Question

2) In ∆ABC seg MN || side AC, seg MN divides ∆ABC into two parts of equal area. Determine the value of AM / AB​

Answer 1

Solution :-

given ,

→ Area (∆BMN) = Area (Quad.MACN)

→ Area (∆BMN) / Area (Quad.MACN) = 1/1

so,

→ Area (∆BMN) / Area (∆BMN) + Area (Quad.MACN) = 1/(1 + 1) = 1/2

→ Area (∆BMN) / Area (∆BAC) = 1/2

now, in ∆BMN and ∆BAC, we have,

→ ∠BMN = ∠BAC (given that, DE || BC, so , corresponding angles .)

→ ∠BNM = ∠BCA (corresponding angles .)

then,

→ ∆BMN ~ ∆BAC (By AA similarity.)

now, we know that,

• Ratio of areas of two similar ∆'s = Ratio of square of their corresponding sides.

therefore,

→ Area (∆BMN) / Area (∆BAC) = BM²/BA²

→ (1/2) = (BM/BA)²

square root both sides,

→ BM / BA = 1/√2

hence,

→ (BA - BM) / BA = (√2 - 1) / √2

→ AM / BA = (√2 - 1)/√2

→ AM / AB = ((√2 - 1)/√2) * (√2/√2)

→ AM / AB = √2(√2 - 1) / 2

→ AM / AB = (2 - √2)/2 (Ans.)

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Answer 2

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