Srinivasa Ramanujan, the great East Indian mathematician, was invited to Cambridge University in England in 1914 by the English
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Answer:
The area of the parallelogram is;
32 square units
Step-by-step explanation:
The given parameters are;
The coordinates of the parallelogram RSTU = R(-4, 4), S(2, 6), T(6, 2), and U(0, 0)
We note that the area of a parallelogram = Base length × Height
From the drawing of the parallelogram RSTU, we have;
The base length = The length of = The length of
= √((2 - (-4))² + (6 - 4)²) = 2·√10
The height of a parallelogram is perpendicular to its base length = The line
∴ Where, the slope of the base length = m, the slope of the height = -1/m
The slope, 'm' of = (6 - 4)/(2 - (-4)) = 1/3
Therefore, the slope of the height = -1/(1/3) = -3
We note that a point on the height is the point 'T', therefore, the equation of the line in point and slope form is therefore;
y - 0 = -3·(x - 0)
∴ y = -3·x
Therefore, the coordinates of the point 'V' is given by the simultaneous solution of the equations of and
The equation of the line in point and slope form from the point 'R' and the slope 'm = 1/3' is given as follows;
y - 4 = (1/3) × (x - (-4)) = (1/3) × (x + 4)
y = x/3 + 4/3 + 4 = x/3 + 16/3
y = x/3 + 16/3
We then have the coordinate at the point 'V' (x, y) is given as follows;
-3·x = x/3 + 16/3
-9·x = x + 16
-10·x = 16
x = -16/10 = -1.6
x = -1.6
∴ y = -3·x = -3 × -1.6 = -4.8
y = 4.8
The coordinate at the point, V = (-1.6, 4.8)
The length of the line = The height of the parallelogram = √((-1.6 - 0)² + (4.8 - 0)²) = 8/5·√10
The height of the parallelogram = 8/5·√10
The area of the parallelogram, A = Base length × Height
∴ A = 2·√(10) × 8/5·√(10) = (16/5) × 10 = 32
The area of the parallelogram, A = 32 square units.
