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RD Sharma Class 12 Exercise 20.4 Areas Of Bounded Region Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 27, 2022 - 8:35 p.m. IST
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The Class 12 RD Sharma chapter 20 exercise 20.4 solution deals with the chapter of 'Area of bounded region,' which brings out the concepts of a bounded region which is a very essential concept for students if they have to understand the basic maths in a further road of knowledge. The RD Sharma class 12 exercise 20.4 collects the most critical questions important in this chapter.

RD Sharma Class 12 Solutions Chapter20 Areas Of Bounded Region - Other Exercise

Areas of Bounded Regions Excercise:20.4


Area of bounded Region exercise 20.4 question 1

Answer:

\frac{32}{3} sq.units
Hint:
Find the point of intersection of parabola and the given line and then find the required region.
Given:
x=4y-y^{2}and x=2y-3
Solution:
To find the point at intersection of the parabola x=4y-y^{2}and the line x=2y-3

Let us substitute x=2y-3 in equation of the parabola
\begin{aligned} &2 y-3=4 y-y^{2} \\ &y^{2}-2 y-3=0 \\ &(y+1)(y-3)=0 \\ &y=-1,3 \end{aligned}
Therefore, the point of intersection are D(-1,-5) and A(3,3)
The area of the required region ABCDOA,
A=\int_{-1}^{3}\left | x_{1}-x_{2} \right |dywhere x=4y-y^{2}and x=2y-3
\begin{aligned} &=\int_{-1}^{3}\left(x_{1}-x_{2}\right) d y \quad\left[\because x_{1}>x_{2}\right] \\ &=\int_{-1}^{3}\left[\left(4 y-y^{2}\right)-(2 y-3)\right] d y \\ \end{aligned}
\begin{aligned} &=\int_{-1}^{3}\left[4 y-y^{2}-2 y+3\right] d y \\ &=\int_{-1}^{3}\left[-y^{2}+2 y+3\right] d y \\ \end{aligned}
\begin{aligned} &=\left[\frac{-y^{3}}{3}+y^{2}+3 y\right]_{-1}^{3} \\ &=\left[\frac{-3^{3}}{3}+3^{2}+3(3)-\left(\frac{1}{3}+1-3\right)\right] \\ \end{aligned}
\begin{aligned} &=\left[-3^{2}+3^{2}+9-\frac{1}{3}-1+3\right] \\ &=11-\frac{1}{3} \\ &=\frac{32}{3} s q \cdot \text { units } \end{aligned}

Area of bounded Region exercise 20.4 question 3 sub question 1

Answer:

9 sq.units
Hint:
Find the point of intersection of parabola and the given line and then integrate it to find the required region.
Given:
y=2x-4 and y^{2}=4x
Solution:

Substitute y=2x-4 in y^{2}=4x
\begin{aligned} &(2 x-4)^{2}=4 x \\ &4 x^{2}+16-16 x=4 x \\ &4 x^{2}-20 x+16=0 \\ \end{aligned}
\begin{aligned} &x^{2}-5 x+4=0 \\ &(x-1)(x-4)=0 \\ &x=1,4 \\ &y=-2,4 \end{aligned}
Therefore point of intersection are (1,-2) and (4,4)
Using horizontal strips:
The required area is the shaded region
\begin{aligned} A &=\int_{-2}^{4}\left(x_{1}-x_{2}\right) d y \end{aligned} where \begin{aligned} x_{1}=\frac{y+4}{2} \end{aligned} and \begin{aligned} x_{2}=\frac{y^{2}}{4} \\ \end{aligned}
\begin{aligned} &=\int_{-2}^{4}\left[\left(\frac{y+4}{2}\right)-\left(\frac{y^{2}}{4}\right)\right] d y \\ \end{aligned}
\begin{aligned} &=\left[\frac{y^{2}}{4}+2 y-\frac{y^{3}}{12}\right]_{-2}^{4} \\ \end{aligned}
\begin{aligned} &=\left[\frac{4^{2}}{4}+2(4)-\frac{4^{3}}{12}\right]-\left[\frac{(-2)^{2}}{4}+2(-2)-\frac{(-2)^{3}}{12}\right] \\ \end{aligned}
\begin{aligned} &=12-\frac{16}{3}+3-\frac{2}{3} \\ &=9 \mathrm{sq} \cdot \text { units } \end{aligned}




Area of bounded Region exercise 20.4 question 4

Answer:

18 sq.units
Hint:
Find the point of intersection of parabola and the given line and then integrate it to find the required region.
Given:
x-y=4 and y^{2}=2x
Solution:


The parabola y^{2}=2x opens towards the positive x-axis and its focus is \left ( \frac{1}{2},0 \right )
The straight linex-y=4 passes through (4,0) and (0,-4)
Solving x-y=4 and y^{2}=2x, we get
\begin{aligned} &y^{2}=2(y+4) \\ &y^{2}-2 y-8=0 \\ &(y-4)(y+2)=0 \\ &y=4,-2 \end{aligned}
So, the points of intersection of given parabola and the line are(8,4) and (2,-2)
Required area,
\begin{aligned} &=\int_{-2}^{4} x_{\text {line }} d y-\int_{-2}^{4} x_{\text {parabola }} d y \\ \end{aligned}
\begin{aligned} &=\int_{-2}^{4}(y+4) d y-\int_{-2}^{4} \frac{y^{2}}{2} d y \\ \end{aligned}
9\begin{aligned} &=\frac{(y+4)^{2}}{2}-\left[\frac{1}{2}\left(\frac{y^{3}}{3}\right)\right]_{-2}^{4} \\ \end{aligned}\begin{aligned} &=\frac{1}{2}(64-6)-\frac{1}{6}[64-(-8)] \\ &=30-12 \\ &=18 \mathrm{sq} \cdot \text { units } \end{aligned}


The RD Sharma class 12th exercise 20.4 consists of a total of 5 questions that are short and easy, covering up the effectively important concepts of this chapter that are profitable for board exams. The concepts covered in the RD Sharma class 12 solution of Area of bounded region exercise 20.4 are-

  • The line segment of the curve

  • Area of the special bounded region

  • Area of the region that is bounded by the curve

  • Area of the region that is bounded by a parabola

The RD Sharma class 12th exercise 20.4 Is trusted by thousands of students and teachers as well when it comes to the subject of mathematics. Students who have passed the Board examinations Praise the RD Sharma solutions and recommend to all other students To make the best use of the solution. The RD Sharma class 12th exercise 20.4 Is prepared by experts who are famous in the field of mathematics to ensure that every student gets the best of the solutions of maths and can easily Ace the subject.

The RD Sharma class 12 chapter 20 exercise 20.4 Updated regularly and every year and as always presented with the latest version that is being used in the academic session. It also matches up with the syllabus of NCERT hence any student who is to prepare for the 12th board exams are any public examinations can very easily go through the RD Sharma class 12th exercise 20.4 and make sure that the self practice and evaluate the performance accordingly.

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