Re: What is the area of the quadrilateral shown above?
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23 Jan 2019, 07:43
Yes it can be considered as a trapezoid, since the two angles alpha are the same and are connected to the longer of the two bases.
Area: 0.5(Base 1 + Base 2) * height.
How do we get the height? The shorter base (length 2) must have its center where the longer base has its center due to the fact that both angles are equal. Thus, we derive that the longer base just extends the shorter base by (4-2 = 2). Split equally on each side, we can see the longer base composition of lengths 1 + 2 + 1.
If we look at the left part of the figure we have the upward sloping line with length 2. If we let fall a perpendicular from the connection of the upward sloping line and its vertex with the shorter base, we arrive exactly at the first part of the 1 + 2 + 1 composition of the longer base.
Thus we have a created triangle with a base 1 and hypotenuse 2. Since we also know that it has a 90-degree angle, we can deduct it must be a 30 - 60 - 90 triangle. (Remainder: Side lengths of a 30 - 60 - 90 triangle are 1:2:sqrt(3). Thus the height of the triangle is sqrt(3) which equals the height of the trapezoid.
Plug in the formula.
Area : 1/2*6* √3 = 3√3