How To Solve: Similar Triangles

Attached pdf of this Article as SPOILER at the top! Happy learning!

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

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.



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.



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.



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 = \(k^2\)

Hope it helps!
Good Luck!