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

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)

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.

Posted by

Safeer PP

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