CBSE BOARD XII, asked by eshadube0509, 1 month ago

If vertices of a parallelogram are respectively (0,0), (1, 0), (2, 2) and (1, 2), then angle between diagonals is (1) π/3 (2) π/2 (3) π/2 (4) π/4.



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Answers

Answered by banerjeeswarnali14
0

Answer:

= π/4

Explanation :Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)= (2 – 0)/(2 – 0)

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)= (2 – 0)/(2 – 0)= 1

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)= (2 – 0)/(2 – 0)= 1So θ1 = tan-1(1) = 450 = π/4

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)= (2 – 0)/(2 – 0)= 1So θ1 = tan-1(1) = 450 = π/4Slope of second diagonal, m2, = (y2 – y1)/(x2 – x1)

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)= (2 – 0)/(2 – 0)= 1So θ1 = tan-1(1) = 450 = π/4Slope of second diagonal, m2, = (y2 – y1)/(x2 – x1)= (2 – 0)/(1 – 1)

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)= (2 – 0)/(2 – 0)= 1So θ1 = tan-1(1) = 450 = π/4Slope of second diagonal, m2, = (y2 – y1)/(x2 – x1)= (2 – 0)/(1 – 1)= ∞

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)= (2 – 0)/(2 – 0)= 1So θ1 = tan-1(1) = 450 = π/4Slope of second diagonal, m2, = (y2 – y1)/(x2 – x1)= (2 – 0)/(1 – 1)= ∞So θ2 = 900 = π/2

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)= (2 – 0)/(2 – 0)= 1So θ1 = tan-1(1) = 450 = π/4Slope of second diagonal, m2, = (y2 – y1)/(x2 – x1)= (2 – 0)/(1 – 1)= ∞So θ2 = 900 = π/2Angle between the diagonals = θ2 – θ1

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)= (2 – 0)/(2 – 0)= 1So θ1 = tan-1(1) = 450 = π/4Slope of second diagonal, m2, = (y2 – y1)/(x2 – x1)= (2 – 0)/(1 – 1)= ∞So θ2 = 900 = π/2Angle between the diagonals = θ2 – θ1= 90 – 45

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)= (2 – 0)/(2 – 0)= 1So θ1 = tan-1(1) = 450 = π/4Slope of second diagonal, m2, = (y2 – y1)/(x2 – x1)= (2 – 0)/(1 – 1)= ∞So θ2 = 900 = π/2Angle between the diagonals = θ2 – θ1= 90 – 45= 450

Given vertices of the parallelogram are (0,0), (1,0), (2,2) and (1,2)Slope of first diagonal, m1, = (y2 – y1)/(x2 – x1)= (2 – 0)/(2 – 0)= 1So θ1 = tan-1(1) = 450 = π/4Slope of second diagonal, m2, = (y2 – y1)/(x2 – x1)= (2 – 0)/(1 – 1)= ∞So θ2 = 900 = π/2Angle between the diagonals = θ2 – θ1= 90 – 45= 450= π/4

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