The perimeter of a triangle with vertices `(0,4)a n d(0,0)a n d(3,0)` is `7+sqrt(5)` (b) `5` (c) `10` (d) `12`

Explanation

Let `ABC` be a triangle having `A(0,4);B(0,0);C(3,0) `

Using distance formula, we get

`AB=sqrt((0−0)^2+(0−4)^2)`
`=sqrt(16)=4`

​`BC=sqrt((3−0)^2+(0−0)^2)`
​`=sqrt(9)=3`

​`CA=sqrt((0−3)^2+(4−0)^2)`
​`=sqrt(25)`
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