{2}\right]_{1}^{3}-\frac{1}{4}\left[(x+5)^{2}\right]_{-1}^{3} \\ &=7+9-12 \\ &=4 \text { sq.units } \end{aligned}

 

(ii)

Equation of,

AB is:  y =\frac{3}{2}x+4

BC is:   y =4-\frac{x}{2}

AC is: y =\frac{1}{2}x+2

Required area

\begin{aligned} &=\int_{-2}^{0}\left(\frac{3}{2} x+4\right) d x+\int_{0}^{2}\left(4-\frac{x}{2}\right) d x-\int_{-2}^{2}\left(\frac{1}{2} x+2\right) d x \\ &=\left[\frac{3 x^{2}}{4}+4 x\right]_{-2}^{0}+\left[4 x-\frac{x^{2}}{4}\right]_{0}^{2}-\left[\frac{x^{2}}{4}+2 x\right]_{-2}^{2} \\ &=5+7-8 \\ &=4 \; \text{sq. units} \end{aligned}

 

(III)

 

Equation of ,

AB is:  y =x+3

BC is : y =\frac{34-5x}{2}

AC is: y =\frac{26-3x}{4}

Required area

\begin{aligned} &=\int_{2}^{4}(x+3) d x+\int_{4}^{6}\left(\frac{34-5 x}{2}\right) d x-\int_{2}^{6}\left(\frac{26-3 x}{4}\right) d x \\ &=\left[\frac{x^{2}}{2}+3 x\right]_{2}^{4}+\left[\frac{34 x}{2}-\frac{5 x^{2}}{4}\right]_{4}^{6}-\left[\frac{26 x}{4}+\frac{3 x^{2}}{8}\right]_{2}^{6} \end{aligned}

\begin{aligned} &=\left[(8+12)-(2+6)+(102-45)-(68-20)-\left(39-\frac{27}{2}\right)-\left(13-\frac{3}{2}\right)\right] \\ &=20-8+57-48-26+12 \\ &=7 \text { sq.units } \end{aligned}

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