In △ABC, DE¯¯¯¯¯¯¯¯ bisects BA¯¯¯¯¯¯¯¯ and BC¯¯¯¯¯¯¯¯ at D and E, respectively. If BA=5, BC=2, and AC=6, explain whether DE=6.(1 point)
The length of DE¯¯¯¯¯¯¯¯ is not 6 because DE¯¯¯¯¯¯¯¯ is the midsegment of BC¯¯¯¯¯¯¯¯. So DE=1.
The length of DE¯¯¯¯¯¯¯¯ is not 6 because DE¯¯¯¯¯¯¯¯ is the midsegment of AC¯¯¯¯¯¯¯¯. S, DE=3.
The length of DE¯¯¯¯¯¯¯¯ is 6 because AC¯¯¯¯¯¯¯¯ is the midsegment of DE¯¯¯¯¯¯¯¯.
The length of DE¯¯¯¯¯¯¯¯ is not 6 because DE¯¯¯¯¯¯¯¯ is the midsegment of BA¯¯¯¯¯¯¯¯. So DE=2.5.
please help me
To determine whether DE is equal to 6, we can use the Midsegment Theorem. The Midsegment Theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length.
In this triangle, DE is the midsegment connecting the midpoints of BA and BC. According to the Midsegment Theorem, DE is parallel to AC and is half its length.
Let's calculate the length of DE to check if it is equal to 6 or not.
The length of BC is given as 2, and DE bisects BC at point E. So, BE = CE = 1, since DE is the midsegment.
Now, let's consider the length of BA. BA is given as 5, and DE bisects BA at point D. So, BD = DA = 5/2 = 2.5, again using the fact that DE is the midsegment.
We have now determined the lengths of BE, CE, BD, and DA. We can calculate the length of DE by applying the Pythagorean Theorem to the right triangle BDE.
Using the Pythagorean Theorem: (BD)^2 + (DE)^2 = (BE)^2
Substituting the known values: (2.5)^2 + (DE)^2 = (1)^2
Simplifying: 6.25 + (DE)^2 = 1
Rearranging the equation: (DE)^2 = 1 - 6.25
Simplifying further: (DE)^2 = -5.25
Since we cannot have a negative length, DE is not equal to 6. Therefore, the correct statement is: The length of DE¯¯¯¯¯¯¯¯ is not 6 because DE¯¯¯¯¯¯¯¯ is the midsegment of BA¯¯¯¯¯¯¯¯. So DE=2.5.