Answer:
(AD)=SQRT(20n^2+34-4n)
Step-by-step explanation:
let the point of intersection between AC and DB be O. Then the length of OA is the same as that of OC which is=2n+5
Similarly the length of DO is the same as that of OB which is=4n-3
Now from Pathagoras theorem we get
(AD)=SQRT((DO)^2+(OA)^2)
(AD)=SQRT((4N-3)^2+(2N+5)^2)
(AD)=SQRT((16n^2+9-24n)+(4n^2+25+20n))
(AD)=SQRT(20n^2+34-4n)
