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Hi All,
I have recently uploaded a video on YouTube to discuss Similar Triangles in Detail:
Following is covered in the video
¤ Definition of Similar Triangles ¤ Properties of Similar Triangles ¤ Relationship of Perimeter of two Similar Triangles ¤ Relationship of Area of two Similar Triangles
Definition of Similar Triangles
Two triangles are similar if at least two of their corresponding angles are equal.
=> If two angles are equal then the third angle will also be equal (As sum of the angles is 180°) => If all three corresponding angles of two triangles are equal then they are similar triangles
Attachment:
Image-1.jpg [ 10.69 KiB | Viewed 226 times ]
In above Figure △ ABC and △ DEF are similar because ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
Properties of Similar Triangles
If two triangles are similar, then their corresponding sides will be in the same ratio.
Attachment:
Image-1.jpg [ 10.69 KiB | Viewed 226 times ]
In above Figure △ ABC and △ DEF are similar => \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\)
Relationship of Perimeter of two Similar Triangles
Ratio of Perimeter of two similar triangles is equal to the ratio of their sides.
Attachment:
Image-1.jpg [ 10.69 KiB | Viewed 226 times ]
In above Figure △ ABC and △ DEF are similar => \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\) = k (assume) => AB = k*DE => BC = k*EF => AC = k*DF
=> Perimeter of △ ABC / Perimeter of △ DEF = \(\frac{AB + BC + AC }{ DE + EF + DF}\) = \(\frac{k*DE + k*EF + k*DF }{ DE + EF + DF}\) = \(\frac{k * ( DE + EF + DF ) }{ DE + EF + DF}\) = k = \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\)
Relationship of Area of two Similar Triangles
Ratio of Area of two similar triangles is equal to square of ratio of their sides.
Attachment:
Image-2.jpg [ 11.85 KiB | Viewed 216 times ]
In above Figure △ ABC and △ DEF are similar and AG is perpendicular(⊥) to BC and DH ⊥ EF
If we consider △ AGB and △ DHE, then ∠B = ∠E, ∠G = ∠H = 90° => ∠GAB = ∠HDE => △ AGB and △ DHE => Their sides will be in the same ratio => \(\frac{AG}{DH}\) = \(\frac{GB}{HE}\) = \(\frac{AB}{DE}\) ...(1)
And we already know that △ ABC and △ DEF => \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\) = k ...(2)
From (1) and (2) we get \(\frac{AG}{DH}\) = \(\frac{GB}{HE}\) = \(\frac{AB}{DE}\) = \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\) = k
=> Area of △ ABC / Area of △ DEF = (\(\frac{1}{2}\) * BC * AG) / (\(\frac{1}{2}\) * EF * DH) = \(\frac{BC * AG }{ EF * DH}\) = \(\frac{BC}{EF}\) * \(\frac{AG}{DH}\) = k * k = \(k^2\)