Math, asked by Angroshashy, 1 year ago

If the aeas of two similar triangles are equal prove that they are congruent

Answers

Answered by zoony
0
If two triangles are similar then ratio of square of the sides is equal to the ratio of areas of the triangles.
Now we will write that the areas of triangles are equal therefore their ratios will be equal to 1. now again square of sides ratio will be equal to 1.
and therefore each side will be equal and angles of similar triangles are equal.
Answered by Anonymous
0

Step-by-step explanation:

Given :-

→ ∆ABC ~ ∆DEF such that ar(∆ABC) = ar( ∆DEF) .

➡ To prove :-

→ ∆ABC ≅ ∆DEF .

➡ Proof :- ----

→ ∆ABC ~ ∆DEF . ( Given ) .

 \begin{lgathered}\tiny \sf \implies \frac{ar( \triangle ABC )}{ ar( \triangle D EF )} = \frac{AB^2}{DE^2} = \frac{AC^2}{DF^2} = \frac{BC^2}{EF^2} .........(1) . \\\end{lgathered}

Now, ar(∆ABC) = ar( ∆DEF ) [ given ] .

 \begin{lgathered}\sf \implies \frac{ar( \triangle ABC )}{ ar( \triangle D EF )} = 1..........(2). \\\end{lgathered}

▶ From equation (1) and (2), we get

 \begin{lgathered}\sf \implies \frac{AB^2}{DE^2} = \frac{AC^2}{DF^2} = \frac{BC^2}{EF^2} = 1 . \\\end{lgathered}

⇒ AB² = DE² , AC² = DF² , and BC² = EF² .

[ Taking square root both sides, we get ] .

⇒ AB = DE , AC = DF and BC = EF .

 \large\pink{ \boxed{ \tt \therefore \triangle ABC \cong \triangle D EF .}}

[ by SSS-congruency ] .

Hence, it is proved.

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