Q. 27 A5.0( 2 Votes )

# Using the method

Here required area

= Area of trapezium OABD + Area of trapezium DBCE

- Area of trapezium OACE

Now equation of line AB is

(y-1) = (5/2)(x-4)

2y - 2 = 5x - 20

y = (5x-18)/2 ...(1)

Equation of line BC is

(y-6) = (-2/2)(x-6)

(y-6) = -(x-6)

y = -x + 12 ...(2)

Similarly equation of line AC is

(y-4) = (3/4)(x-8)

4y-16 = 3x-24

4y = 3x-8

y = (3x-8)/4 ...(3)

Hence required area is

= (1/2)[90-108-40+72] + [96-32-72+18]

- (1/4)[96-64-24+32]

= (1/2)[14] + [10] - (1/4)[40]

= 7 + 10 - 10

= 7 sq. units

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