Question 39430
In rectangle ABCD, diagonal line BD=D SAY is one more than twice lined BC=W SAY. Line CD=L SAY is 7 more than line BC. find line AB.
LET AB=CD=L
BC=DA=W
HENCE BD^2=D^2=L^2+W^2
WE ARE GIVEN THAT D=2W+1 AND L=W+7...SO
(2W+1)^2=(W+7)^2+W^2
4W^2+1+4W=W^2+49+14W+W^2=2W^2+14W+49
2W^2-10W-48=0
W^2-5W-24=0
W^2-8W+3W-24=0
W(W-8)+3(W-8)=0
(W-8)(W+3)=0
W=8=BC
AB=L=W+7=15