∆ABC and ∆DEF are equilateral triangles. If A(∆ABC) : A(∆DEF) = 1 : 2 and AB = 4, find DE.
Practice Set 1.4 | Q 6 | Page 25
∆ABC and ∆DEF are equilateral triangles. If A(∆ABC) : A(∆DEF) = 1 : 2 and AB = 4, find DE.
In ∆ABC and ∆DEF,
`{:(∠"A" ≅ ∠"D"),(∠"B" ≅ ∠"E"):} ...}("Each angle is of measure 60°")`
∴ ∆ABC ∼ ∆DEF ...(AA test of similarity)
By the Theorem of areas of similar triangles,
∴ `"A(∆ABC)"/"A(∆DEF)" = "AB"^2/"DE"^2`
∴ `1/2 = 4^2/"DE"^2`
∴ DE2 = 42 × 2
Taking square root of both sides,
∴ DE = 4√2 units
Answer:-
Given that ∆ABC and ∆DEF are equilateral triangles with AB = 4 and A(∆ABC) : A(∆DEF) = 1 : 2.
Since both triangles are equilateral, all their angles are 60 degrees, and we can use the AA (angle-angle) test of similarity to conclude that ∆ABC ~ ∆DEF.
According to the theorem of areas of similar triangles, the ratio of their areas is equal to the square of the ratio of corresponding sides:
A(∆ABC) : A(∆DEF) = AB^2 : DE^2
Substituting the given values, we get:
1 : 2 = 4^2 : DE^2
Simplifying this equation, we get:
DE^2 = 4^2 × 2
DE = 4√2
Therefore, the corresponding side of the bigger equilateral triangle is 4√2 units.
Chapter 1. Similarity- Practice Set 1.4 – Page 25
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