Ratio of areas of similar triangles

[Dr. Ashani Dasgupta]

Suppose triangles ABC and XYZ are similar with AB and XY being corresponding sides. Then show that $$ \frac {[ABC]}{[XYZ]} = \frac{AB^2} {XY^2}  $$

[Saarang Srini]

ABC ~ A'B'C'

Draw perpendiculars in both triangles at D and D' from A and A' respectively.

ABD ~ A'B'D'

AB/A'B' = AD/A'D' = BD /B'D'

Ar ABC = 1/2 * AC * BD

Ar A'B'C' = 1/2 * A'C' * B'D'

Ar ABC/ Ar A'B'C' = (AC * BD) / (A'C' * B'D') = AC/A'C'  * BD/B'D'

In similar triangles, ratio of all the sides are equal.

Therefore ratio of areAs is equal to AC^2/A'C'^2 = BD^2/B'C'^2 = CA^2/C'A'^2

[swastik pramanik]

in similar triangle the corresponding sides and altitudes are in proportion

Let height of \(\Delta ABC\) be \(h\) and \(\Delta XYZ\) be \(h_1\)

Therefore, \(\frac{h}{h_1}=\frac{AB}{XY}\)

\(\frac{[ABC]}{[XYZ]}=\frac{\frac{1}{2}\times AB\times h}{\frac{1}{2}\times XY\times h_1}\)

\(\frac{[ABC]}{[XYZ]}=\frac{AB}{XY}\times\frac{h}{h_1}\)

Therefore\(\frac{[ABC]}{[XYZ]}=\frac{AB}{XY}\times \frac{AB}{XY}=\frac{AB^2}{XY^2}\)

 

[Jayanti Deb]

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