Rectangle upper A upper B upper C upper D is diagonally intersected by a line that connects upper B to upper D, forming 2 triangles. Sides upper A upper B and upper C upper D are marked at their centers by a single tick mark.

if rectangle ADCB
has DB=26
and AB=10
, how long is CB¯¯¯¯¯¯¯¯
?

Option #1: 16

Option #2: 24

Option #3: 28

Option #4: 32

(1 point)
The correct response is Option #
.

1 answer

To solve for the length of \( CB \) in rectangle \( ABCD \), we need to understand the arrangement of the rectangle and the meaning of the marked distances.

Given:

  • \( DB = 26 \) (which means the distance from point D to point B along the diagonal \( BD \) is 26)
  • \( AB = 10 \) (the height of the rectangle or the length of side \( AB \))

In rectangle \( ABCD \):

  • \( AB \) is perpendicular to \( AD \).
  • \( AB = AD = 10 \) (since opposite sides of a rectangle are equal).
  • \( DB \) is the diagonal from point D to point B.

Using the Pythagorean theorem in right triangle \( ABD \) (with right angle at \( A \)): \[ DB^2 = AB^2 + AD^2 \] Substituting in the known values: \[ 26^2 = 10^2 + AD^2 \] \[ 676 = 100 + AD^2 \] \[ AD^2 = 676 - 100 \] \[ AD^2 = 576 \] \[ AD = \sqrt{576} = 24 \]

Since \( CB \) is equal to \( AD \) (the height of the rectangle), \[ CB = 24 \]

Therefore, the correct response is Option #2: 24.

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