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brainiest for correct answer.

given: △abc, bk=10, ac=30, m∠nmo=90°, mn = mo,
bk⊥ac, no∥ac, m∈ac

find: no

fun one - thanx 4 posting!!!

let the interaction of BK and NO be P

and NO be x

triangle NBO n ABC are similar by AAA as NO // AC

so BP/BK=NO/AC=x/30

BP=10*x/30=x/3

PK=height of triangle NMO

cuz' NM=MO n m∠NMO=90°

height of triangle NMO=NO/2=x/2

PK=x/2

area of triangle NBO + area ot trapezoid NOCA = area of triangle ABC

BP*NO/2 + (NO+AC)*PK/2 = BK*AC/2

(x/3)*x/2+(x+30)*x/2 = 10*30/2

simplifying x^2/6+x^2/4+30x/4=150

5x^2+90x-1800=0

solving x=12 or x=-30

as length cannot be negative

NO=12

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