Find area of triangle whose base is 2 √3 and height is 3 √3
1 answer:
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Answer:
A. Parallelogram
Step-by-step explanation:
The given points are;
(-10, -4), (-7, -8), (-5, -4), (-2, -8)
Let 'A' represent the point (-10, -4), let 'B' represent the point (-7, -8), let 'C' represent the point (-5, -4), and let 'D' represent the point (-2, -8), we have;
The (horizontal) distance between points A and C = -5 - (-10) = 5
The (horizontal) distance between points B and D = -2 - (-7) = 5
Therefore;
The length of = The length of
The length of = √((-10 - (-7))² + (-4 - (-8))²)) = 5
The length of = √((-5 - (-2))² + (-4 - (-8))²)) = 5
∴ The length of = The length of
is parallel to the horizontal axis
is parallel to the horizontal axis
∴ ║
Given that =
, then
║
Therefore, the quadrilateral ABCD is a parallelogram
(The angle ∠ABD = 180 - arctan (4/3) ≈ 126.9 > 90, therefore, the parallelogram ABCD is not a rectangle.)
