Question

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

Answer 1

Answer: 12 unit.    

Step-by-step explanation:

Given : In triangle ABC,

m∠NMO=90°, MN=MO, BK⊥AC, NO∥AC, M∈AC,  BK=10, AC=30,

We have to find : NO

Since, NO∥AC,

By the alternative interior angle theorem,

$\angle BNO\cong \angle BAC$

$\angle BON\cong \angle BCA$

Also,

$\angle NBO\cong \angle ABC$

Thus, by AAA similarity postulate,

$\triangle NBO\cong \triangle ABC$

Let S ∈ NO such that BS ⊥ NO,

By the property of similar triangles,

$\frac{BS}{BK}=\frac{NO}{AC}$

$\implies \frac{BK-SK}{BK}=\frac{NO}{AC}$   -------- (1),

Now, m∠NMO=90° and MN=MO,

Let J ∈ NO, such that MJ⊥NO

⇒ Triangle NMO is a isosceles triangle,

⇒ ∠MNJ = 45°,

$\implies tan 45^{\circ} = \frac{MJ}{NJ}$

$\implies MJ = NJ = SK$

$\implies NO = 2 SK$ -------(2)

From equation (1),

$\implies \frac{BK-SK}{BK}=\frac{2 SK}{AC}$

Since, BK=10, AC=30

$\implies \frac{10-SK}{10}=\frac{2 SK}{30}$

$\implies 300 - 30 SK = 20 SK \implies 50 SK = 300\implies SK = 6$

From equation (2),

NO = 2 × 6 = 12 unit.

Answer 2

Answer:

the answer is 12

Step-by-step explanation:

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