Question

How do you solve this

Answer 1

(AD)=SQRT(20n^2+34-4n)

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)