Question

If i am being honest i dont know how to start

Answer 1

`\text{Here, we need to calculate the lengths of each sides of both triangles. If all the }\\\text{corresponding pairs of sides of these triangles are equal, then the triangles}\\\text{will be congruent by S.S.S. axiom.}`

`\text{We will here use the distance formula to calculate the side lengths of the}\\\text{triangles. Distance formula between two points }(x_1,y_1)\ \text{and }(x_2,y_2)\text{ is given}\\\text{by:}\\\text{D = }√((x_2-x_1)^2+(y_2-y_1)^2).`

`\text{In }\triangle \text{BCD,}\\\text{CB}^2=(-3+6)^2+(4-1)^2=9+9=18\\\text{or, CB}=√(18)\ \text{units}\\\\\text{DB}^2=(-1+6)^2+(2-1)^2=25+1=26\\\text{or, DB}=√(26)\ \text{units}\\\\\text{CD}^2=(-3+1)^2+(4-2)^2=4+4=8\\\text{or, CD}=√(8)\ \text{units}`

`\text{In }\triangle \text{JKL,}\\\text{JL}^2=(-1-0)^2+(2-1)^2=25+1=26\\\text{or, JL}=√(26)\ \text{units}\\\\\text{JK}^2=(-1-2)^2+(8-5)^2=9+9=18\\\text{or, JK}=√(18)\ \text{units}\\\\\text{KL}^2=(2-0)^2+(5-3)^2=4+4=8\\\text{or, KL}=√(8)\ \text{units}`

`\text{Here, DB = JL, CD = KL and JK = CB. All the corresponding sides of }\triangle\text{s}\\\text{JKL and BCD are equal. So }\triangle \text{BCD}\cong\triangle\text{JKL}.`