Using integration, find the area of the triangle ABC, coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4)

Question

SHAIK AASIF AHAMED · Answer

Hello student,

i)Area of an enclosed region bounder by the curve y = f(x), x-axis and the boundaries,x = a to b is given by A = ∫f(x) dx in [x = a to b]
ii) Hence, here the area of the triangle ABC is enclosed by the lines AB, BC & CA; its area by integration is given by Area under AB + Area under BC - Area under AC
iii) Using two point form equation of AB, Bc & CA are respectively:
y = (5x - 18)/2; y = (12 - x) and y = (3x - 8)/4
iv) Area under AB = ∫(5x - 18)/2 dx in [4 to 6] = (1/2[5x²/2 - 18x] in [4 to 6]
Evaluating for the limits, A1 = (1/2)[(90 - 108) - (40 - 72)] = 7
Similarly area under BC = (12x - x²/2) in [6 to 8] = 10
Area under AC = (1/4)(3x²/2 - 8x) in [4 to 8] = 10
Hence net area = 7 + 10 - 10 = 7 sq units.

Thanks and Regards

Shaik Aasif

askIITians faculty

ankit singh · Answer

The equations of sides of the triangles are y = 2 x + 1 , y = 3 x + 1 , and x = 4 . On solving these question, we obtain the vertices of triangle as A ( 0 , 1 ) , B ( 4 , 1 3 ) , and C ( 4 , 9 ) . It can be observed that, A r e a ( Δ A C B ) = A r e a ( O L B A O ) − A r e a ( O L C A O ) = ∫ 0 4 ​ ( 3 x + 1 ) d x − ∫ 0 4 ​ ( 2 x + 1 ) d x = [ 2 3 x 2 ​ + x ] 0 4 ​ − [ 2 2 x 2 ​ + x ] 0 4 ​ = ( 2 4 + 4 ) − ( 1 6 + 4 ) = 2 8 − 2 0 = 8 sq. units.

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Using integration, find the area of the triangle ABC, coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4)

Using integration, find the area of the triangle ABC, coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4)

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Using integration, find the area of the triangle ABC, coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4)

Using integration, find the area of the triangle ABC, coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4)

2 Answers

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