ABC & BDE R THE 2 EQUILATERAL TRIANGLES SUCH THAT D IS THE MID POINT OF BC .RATIO OF THE AREAS OF TRIANGLES ABC & BDE

 

Given : ABC and BDE are equilateral triangles and D is the mid point of BC

Let the side of ABC = a

Then the side of BDE = 

 

We know that area of equilateral triangle 

 

Putting the value of  from equation (1) in equation (2) we get

 

 

  • 77

FIrst of all, Triangle ABC is similar to Triangle BDE ---> All equilateral triangles are similar

Ar (ABC) / Ar (BED) = (BC / ED)2 
Since BED is an equilateral triangle, ED = BD
So,
Ar (ABC) / Ar (BED) = (BC / BD)2

(Ratio of areas of 2 similar triangles is equal to the square of the ratios of their corresponding sides)

BC = 2 BD (D midpoint)

So, Ar (ABC) / Ar (BED) = (BC / 2BC)2  = (1/2)2 = 1/4

Ratio
Ar (ABC) : Ar (BED) = 1 : 4

  • 26

Sorry, I made a mistake in between...

BC = 2BD
So, Ar (ABC) / Ar (BED) = (2BD / BD)2 = 22 = 4

Ratio
Ar (ABC) : Ar (BED) = 4 : 1
 

  • 23

it's alright

  • 2
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