Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

 

Answers (1)
S safeer

Here \bigtriangleup ABC \sim \bigtriangleup D\!E\!F

To prove :

\frac{ar(\bigtriangleup ABC)}{ar(\bigtriangleup DEF)}=\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}

\frac{ar(\bigtriangleup ABC)}{ar(\bigtriangleup DEF)}=\frac{\frac{1}{2}\times BC \times AG}{\frac{1}{2}\times EF \times DH}

\frac{ar(\bigtriangleup ABC)}{ar(\bigtriangleup DEF)}=\frac{ BC }{EF} \times \frac{AG}{DH} \;\;\;\;\;\;\; -(1)

Now in the \bigtriangleup ABG  and  \bigtriangleup D\!E\!H

\angle B = \angle E

\angle AG\!B = \angle D\!E\!H   (each 90^{\circ}

\therefore \bigtriangleup AG\!B \sim \bigtriangleup D\!E\!H

\therefore \frac{AB}{DE}= \frac{BG}{EH} = \frac{AG}{DH}

\text{But}\;\; \frac{AB}{DE}= \frac{BC}{EF} \;\; (\bigtriangleup ABC \sim \bigtriangleup DEF )

\therefore \frac{AG}{DH}= \frac{BC}{EF}\;\;\;\;\;\; -(2)

From eqns (1) and (2), we get 

\frac{ar(\bigtriangleup ABC)}{ar(\bigtriangleup DEF)}= \frac{BC}{EF} \times \frac{BC}{EF} = \frac{BC^2}{EF^2}

Similarly, we can prove 

\frac{ar(\bigtriangleup ABC)}{ar(\bigtriangleup DEF)}=\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}

Hence proved.

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