In the parallelogram shown, AE = t + 2, CE = 3t − 14, and DE = 2t + 8.What is the length of line segment DB?

20 units
24 units
48 units
68 units

Answers

Answer 1

Answer: Answer: 48 units

Given,
ABCD is a parallelogram
The diagonals bisect at E
Line segment AE = t + 2
Line segment CE = 3t – 14
Line segment DE = 2t + 8

Line segment DB is unknown

Since ABCD is a Parallelogram, Midpoint AC = Midpoint DB
The midpoint is E
Therefore
AE = CE … (Equation I)
DE = BE … (Equation II)

DB = DE + BE
Since DE = BE (Equation II),
DB = 2DE = 2BE

Given, DE = 2t + 8
DB = 2(2t + 8)
DB = 4t + 16 … (Equation III)

To solve for the value of ‘t’
AE = CE … (Equation I)
Line segment AE = t + 2
Line segment CE = 3t – 14
Therefore, t + 2 = 3t – 14

Subtract t from both sides of the equation
3t – 14 – t = t + 2 – t
2t – 14 = 2

Add 14 to both sides of the equation
2t – 14 + 14 = 2 + 14
2t = 16
t = 8

Substitute 8 for t in Equation III to solve for the length of line segment DB
DB = 4t + 16 … (Equation III)
DB = (4*8) + 16
DB = 32 + 16
DB = 48

Therefore, the length of line segment DB is 48 units

Answer 2

Answer: the answer is 48 units

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