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There is a parallelogram that has a side length of 8 and 7. Which of the following can be the area of the parallelogram?

I. 47 II. 55 III. 56

A. I. only
B. II. only
C. III. only
D. I and II only
E. I, II and III
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Area of the parallelogram = \(Base \ (b)\) x \(Height \ (h)\)

If the \(Height \ (h) = Side \ (a)\) , maximum possible value of the parallelogram is 56

So, the possible values of the parallelogram will be less than or equal to 56, thus all the three options satisfies, correct answer will be (E)
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==>The area of parallelogram is base*height. However, since a rectangle is a parallelogram and if one side length is 8 and the other is 7, you get the area of 8*7=56. The one side length of the parallelogram, which is 8, remains the same and the height is less than or equal to 7, so the area of the parallelogram becomes ≤56.

Therefore, the answer is I. 47 II. 55 III. 56 all possible.

The answer is E.
Answer: E
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rectangle and square are also counted as parallelogram. Test takers should raise alertness!
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Area of a parallelogram is absin(@)
where a and b are the sides of triangle and sin theta is the angle between them.

The maximum value of sin theta is 1 and minimum is 0.

So the area of a parallelogram can be little over 0 till max value 56 as sin theta would be 1 for the max value.

Hence, E is the answer.
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If you understand the concept, you can logically get to the answer pretty quickly.


Concept: whenever you are given two sides of a triangle, you can maximize the area of that triangle by making the 2 given sides perpendicular to each other. Out of all the possible triangles that can be drawn, the one with the maximum area will be when the 2 given sides are perpendicular to create a right triangle.


This concept extends to a parallelogram.

A parallelogram is defined as a quadrilateral in which each pair of opposite sides are parallel to each other. (Also leads to the property that every parallelogram has diagonally opposite angles equal in measure as well as opposite sides equals in length)


A square is simply a form of a parallelogram that has the above three features of any parallelogram and also:

-each adjacent angle is equal (i.e., each of the 4 angles are equal to 90 degrees)


Concept: given 2 side lengths of a parallelogram, we can maximize the Area of that parallelogram by placing the 2 different side lengths perpendicular to each other, in effect creating a Rectangle

In this case, the maximum area of the parallelogram will occur when we have a Rectangle with the Length = 8 and the Width = 7

Maximum Area = (8) (7) = 56

III is certainly possible


Furthermore, we can keep lowering the area by “pulling” the Rectangle by diagonal opposite Vertices, in effect “pressing down” the Rectangle into the “usual” Parallelogram figure we are accustomed to seeing.

This can be done include the Areas included in I and II

We can use 8 as the base in both cases and keep “pressing down” the rectangle and stretching it apart by its diagonal opposite vertices such that the Height as measured by the perpendicular distance between the two opposite sides of length 8 is:

Height = 55/8

Or

Height = 47/8

Area of any parallelogram = (Base) * (Perpendicular Height between the Base and its Parallel, Equal Opposite Side)


These Area can be:

(8) * (55/8) = 55

Or

(8) * (47/8) = 47


Which means all three Roman numerals are possible areas

I, II, and III

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