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| 1. |
Find the perimeter of triangle with vertices (0,4)(0,0)and (3,0) |
| Answer» let the three vertices of a triangle be A(0,4) , B (0,0) and C (3,0).To find perimeter of a triangle , we have to find the length of sides AB, BC and AC.Length of AB ={tex}√(0-0)^2+(4-0)^2{/tex}AB =\xa0{tex}√(0)^2+(4)^2{/tex}AB =\xa0{tex}√0+16{/tex}AB=\xa0{tex}√16{/tex}AB=4\xa0Similarly ,Length of BC =\xa0{tex}√(3-0)^2+(0-0)^2{/tex}BC= √9BC=3Length of AC = √({tex}0-3)^2+(4-0)^2{/tex}AC= √{tex}(3)^2+(4)^2{/tex}AC =\xa0{tex}√9+16{/tex}AC = √25AC= 5So, perimeter = Sumof lengths of AB, BC and ACperimeter = AB+BC+ACperimeter = 4+3+5perimeter = 12\xa0 | |