Answer :
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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