Question

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ANS :

$\bold{ \frac{ \sqrt{2 - 1} }{ \sqrt{2} } }$
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Answer 1

Question:

In Δ ABC , seg MN ║ side AC . Seg MN divides Δ ABC into 2 parts equal in area . Determine AM/MB .

Answer:

$\huge\boxed{\boxed{\red{\sqrt{2}-1}}}$

Step-by-step explanation:

➡ Seg MN divides ABC into 2 equal parts of equal area .

➡ Ar ( Δ MNB ) = Ar ( quadrilateral ACMN )

Ar ( Δ MNB ) + Ar ( quadrilateral ACMN ) = Ar ( Δ ABC )

⇒ 2 Ar ( Δ MNB ) = Ar ( Δ ABC )

⇒ Ar ( Δ MNB ) / Ar ( Δ ABC ) = 1/2  

In Δ MNC and Δ ABC

➡ MN║ AC

➡ ∠BMN = ∠ BAC  

➡ ∠B = ∠B [ Common ]

Δ MNC ≈ Δ ABC [ A.A ]

So BM²/AB² = Area of Ar ( Δ BMN ) / Area of Ar ( Δ ABC )

⇒ BM²/AB² = 1/2

⇒ BM/AB = 1/√2

AB/MB = √2

⇒ AB/MB - 1 = √2 - 1  

⇒ ( AB - MB )/( MB ) = √2 - 1

⇒ AM/MB = √2 - 1

Hence the required ratio is √2 - 1 .

Answer 2

Question:

In Δ ABC , seg MN parallel to side AC . Seg MN divides Δ ABC into 2 parts equal in area . Determine AM/MB .

Answer:

Given:

In triangle ABC =>

MN parallel to side AC.

MN divides triangle ABC into two equal parts such that=>>

Area of triangle ABC

= Area of traingle BNM + Area of Quadrilateral MNCA.

To find: Ratio of AM to MB

Or. AM/MB = ?

Now,

In traingle ABC and traingle BMN=>

✓Angle ABC = Angle MBN

(common angle in both triangles)

✓Angle BAC = Angle BMN

(Corresponding angles as MN parallel to AC)

So,

Traingle ABC similar to traingle MBN.

(By angle-angle similarly criteria).

Now, as we know that if two triangles are similar then their ratio of areas can be defined as the square of the ratio of their corresponding sides.

Now,

Area(ABC)/Area(BMN) = (AB^2)/(BM^2)

---------(i)

But,

Area of ABC = 2 × Area of BMN

(since area of BMN is equal to MNAC)

----------------(ii)

From (i) and (ii) we get =>

=> (AB^2)/(BM^2) = 2/1

On rooting both sides we get =>>

=>AB/BM = (root 2)/1

=>( AM + BM) /BM = (root 2)/1

=>AM/BM + 1 = (root 2)/1

=> AM/BM = root 2 - 1

Hence, ratio of AM to BM is root 2 - 1.

Thanks