215k views
0 votes
In Parallelogram EFGH, diagonals (HF) and (EG) intersect at point D. Give HD=x^2-34 and DF=x^2-4x

In Parallelogram EFGH, diagonals (HF) and (EG) intersect at point D. Give HD=x^2-34 and-example-1

2 Answers

1 vote

Answer:

We can use the properties of the diagonals of a parallelogram to find the value of x.


In a parallelogram, the diagonals bisect each other, so we have:HD = DF


Substituting the given expressions for HD and DF, we get:x^2 - 34 = x^2 - 4xSimplifying this equation, we can cancel out the x^2 terms on both sides and get:-34 = -4xDividing both sides by -4, we get:x = 8.5


Therefore, the value of x is 8.5. We can use this value to find the lengths of HD and DF:


HD = x^2 - 34 = 8.5^2 - 34 = 44.75

DF = x^2 - 4x = 8.5^2 - 4(8.5) = 44.75


Therefore, HD = DF = 44.75.

User Rufflewind
by
7.0k points
2 votes

Answer:

76.5

Explanation:

In Parallelogram EFGH, diagonals (HF) and (EG) intersect at point D. Give HD=x^2-34 and-example-1