Question

16. Find the perimeter of the triangle obtained by joining (3 , 1) , (6 , 4) and (8 , 6).

Answer 1

The three vertices of the triangle are (3,1), (6,4) and (8,6).

Let, A, B, and C are those vertices respectively.

So, the length of AB = $\sqrt{(6-3)^{2} + (4-1)^2}=3\sqrt{2}$ units.

Length of BC =  $\sqrt{(8-6)^2+(6-4)^2} = 2\sqrt{2}$units.

And the length of AC =  $\sqrt{(8-3)^2+(6-1)^2} = 5\sqrt{2}$units.

Therefore, the perimeter of the triangle will be

$(5\sqrt{2}+2\sqrt{2}+3\sqrt{2})$units

= $10\sqrt{2}$ units. (Answer)

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