Question

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 #
.

Answer 1

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.