Given:
In rectangle ABCD, diagonals AC and BD intersect at E, AE=3x-28 and DE=.5x+12.
To find:
The length of AC.
Solution:
We know that, diagonals of a rectangle are equal and they bisect each other.
In rectangle ABCD, diagonals AC and BD intersect at E, so
[tex]AE=BE=CE=DE[/tex] ...(i)
Taking [tex]AE=DE[/tex], we get
[tex]3x-28=0.5x+12[/tex]
[tex]3x-0.5x=28+12[/tex]
[tex]2.5x=40[/tex]
Divide both sides by 2.5.
[tex]x=\dfrac{40}{2.5}[/tex]
[tex]x=16[/tex]
Now,
[tex]AE=3x-28[/tex]
[tex]AE=3(16)-28[/tex]
[tex]AE=48-28[/tex]
[tex]AE=20[/tex]
Using segment addition property,
[tex]AC=AE+CE[/tex]
[tex]AC=AE+AE[/tex] [Using (i)]
[tex]AC=20+20[/tex]
[tex]AC=40[/tex]
Therefore, the length of AC is 40 units.
