We will discuss here about the different criteria of similarity between triangles with the figures.
1. SAS criterion of similarity:
If two triangles have an angle of one equal to an angle of the other and the sides including them are proportional, the triangles are similar.
In ∆XYZ and ∆PQR, if ∠Y = ∠Q and \(\frac{XY}{PQ}\) = \(\frac{YZ}{QR}\) then ∆XYZ ∼ ∆PQR.
Similarly, if ∠X = ∠P and \(\frac{XY}{PQ}\) = \(\frac{XZ}{PR}\) then ∆XYZ ∼ ∆PQR.
Also, if ∠Z = ∠R and \(\frac{XY}{PR}\) = \(\frac{YZ}{QR}\) then ∆XYZ ∼ ∆PQR.
2. AA criterion of similarity:
If two triangles have two angles of one equal to two angles of the other, the triangles are similar.
In ∆XYZ, if ∠X = ∠P and ∠Y then ∆XYZ ∼
∆PQR.
If in two triangles, two angles of one are equal to two angles of the ther, then the third angle of the first triangle is also equal to the third angle of the other because the sum of the three angles in a triangle is 180°.
Thus, similar triangles are equiangular.
3. SSS criterion of similarity:
If in two triangles, three sides of one are proportional to the three sides of the other, the triangles are similar.
In ∆XYZ and ∆PQR, \(\frac{XY}{PQ}\) = \(\frac{YZ}{QR}\) = \(\frac{ZX}{RP}\) then ∆XYZ ∼ ∆PQR.
Theorem on Similarity between Triangles
If ∆XYZ is similar to ∆PQR and XM, PN are corresponding medians of the triangles respectively, show that \(\frac{XY}{PQ}\) = \(\frac{XM}{PN}\).
Solution:
In ∆XYM and ∆PQN,
∠Y = ∠Q and \(\frac{XY}{PQ}\) = \(\frac{YM}{QN}\), (since, ∆XYZ ∼ ∆PQR and YM = \(\frac{1}{2}\)YZ, QN = \(\frac{1}{2}\)QR)
Therefore, ∆XYM ∼ ∆PQN
Therefore, \(\frac{XY}{PQ}\) = \(\frac{XM}{PN}\) (Proved)
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