∆ABC and ∆DEF are equilateral triangles. If A(∆ABC) : A (∆DEF) = 1 : 2 and AB = 4, find DE.

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Solution

Consider, A(∆ABC) : A (∆DEF) = 1 : 2
$\Rightarrow \frac{A (∆ ABC)}{A (∆ DEF)} = \frac{1}{2} \Rightarrow \frac{{AB}^{2}}{{DE}^{2}} = \frac{1}{2} \Rightarrow {DE}^{2} = 2 {AB}^{2}$
$\Rightarrow {DE}^{2} = 2 \times 4^{2} [∵ AB = 4] \Rightarrow DE = \sqrt{32} \Rightarrow DE = 4 \sqrt{2}$

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State true/false:

If in $Δ A B C$ and $Δ D E F$ , AB = DE, BC = DE, BC = EF and $∠ B = ∠ E$ , then both the triangles are congruent.

If $Δ A B C \sim Δ D E F$ such that $A B = 1.2$ cm and
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Q. Let

Δ A B C \sim Δ D E F

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and

D E = 7 c m

, then find

A B

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∆ABC and ∆DEF are equilateral triangles. If A(∆ABC) : A (∆DEF) = 1 : 2 and AB = 4, find DE.

Consider, A(∆ABC) : A (∆DEF) = 1 : 2 ⇒A∆ABCA∆DEF=12⇒AB2DE2=12⇒DE2=2AB2​ ⇒DE2=2×42 ∵AB=4⇒DE=32⇒DE=42

Consider, A(∆ABC) : A (∆DEF) = 1 : 2
$\Rightarrow \frac{A (∆ ABC)}{A (∆ DEF)} = \frac{1}{2} \Rightarrow \frac{{AB}^{2}}{{DE}^{2}} = \frac{1}{2} \Rightarrow {DE}^{2} = 2 {AB}^{2}$
$\Rightarrow {DE}^{2} = 2 \times 4^{2} [∵ AB = 4] \Rightarrow DE = \sqrt{32} \Rightarrow DE = 4 \sqrt{2}$