A trapezoid has 8.2 millimeters base and 5.2 millimeters base, as well as 10 millimeters height. What is the area?
1 answer:
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<h3>Answer: Choice B</h3>
Explanation:
Cosine is positive in quadrants I and IV, but quadrant IV isn't shaded in so we can rule out choice A.
Sine is positive in quadrants I and II. So far it looks like choice B could work. In fact, it's the answer because sin(pi/6) = 1/2 and sin(5pi/6) = 1/2. So if 0 ≤ sin(x) < 1/2, then we'd shade the region between theta = 0 and theta = pi/6; as well as the region from theta = 5pi/6 to theta = pi.
Choice C is ruled out because tangent is positive in quadrants I and III, but quadrant III isn't shaded.
Choice D is ruled out for similar reasoning as choice A. Recall that
Part A:
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by
The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same
Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by
The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same
Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>