Question

In rectangle ABCD AB+BC=28 and AC+BD=40, what is AB?

Answer 1

here AC+BD=40
in a rectangle ABCD we have AC=BD
Hence AC=BD=40/2=20
Let AB=x and BC=28-x
Now AB^2+BC^2=Ac^2
Or, x^2+(28-x)^2=(20)^2
Or, x^2+784-56x+x^2=400
Or, 2x^2-56x+784-400=0
Or, 2x^2-56x+384=0
Or, 2(x^2-38x+192)=0
Or, x^2-28x+192=0
Or, x^2-16x-12x+192=0
Or, x(x-16)-12(x-16)=0
Or, (x-16)(x-12)=0
Hence either x-16=0
Or, x=16
/OR x-12=0
Or, x=12
Therefore AB=16,12