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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 41 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 41 Maths textbook solution.

Answers (1)

Answer:

\frac{15}{2}  sq. units

Hints:

Use concept.

Given:

Lines y=4x+5,y=5-x  and 4y=x+5

Solution:

To find area bounded by lines

 y=4x+5 {Say AB}

y=x-5  {Say BC}

4y=x+5  {Say AC}

By solving equation (1) and (2) , we get B(0,5)

By solving equation (2) and (3) , we get C(3,2)

By solving equation (1) and (3) , we get A(-1,1)

A rough sketch of the curve is as under:-

Shaded area ABC is the required area.

Required area = ar( ABD) + ( BDC)

ar( ABD) =\int_{0}^{-1}(y_1-y_2)dx

 

\begin{aligned} &=\int_{-1}^{0}\left(4 x+5-\frac{x}{4}-\frac{5}{4}\right) d x \\ &=\int_{-1}^{0}\left(\frac{15 x}{4}+\frac{15}{4}\right) d x \\ &=\frac{15}{4}\left(\frac{x^{2}}{2}+x\right)_{-1}^{0} \\ &=\frac{15}{4}\left[(0)-\left(\frac{1}{2}-1\right)\right] \\ &=\frac{15}{4} \times \frac{1}{2} \end{aligned}

 

AR ( ABD) = \frac{15}{8} sq. units

AR ( BDC) = \int_{0}^{3}(y_2-y_3)dx

\begin{aligned} &=\int_{0}^{3}\left[(5-x)-\left(\frac{x}{4}+\frac{5}{4}\right)\right] d x \\ &=\int_{0}^{3}\left[5-x-\frac{x}{4}-\frac{5}{4}\right] d x \\ &=\int_{0}^{3}\left(\frac{-5 x}{4}+\frac{15}{4}\right) d x \\ &=\frac{5}{4}\left(3 x-\frac{x^{2}}{2}\right) \\ &=\frac{5}{4}\left(9-\frac{9}{2}\right) \end{aligned}

AR ( BDC) = \frac{45}{8}  sq. units

Using equation (1), (2) and (3),

AR ( ABC) = \frac{15}{8}+\frac{45}{8}

=\frac{60}{8}

ar( ABC) = =\frac{15}{2} sq. units

Posted by

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