If vertices of a quadrilateral are A(0,0), B( | vedclass

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(D) Calculate the side lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.$AB = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9+16} = 5$.$BC

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Rhombus

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(D) Calculate the side lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.$AB = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9+16} = 5$.$BC = \sqrt{(7-3)^2 + (7-4)^2} = \sqrt{4^2+3^2} = 5$.$CD = \sqrt{(4-7)^2 + (3-7)^2} = \sqrt{(-3)^2+(-4)^2} = 5$.$AD = \sqrt{(4-0)^2 + (3-0)^2} = \sqrt{4^2+3^2} = 5$.Since all sides are equal,it is a rhombus or a square.Calculate the diagonals:$AC = \sqrt{(7-0)^2 + (7-0)^2} = \sqrt{49+49} = \sqrt{98} = 7\sqrt{2}$.$BD = \sqrt{(4-3)^2 + (3-4)^2} = \sqrt{1^2+(-1)^2} = \sqrt{2}$.Since $AC 
eq BD$,the diagonals are not equal.Therefore,$ABCD$ is a rhombus.

If vertices of a quadrilateral are $A(0,0), B(3,4), C(7,7)$ and $D(4,3)$,then the quadrilateral $ABCD$ is

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