Triangles Test - 9

Since, AP  bisects ∠BAC.Hence the point P on BC also bisects it as it the opposite side of ∠BAC.

Hence, we can conclude that, BA > BP.

∠A is included between AB and AC and ∠P is included between PQ and PR and corresponding sides must be equal . Since AB = PR, hence AC=PQ for the given triangles to be congruent by SAS axiom.

If △ABC ≅ ΔPQR by SSS congruence rule, then the correspondig sides must be equal i.e AB=PQ ,BC=QR and AC=PR

In △PQR, ∠P=60^∘, ∠Q=50^∘

Now , by angle sum property , ∠ P +∠ Q +∠ R =180^∘

60^o + 50^o + ∠ R = 180^o

or, ∠ R = 180^o - 110^o = 70^o

So, ∠R is the largest angle and the side opposite to it ,i.e, PQ will be the longest side.

By angle sum property,we have,

∠A + ∠B + ∠C = 180^o

=> 50^o + 60^o + ​​​​​​​∠C = 180^o

=>∠C =180^o - (50^o+ 60^o) = 70^o

Therefore, ∠C is the largest angle in the triangle and  the side opposite to it  i.e.  AB is the longest side

According to SAS criterion, if the corresponding sides and their included angles are equal, then the triangles are congruent. Here, in  △AOC and △XYZ, AO = XY,and AC = XZ are the corresponding sides and ∠A=∠X are included angles, Hence,△AOC ≅ △XYZ by SAS.

In a triangle, the sum of any two sides must be  greater than the third side and here 2.3 + 3.1 = 5.4 and hence, the triangle is not possible with the given measurements.