Question

Given: △ABC, BK=10, AC=30, m∠NMO=90°, MN = MO, BK ⊥ AC, NO ∥ AC, M∈ AC
Find: NO

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Answer 1

Answer:

12

Step-by-step explanation:

Let P be the point of interaction of BK and NO.

Let BP = x

Δ NBO similar to Δ ABC, by AAA as NO // AC

BP/BK = NO/AC

∴ BP/BK = x/30

∴ BP=10*x/30=x/3

PK ⊥ NO since NM=MO & ∠NMO=90°

height of Δ NMO(PK) = NO/2 = x/2

∴ PK=x/2

area of Δ NBO + area of trapezoid OCAN = area of Δ ABC

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

(x/3)*x/2+(x+30)*x/2 = 10*15

∴ 5x^2+90x-1800=0

solving x=12 or x=-30

since length cannot be negative, so answer is 12.

∴ NO=12