Question

$\triangle ABC and \triangle BDE$ are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangle ABC and BDE is
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4

Answer 1

Answer:

The ratio of the ar (ΔABC) : ar(ΔBDE) is 4 : 1.

Among the given options option (c) is 4 : 1  is the correct answer.

Step-by-step explanation:

Given:

ΔABC and ΔBDE are two equilateral triangles and D is the midpoint of BC.

 

Since, ΔABC and ΔBDE are equilateral ∆’s Hence they are equiangular, ΔABC ~ ΔBDE.

 ar (ΔABC) / ar(ΔBDE) = (AB/BD)²

[The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides]

ar (ΔABC) / ar(ΔBDE) = (BC/BD)²

[AB = BC = AC , sides of equilateral triangle]

ar (ΔABC) / ar(ΔBDE) = (BD + DC)² /(BD)²

ar (ΔABC) / ar(ΔBDE) = (BD + BD)² /(BD)²

[BD = DC, D is the midpoint of BC]

ar (ΔABC) / ar(ΔBDE) = (2BD)²/BD²  

ar (ΔABC) / ar(ΔBDE) = 4 BD² / BD²

ar (ΔABC) / ar(ΔBDE) = 4/1

ar (ΔABC) : ar(ΔBDE) = 4 : 1

Hence, the ratio of the ar (ΔABC) : ar(ΔBDE) is 4 : 1.

HOPE THIS ANSWER WILL HELP YOU…

 

Answer 2

Equilateral triangles

ΔABC and Δ DEF

$\frac{de}{ab} = \frac{12}{6} = 2$

$\frac{ef}{bc} = \frac{12}{6 } = 2$

$\frac{df}{ac} = \frac{12}{6} = 2$

SSS triangle similarities

ΔABC is congruent to ΔDEF

here,

Δ ABC and BCE is equilateral BD=

$\frac{1}{2} \: bc \: and \: d \: is \: the \: midpoint$

Now,

$\frac{abc}{bbc}$

here,

Δ ABC and BDE are equaletrial both Δ sides ration are the same and ΔABC ~ ΔBDE

here,

$\frac{area \: ofthe \: triangle(bc) {}^{2} }{area \: of \: the \: triangle(bd) {}^{2} }$

$= (bc) {}^{2}$

divided by

$(bc) {}^{2}$

divided by 4

=4 (BC)²/(BC) ²

$= \frac{4}{1}$

So,

answer =4:1