The area of the considered trapezoid PQRS in simplest radical form is given by: Option c: 256 + 128√3 (in sq. units)
The area of a trapezoid is the half of the product of sum of its parallel sides to the height of that trapezoid (the distance between those parallel sides).
Thus, if we have:
Length of its parallel sides = 'a' and 'b' units respectively
Perpendicular distance between its parallel sides = 'h' units,
Then, we get:
[tex]A = \dfrac{1}{2}\times(a+b) \times h \: \rm unit^2[/tex]
Consider the figure attached below, which is a bit more labeled figure of the given trapezoid.
We constructed PA perpendicular to the level of SR, and therefore, as SR is parallel to PQ, we have PA perpendicular to PQ too.
Similarly, QB is constructed perpendicular to both PQ and SR.
Thus, for PQBA, all pair of adjacent sides are perpendicular to each other, therefore its a rectangle.
Sum of parallel sides = Length of PQ + Length of SR
= 24 + |SR| = 24 + |SB| + |BR| = 24 + |AS| - |AS| + |SB| |+ |BR|
= 24 + |AS| + |SB| + |BR| - |AS| = 24 + |AB| + |BR| - |AS|
Due to PABQ being a rectangle (its four sides are such that adjacent sides are perpendicular to each other), we have |PQ| = |AB| = 24 units.
Thus, we get:
Sum of parallel sides = 24 + |AB| + |BR| - |AS| = 48 + |BR| - |AS|
Perpendicular distance between its parallel sides = |PA| = |QB|
Thus, we need to find the values of 3 line segments, which are:
|BR|, |AS|, and |PA| (instead of |PA|, we can find |QB| too, since both are of same length).
From the perpective of angle R(interior) in right triangle RBQ, we know the side RQ's length (32 units), and want length of BR.
Thus, its hypotenuse and base from the perspective of angle R (interior).
Thus, we can use cosine ratio or secant ratio. Let we use cosine ratio.
Then, we get:
[tex]\cos(30^\circ) = \dfrac{|BR|}{|RQ|} \\\\|BR| = 32 \times \cos(30^\circ) = 32 \times \dfrac{\sqrt{3}}{2} = 16\sqrt{3}\: \rm units[/tex]
For the same triangle as in case 1 (ie RBQ), but now taking the sine ratio, we get:
[tex]\sin(30^\circ) = \dfrac{|BQ|}{|RQ|} \\\\|BQ| = 32 \times \sin(30^\circ) = 16 \: \rm units[/tex]
This is same as |PA| because of PQBA being a rectangle.
The angle PSA is supplement of angle PSR as both joined make a straight line ASR. Thus, we get:
[tex]m\angle PSA + m\angle PSR = 180^\circ\\m\angle PSA = 180^\circ-135^\circ = 45^\circ[/tex]
Thus, for the triangle PSA, similar to the first case or second case, but now using tangent ratio as we've got |PA| = |QB| = 16 units, we get:
[tex]\tan(m\angle PSA) = \dfrac{|PA|}{|AS|}\\\tan(45^\circ) = \dfrac{|PA|}{|AS|} \\\\|AS| = 1 \times 16 = 16 \: \rm units[/tex]
Thus, we get:
Sum of parallel sides = 48 + |BR| - |AS| = 48 + 16√3 - 16 = 32 + 16√3 units
Perpendicular distance between its parallel sides = |PA| = |QB| = 16 units.
Thus, Area of the trapezoid PQRS is evaluated as:
[tex]A = \dfrac{1}{2} \times \text{Sum of parallel sides} \times \text{Distance between those sides}\\A =\dfrac{1}{2} \times (32 + 16\sqrt{3}) \times 16 \\\\A = 256 + 128\sqrt{3} \: \rm unit^2[/tex]
Thus, the area of the considered trapezoid PQRS in simplest radical form is given by: Option c: 256 + 128√3 (in sq. units)
Learn more about inverse trigonometric functions here:
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