Question

.. ∆ABC and ∆DEF are equilateral triangles, A(∆ABC) : A(∆ DEF) = 1:3. If AB =5 then What is length of DE ? *

3√3

) 5√3

5

25​

Answer 1

Given :

• ∆ABC and ∆DEF are equilateral triangles, A(∆ABC) : A(∆ DEF) = 1:3. If AB =5

To find :

• length of DE

Solution :

In ΔABC and ΔDEF,

∠A ≌ ∠D [Each angle is of measure 60°]

∠B ≌ ∠E [Each angle is of measure 60°]

AΔBC = ΔDEF [AA test of similarity]

A(ΔABC)/A(ΔDEF)= AB²/DE² (Theorem of areas of similar triangles]

⅓ = 5²/DE²

DE² = 5²x3

DE = 5√3 units [Taking square root of both sides]

Answer 2

Given

• A(∆ABC) : A(∆ DEF) = 1:3

We Find

• length of DE

We know That

• ∆ABC and ∆DEF are equilateral triangles

According to the question

We knows,

∠ A ≈ ∠ D

∠ B ≈ ∠ E

( It's measures are 60 Degree angle )

So ,

= A(∆ABC) / A(∆ DEF) ≈ ∠AD²/∠BE²

= 1/3 ≈ 5² / 3

= DE² = 5² / 3

= DE = 5√3

So, DE = 5√3

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