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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 6 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 6 Maths textbook solution.

Answers (1)

Answer:

4 square units

Hint:

Given:

A(2,1),B(3,4)and C(5,2)

Solution:

https://www.sarthaks.com/?qa=blob&qa_blobid=16818331701441575165

The equation of AB,

\begin{aligned} &y-y_{1}=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\left(x-x_{1}\right) \\ &y-1=\left(\frac{4-1}{3-2}\right)(x-2) \\ &y-1=\frac{3}{1}(x-2) \\ &y=3 x-5 \ldots \text { (1) } \end{aligned}

The equation of BC,

\begin{aligned} &y-4=\left(\frac{2-4}{5-3}\right)(x-3) \\ &=\frac{-2}{2}(x-3) \\ &y=-x+7 \ldots(2) \end{aligned}

The equation of AC,

\begin{aligned} &y-1=\left(\frac{2-1}{5-2}\right)(x-2) \\ &y-1=\frac{1}{3}(x-2) \\ &y=\frac{1}{3} x-\frac{2}{3}+1 \\ &y=\frac{1}{3} x+\frac{1}{3} \ldots . .(3) \end{aligned}

Now the required area (A)=[(Area between line AB and x-axis)-(Area between line AC and x-axis)from x=2 to x=3]

+[(Area between line BC and x-Axis )-(Area between line AC and x-Axis )from x=3 to x=5]

\begin{aligned} A &=\int_{2}^{3}\left(y_{1}-y_{3}\right) d x+\int_{3}^{5}\left(y_{2}-y_{1}\right) d x \\ =& \int_{2}^{3}\begin{aligned} \left[(3 x-5)-\left(\frac{1}{3} x+\frac{1}{3}\right) ]d x\right.\\ +\int_{3}^{5}\left[(-x+7)-\left(\frac{1}{3} x+\frac{1}{3}\right)\right] d x \end{aligned} \\ =& \int_{2}^{3}\left[3 x-5-\frac{1}{3} x-\frac{1}{3}\right] d x \\ +& \int_{3}^{5}\left[-x+7-\frac{1}{3} x+\frac{1}{3}\right] d x \end{aligned}

\begin{aligned} &=\int_{2}^{3}\left(\frac{8 x}{3}-\frac{16}{3}\right) d x+\int_{3}^{5}\left(-\frac{4}{3} x+\frac{20}{3}\right) d x \\ &=\frac{8}{3}\left(\frac{x^{2}}{2}-2 x\right)_{2}^{3}-\left(\frac{4 x^{2}}{6}-\frac{20}{3}\right)_{3}^{5} \\ \; \; \; \; =& \frac{8}{3}\left[\left(\frac{9}{2}-6\right)-(2-4)\right]-\left[\left(\frac{50}{3}-\frac{100}{3}\right)(6-20)\right] \\ &=\frac{8}{3}\left[-\frac{3}{2}+2\right]-\left[-\frac{50}{3}+14\right] \\ &=\frac{4}{3}-\left[-\frac{8}{3}\right] \end{aligned}

=4 square unit

The area of the region bounded by the triangle whose vertices are (2,1),(3,4) and (5,2) is 4 sq. units

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