33. Using the method of integration, find the area of the triangle whose vertices are A (2, 0), B (4, 5) and C (6, 3).

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Vishal Baghel | Contributor-Level 10

4 months ago

The given vertices of the triangle are A(2,0),B(4,5)and C(6,3)

So, equation of line AB is $y - 0 = \frac{5 - 0}{4 - 2} (x - 2)$

$= y = \frac{5}{2} (x - 2)$

Similarly equation of BC is

$\begin{array}{l} y - 5 = \frac{3 - 5}{6 - 4} (x - 4) \\ = y = 5 + (- x + 4) \\ y = 9 - x \end{array}$

And equation of AC is $y - 0 = \frac{3 - 0}{6 - 2} (x - 2)$

= $y = \frac{3}{4} (x - 2)$

$∴$ Area of $? A B C$

= $a r e a (? A B E) + a r e a (B C D E) - a r e a (? A C D)$

$\begin{array}{l} = \int_{2}^{4} y_{A B} d x + \int_{4}^{6} y_{B C} d x - \int_{2}^{6} y_{B C} d x \\ = \int_{2}^{4} \frac{5}{2} (x - 2) d x + \int_{4}^{6} (9 - x) d x - \int_{2}^{6} \frac{3}{4} (x - 2) d x \\ = \frac{5}{2} {[\frac{x^{2}}{2} - 2 x]}_{2}^{4} + {[9 x - \frac{x^{2}}{2}]}_{4}^{6} - \frac{3}{4} {[\frac{x^{2}}{2} - 2 x]}_{2}^{6} \end{array}$

$\begin{array}{l} = \frac{5}{2} [(\frac{4^{2}}{2} - 2.4) - (\frac{2^{2}}{2} - 2.2)] + [(9.6 - \frac{6^{2}}{2}) - (9.4 - \frac{4^{2}}{2})] - \frac{3}{4} [(\frac{6^{2}}{2} - 2.6) - (\frac{2^{2}}{2} - 2.2)] \\ = \frac{5}{2} [(8 - 8) - (2 - 4)] + [54 - 18 - 36 + 8] - \frac{3}{4} [18 - 12 - 2 + 4] \\ = 5 + 8 - 6 = 7 u n i t^{2} \end{array}$

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