Find the perimeter of the rectangle formed by the joining the points $A (- 3, - 1), B (5, - 1), C (5, 3)$ and $D (- 3, 3)$ in sides.

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Solution

Using distance formula $= \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$
Perimeter of rectangle $A B C D = A B + B C + C D + D A$
$A B = \sqrt{{(5 - (- 3))}^{2} + {(- 1 - (- 1))}^{2}} = \sqrt{{(5 + 3)}^{2} + {(- 1 + 1)}^{2}} = \sqrt{64 + 0} = 8$ units
$B C = \sqrt{{(5 - 5)}^{2} + {(3 - (- 1))}^{2}} = \sqrt{0 + {(3 + 1)}^{2}} = \sqrt{0 + 16} = 4$ units
$C D = \sqrt{{(- 3 - 5)}^{2} + {(3 - 3)}^{2}} = \sqrt{{(- 8)}^{2} + 0} = \sqrt{64} = 8$ units
$D A = \sqrt{{(- 3 - (- 3))}^{2} + {(3 - (- 1))}^{2}} = \sqrt{{(- 3 + 3)}^{2} + {(3 + 1)}^{2}} = \sqrt{0 + {(4)}^{2}} = \sqrt{16} = 4$ units
$∴$ Perimeter of $A B C D = 8 + 4 + 8 + 4 = 24$ units

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