Find out if the triangles with vertices (2,6) , (1,1) (3,4) and (1,3), (4,8) and ( 2,4) are similar or not.

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william1941 | College Teacher | (Level 3) Valedictorian

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For the triangle with the vertices (2, 6), (1, 1) and (3, 4), the length of the sides are

sqrt [(2-1) ^2+ (6-1) ^2] = sqrt [1+25] = sqrt 26

sqrt [(3-1) ^2+ (4-1) ^2] = sqrt [4+9] = sqrt 13

sqrt [(3-2) ^2+ (6-4) ^2] = sqrt [1+4] = sqrt 5.

So the lengths of the sides are sqrt 26, sqrt 13 and sqrt 5

For the triangle with the vertices (1, 3), (4, 8) and (2, 4), the length of the sides are

sqrt [(4-1) ^2+ (8-3) ^2] = sqrt [9+25] = sqrt 34

sqrt [(2-1) ^2+ (4-3) ^2] = sqrt [1+1] = sqrt 2

sqrt [(4-2) ^2+ (8-4) ^2] = sqrt [4+16] = sqrt 20.

Now, we do not find the ratio of the length any two sides of the triangles equal to the ratio of any two other sides. For similar triangles the ratio of the lengths of the sides is equal for the two triangles, therefore we can say that the two triangles are not similar.

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neela | High School Teacher | (Level 3) Valedictorian

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Triangles with vertices A(2,6) ,  B(1,1)  C(3,4) and P(1,3), Q(4,8) and R( 2,4) to be verified whether similar or not.

We find the lengths of the sides . We write them in increasing orderof  the sides according to their lengths in each of the triangles. We see whether there is the same  ratio between length of each side of the one to that of another triangle:

AB^2 = (2-1)^2+(6-1)^2  = 1+25=  26. So AB = sqrt 26.

AC^2 = (2-3)^2+(6-4)^2 = 1+4 = 5 , AC = sqrt 5

BC^2 = (1-3)^2+(1-4)^2 = 4+9 = 13. BC = sqrt13.

PQ^2 = (1-4)^2+(3-8)^2 = 9+25=34, PQ = sqrt34

PR^2 = (1-2)^2+(3-4)^2 = 1+1 =2. PR = sqrt 2

QR^2 = (4-2)^2+(8-4)^2 = 4+16 = 20, QR = sqrt20.

Therefore , sides of the rwo triangles in the incresing orderare:

(sqrt5 , sqrt13 , sqrt26) and (sqrt2 , sqrt20 , sqrt34).

For similarity, the correspoding sides of the triangles  should bear the  same ratio.

 sqrt5/sqrt2  and  sqrt13/sqrt20  and  sqrt20/sqrt34 are all different.

So the triangles are not similar.

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