Try drawing this figure. ABCD would actually be a triangle, rather than the common quadrilateral you would expect - considering there are 4 points plotted. Therefore, we would have to find the distance between respective points ( 0, 1 ) and ( 2, 5 ) / ( 2, 5 ) and ( 7, 0 ) / ( 0, 1 ) and ( 7, 0 ). Let's apply the distance formula and calculate the distance between each point, therefore determining the perimeter.

Distance between points ( 0, 1 ) and ( 2, 5 )

= $\sqrt{(2-0)^2+(5-1)^2}$

= $\sqrt{(2)^2+(4)^2}$

= $\sqrt{4+16}$

= $\sqrt{20}$ = $2\sqrt{5}$

Distance between points ( 2, 5 ) and ( 7, 0 )

= $\sqrt{(7-2)^2+(0-5)^2}$

= $\sqrt{5^2+(-5)^2}$

= $\sqrt{25+25}$

= $\sqrt{50}$ = $5\sqrt{2}$

And you can calculate the distance between points ( 0, 1 ) and ( 7, 0 ) to be exactly $5\sqrt{2}$ as well. Therefore, you can say that the perimeter of this triangle is $2(5\sqrt{2}) + 2\sqrt{5}$ = $10\sqrt{2} + 2\sqrt{5}$ . Note that this is similar to the form we are given. Thus, a = 10, and b = 2 - making it's sum 10 + 2 = 12.

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4 years ago