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Explain solution RD Sharma Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 33 Maths.

Answers (1)

Answer:

7sq. Units.

Hint:

Use conceptual part.

Given:

Triangle ABC, coordinates of whose vertices are A(4, 1), B(6, 6) and C (8,4).

Solution:

Equation of AB is given by

y=-\frac{5}{2}x-9

                Equation of BC is given by

y=-x+12

Equation of AC is given by

y=\frac{3}{4}x-2

Area of  ABC

\begin{aligned} &=\int_{4}^{6}\left(y_{A B}-y_{A C}\right) d x+\int_{6}^{8}\left(y_{B C}-y_{A C}\right) d x \\ &=\int_{4}^{6}\left(\frac{5}{2} x-9-\frac{3}{4} x+2\right) d x+\int_{0}^{8}\left(-x+12-\frac{3}{4} x+2\right) d x \\ &=\int_{4}^{6}\left(\frac{7}{4} x-7\right) d x+\int_{6}^{8}\left(-\frac{7 x}{4}+14\right) d x \\ &=\left[\frac{7 x^{2}}{8}-7 x\right]_{4}^{6}+\left[-\frac{7 x^{2}}{8}+14 x\right]_{6}^{8} \\ &=\left[\left(\frac{63}{2}-42\right)-(14-28)\right]+\left[(-56+112)-\left(\frac{-63}{2}\right)+84\right] \\ &=\left[\frac{63}{2}-42-14+28-56+112+\frac{63}{2}-84\right] \\ &=63-56 \\ &=7 \text { sq.units } \end{aligned}

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