Question

ASAP

Answer 1

Answer:

Step-by-step explanation:

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

Answer 2

Answer:

Step-by-step explanation:

I have a slightly different way to solve this.

Let P be the mid-point of NO.

As MN = MO, MNO is an isosceles triangle,

line MP is ⊥ to NO

As ∠NMO=90°, triangle MNO is similar to triangles PNM and POM.

MP = NP = PO

MP = NO/2

Let Q be the interaction of BK and NO

BQ = BK - QK

As QK = MP

BQ = BK - MP

= 10 - NO/2

Triangle BNO and BAC are similar

BQ/NO = BK/AC

Substituting BQ by 10 - NO/2

(10 - NO/2) / NO = 10 /30

NO = 3*(10 - NO/2)

NO = 30 - 3*NO/2

5*NO/2 = 30

NO = 12