The perimeter of the triangle with vertices $(1, 3), (1, 7)$ and $(4, 4)$ is

3 + \sqrt{2}

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3 \sqrt{2}

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6 + 3 \sqrt{2}

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9 + \sqrt{2}

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Solution

The correct option is B $6 + 3 \sqrt{2}$
Assume $A (1, 3), B (1, 7), C (4, 4)$
Formula of distance $=$ $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2})}$
Distance of $A B =$ $\sqrt{(1 - 1)^{2} + (7 - 3)^{2}}$
$=$ $\sqrt{0 + 4^{2}}$
$= 4$
Likewise calculate the distance for $B C$ and $C A$ .
Distance of $B C =$ $\sqrt{(4 - 1)^{2} + (4 - 7)^{2}} = 3 \sqrt{2}$
Distance of $C A =$ $\sqrt{(1 - 4)^{2} + (3 - 4)^{2}} = 2$
Perimeter $=$ distance $A B$ $+$ distance $B C$ $+$ distance $C A$
$=$ $4 + 3 \sqrt{2} + 2$
$=$ $6 + 3 \sqrt{2}$

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The perimeter of the triangle with vertices (1,3),(1,7) and (4,4) is

The perimeter of the triangle with vertices $(1, 3), (1, 7)$ and $(4, 4)$ is