Practicing Success
A quadrilateral has vertices in the order (0, -1), (6, 7), (-2,3) and (8, 3). The quadrilateral is a : |
Trapezium Square Rhombus Rectangle |
Rectangle |
Let's start naming these points , A( 0, -1), B(6 , 7) , C ( -2, 3 ) and D ( 8 , 3 ) Now, AD = \(\sqrt { ( 8 - 0)² + ( 3 + 1 )² }\) = \(\sqrt { 64+ 16 }\) = \(\sqrt {80 }\) = 4\(\sqrt { 5 }\) BC = \(\sqrt { ( - 2 - 6 )² + ( 3 - 7 )² }\) = \(\sqrt { 64+ 16 }\) = \(\sqrt {80 }\) = 4\(\sqrt { 5 }\) AC = \(\sqrt { ( -2 - 0)² + ( 3 + 1 )² }\) = \(\sqrt { 4+ 16 }\) = \(\sqrt {20 }\) = 2\(\sqrt { 5 }\) And BD = \(\sqrt { ( 8- 6)² + (3 - 7 )² }\) = \(\sqrt { 4 + 16 }\) = \(\sqrt {20 }\) = 2\(\sqrt { 5 }\) That means , AD = BC and AC = BD Hence , ABCD is a parallelogram . Now, AB = \(\sqrt { ( 6 - 0)² + ( 7 + 1 )² }\) = \(\sqrt { 36 + 64 }\) = \(\sqrt {100 }\) = 10 CD =\(\sqrt { (8 + 2 )² + ( 3 - 3 )² }\) = \(\sqrt { 100 + 0 }\) = 10 That means , AB = CD Now, AB² = AD² + DB² 10² = ( 4\(\sqrt { 5 }\) ) ² + ( 2\(\sqrt { 5 }\) ) ² 100 = 80 + 20 100 = 100 ( followed ) Similarly , CD² = CB² + BD² 10² = ( 4\(\sqrt { 5 }\) ) ² + ( 2\(\sqrt { 5 }\) ) ² 100 = 80 + 20 100 = 100 ( followed ) Hence it is concluded that :- ABCD is a rectangle. |