Question

If D is a point on the side AB of ∆ABC such that AD : DB = 3.2 and E is a point on BC such that DE || AC. Find the ratio of areas of ∆ABC and ∆BDE.

Answer 1

SOLUTION :  

Given : In ΔABC, D is appoint on the side AB such that AD:DB = 3:2. E is a point on side BC such that DE || AC.

Let AD = 2x and BD = 3x

In ΔABC and ΔBDE,

∠BDE =∠A     (corresponding angles)

∠DBE =∠ABC (common)

ΔABC∼ΔBDE  [By AA similarity]

We know that the ratio of the two similar triangles is equal to the ratio of the squares of their corresponding sides

arΔABC/ arΔBDE = (AB/BD)²

arΔABC/ arΔBDE = ((BD+DA) /BD)²

[AB = BD + DA]

arΔABC/ arΔBDE = ((3x+2x) /2x)²

arΔABC/ arΔBDE = (5x / 2x)²

arΔABC/ arΔBDE = 25x / 4x

arΔABC/ arΔBDE = 25/4

Hence, the ratio of areas of ΔABC and ΔBDE is 25 : 4.

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Answer 2

Given - AD / DB = 3/2

and DE parallel to AC

AD / BD = 2/3

AD / BD +1 = 2 /3 +1

AD + BD / BD = 2+3/ 2

AB. /BD =5/2
BD /AB = 2/5

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Given,and DE || AC.


In  ΔABC and ΔBDE, we have

∠BDE = ∠BAC  [corresponding angles]

∠BED = ∠BCA  [corresponding angles]

∠B = ∠B  [common]


ABC congruent to BDE ( by AAAcriteria)

Ar ( ABC) /BDE )= (BD/ DB) ²

( 2 /5)²

= 4 /25