The perimeter of a triangle with vertices (0,4), (0,0) and (3,0) is

3 + \sqrt{5}

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\sqrt{13}

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Solution

The correct option is C 12
According to the figure
$O A = 4 u n i t s, O B = 3 u n i t s$
$A B = \sqrt{(O A)^{2} + (O B)^{2}}$
$A B = \sqrt{(4)^{2} + (3)^{2}}$
$A B = \sqrt{16 + 9}$
$A B = \sqrt{25} = 5 u n i t s$
$∴$ Perimeter of the triangle $= 4 + 3 + 5 = 12 u n i t$

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