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The bisectors of the four angles of a parallelogram are drawn. An acute angle of the parallelogram is 60° and the adjacent sides are 8 and 6 units. What is the area of the quadrilateral whose vertices are the points of intersection of the bisectors?
  • a)
    3/2 square units
  • b)
    √3/2 square units
  • c)
    √3 square units
  • d)
    1 square units
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The bisectors of the four angles of a parallelogram are drawn. An acut...
°. What is the measure of the obtuse angle of the parallelogram?

Since the opposite angles of a parallelogram are equal, the other acute angle is also 60°. Thus, the sum of the acute angles is 120°. Since the sum of the interior angles of a parallelogram is 360°, the sum of the obtuse angles is 240°.

Let's call the obtuse angle we're looking for "x". Since the bisector of an angle divides it into two equal parts, each acute angle adjacent to the obtuse angle is also split into two equal parts. This means that each of the two acute angles adjacent to the obtuse angle is 30°.

Now we can set up an equation to solve for x:

2(30°) + x = 240°

Simplifying:

60° + x = 240°

Subtracting 60° from both sides:

x = 180°

Therefore, the measure of the obtuse angle of the parallelogram is 180°.
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The bisectors of the four angles of a parallelogram are drawn. An acute angle of the parallelogram is 60° and the adjacent sides are 8 and 6 units. What is the area of the quadrilateral whose vertices are the points of intersection of the bisectors?a)3/2square unitsb)√3/2square unitsc)√3 square unitsd)1square unitsCorrect answer is option 'C'. Can you explain this answer?
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