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3 years ago

In the parallelogram shown, AE = t + 2, CE = 3t − 14, and DE = 2t + 8.

Mathematics

2 answers:

zalisa [80]3 years ago

6 0

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

Dennis_Churaev [7]3 years ago

6 0

The answer is 48 units

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Statement:

The hypotenuse of a right angled triangle is 2√13 cm. If the smaller side is increased by 2 cm and the larger side is increased by 3 cm, the new hypotenuse will be √117 cm.

To find out:

The length of the larger side of the right angled triangle.

Solution:

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And if the larger side is increased by 3 cm, then the larger side will be (y + 3).

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Now, put the value of x² + y² from equation (I),

or, (2√13)² + 4x + 6y + 13 = (√117)²

or, (2 × 2 × √13 × √13) + 4x + 6y + 13 = (√117 × √117)

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or, 4x + 6y = 52

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or, y = 6

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So, the length of the larger side is 6 cm.

Answer:

6 cm

Hope you could understand.

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