Ex.6.4 Q4 Triangles Solution - NCERT Maths Class 10

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Question

If the areas of two similar triangles are equal, prove that they are congruent.

Diagram

Text Solution

 

Reasoning:

As we know that two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio.

The ratio of the areas of two similar trianglesis equal to the square of the ratio of their corresponding sides.   

As we know if three sides oh one triangle are equal to the three sides of another triangle,then the two triangles are congruent.

Steps:

\(\Delta ABC \sim \Delta DEF\)

\(\begin{align}\Rightarrow \frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{CA}}{{FD}} \end{align}\)(\(SSS\) criterion)

But area of \(\Delta ABC\) \(=\) Area of \(\Delta DEF\)

\[\begin{align}\Rightarrow\,\,\frac{{Area\,of\,\Delta ABC}}{{Area\,of\,\Delta DEF}} = 1\,\,\,\,\,\,\,....(1)\end{align}\]

But,
\[\begin{align}\frac{{Area\,of\,\Delta ABC}}{{Area\,of\,\Delta DEF}} &= \frac{{{{(AB)}^2}}}{{{{(DE)}^2}}} \\= \frac{{{{(BC)}^2}}}{{{{(EF)}^2}}} &= \frac{{{{(CA)}^2}}}{{{{(FD)}^2}}}\end{align}\]

From (1)

\[\begin{align}{\frac{(A B)^{2}}{(D E)^{2}}}&={\frac{(B C)^{2}}{(E F)^{2}}=\frac{(C A)^{2}}{(F D)^{2}}=1} \\ {\Rightarrow \quad \frac{(A B)^{2}}{(D E)^{2}}}&={1} \\ {\Rightarrow \,\,\quad(A B)^{2}}&={(D E)^{2} }\\ {\Rightarrow \;\;\quad A B}&={D E \ldots (2)}\end{align}\]

Similarly,       

 \( \Rightarrow\)   \( BC = EF .... (3)\)

            \( \Rightarrow\)   \(CA =FD …. (4)\)

Now, in \(\Delta ABC,\,\Delta DEF\)

\( \Rightarrow\)\(AB = DE\)                   (form \(2\))

\( \Rightarrow\)\(BC = EF\)                   (form \(3\))

\( \Rightarrow\)\(CA = FD\)                   (form \(4\))

\( \Rightarrow \Delta ABC \cong \Delta DEF\) (SSS congruency)

  
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