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Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 7 Maths textbook solution.

Answers (1)

Answer:

\frac{15}{2} sq. Units

Hint:

Given : A(-1,1),B(0,5),C(3,2)

Solution

We have to find the area of the triangle whose vertices are A(-1,1),B(0,5),C(3,2) as shown below

https://www.sarthaks.com/?qa=blob&qa_blobid=4507152498576551315

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-y_{1}=\left(\frac{5-1}{0+1}\right)(x+1) \\ &y-1=\frac{4}{1}(x+1) \\ &y-1=4 x+4 \\ &Y=4 x+5 \ldots . .(1) \end{aligned}

The equation of BC,

\begin{aligned} &y-5=\left(\frac{2-5}{3-0}\right)(x-0) \\ &=\frac{-3}{3}(x-0) \\ &y-5=-x \\ &y=5-x \ldots(2) \end{aligned}

The equation oa AC,

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

Now the required Area (A)=[(Area between line BC and x-Axis )-(Area between line AC and x-Axis) from x=0 to x=3]

Say ,AreaA =A_1+A_2

\begin{aligned} &A_{1}=\int_{-1}^{0}\left[(4 x+5)-\frac{1}{4}(x+5)\right] d x \\ &=\int_{-1}^{0}\left[4 x+5-\frac{x}{4}-\frac{5}{4}\right] d x \\ &=\int_{-1}^{0}\left(\frac{15}{4} x+\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}{8} \end{aligned}

\begin{aligned} &\text { And, } A_{2}=\int_{0}^{3}\left(y_{2}-y_{3}\right) d x \\ &=\int_{0}^{3}\left[(5-x)-\left(\frac{1}{4} x+\frac{5}{4}\right)\right] d x \\ &=\int_{0}^{3}\left[5-x-\frac{1}{4} x-\frac{5}{4}\right] d x \\ &=\int_{0}^{3}\left(-\frac{5}{4} x+\frac{15}{4}\right) d x \\ &=\frac{5}{4}\left(3 x-\frac{x^{2}}{2}\right)_{0}^{3} \\ &=\frac{5}{4}\left[9-\frac{9}{2}\right] \\ &=\frac{45}{8} \end{aligned}

So the enclosed area of the triangle is \frac{15}{8}+\frac{45}{8}=\frac{15}{2} square units

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