Question

If the area of two similar traingle are equal,prove that they are congruent.

Answer 1

Given areas of the two similar triangles is equal

Let the two triangles be ΔABC and ΔPQR

Since the two triangles are similar,

AB/ PQ = BC/QR = AC/PR ..... (1)

Now, we know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

So

Area. of ΔABC/ Area of ΔPQR= (AB/ PQ)²= (BC/QR)² = (AC/PR )²

But Area of ΔABC= Area of ΔPQR

Which means, area of ΔABC/ Area of ΔPQR = 1

From eq (1),

(AB/ PQ)² = 1 ⇒ AB/PQ = 1 ⇒ AB = PQ ... (2)
(BC/QR)² = 1 ⇒ BC/QR= 1 ⇒ BC= QR..... (3)
(AC/PR)² =1 ⇒ AC/PR= 1 ⇒ AC= PR.... (4)

From eq, (1) (2) and (2)

ΔABC is congruent to ΔPQR (By SSS congruency criteria)

Hence Proved. 

Answer 2

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.