Q. 205.0( 1 Vote )

Using the method

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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Draw a rough sketMathematics - Exemplar

Compute the area Mathematics - Exemplar

Using integrationMathematics - Board Papers

Find the area of Mathematics - Board Papers

Find the area of Mathematics - Exemplar

Evaluate <span laMathematics - Board Papers

The area of the rMathematics - Exemplar

Find the area bouMathematics - Exemplar