Question

If ∆ABC and ∆BDE are equilateral triangles, where D is the mid point of BC, find the ratio of areas of ∆ABC and ∆BDE.

Answer 1

SOLUTION :

Given : ΔABC and ΔBDE are equilateral triangles. D is the point of BC.

ΔABC∼ΔBDE (By AAA criteria of similarity) [all angles of the equilateral triangles are equal]

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

arΔABC/ arΔBDE = (BC/BD)²

BD = DC as D is the mid point of BC.

Hence

arΔABC/ arΔBDE = ((BD+DC)/ BD)²

arΔABC/ arΔBDE = ((BD+ BD )/ BD)²

arΔABC/ arΔBDE = (2BD/BD)²

arΔABC/ arΔBDE = (2/1)²

arΔABC/ arΔBDE = 4/1

arΔABC : arΔBDE = 4 : 1

Hence, the ratio of areas of ΔABC  and ΔBDE is 4 : 1 .  

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Answer 2

$<b>$Solution

triangles. D is the midpoint of BC.

To find: Ar(∆ABC)/Ar(∆BDE)

In ΔABC and ΔBDE

ABC and BDE congruent to each other by ( AAA criteria)

equilateral triangle are equal

Since D is the midpoint of BC, BD : DC = 1.

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

Let DC = x, and BD = x

Therefore BC = BD + DC = 2x

Hence 

ABC / BDE = BC²/ BD²

= ( BD+ DC) ² / BD²

= ( 1x + 1x) ² / 1x

= (2x².)/ 1x²

= 4x² / x²

ABC / BDE = 4/ 1

ABC: BDE = 4:1