Question

find the perimeter of the triangle formed by the points (3, 5), (4, 8) and (5, 6)

• A. $\sqrt{5}(2+\sqrt{2})$
• B. $\sqrt{3}(2+\sqrt{2})$
• C. $\sqrt{2}(5+\sqrt{3})$
• D. $\sqrt{5}(\sqrt{2}+\sqrt{3})$

Answer 1

The correct option is A $\sqrt{5}(2+\sqrt{2})$

Let $A(3,5);B(4,8);C(5,6)$
Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{{(x_2-x_1)}^2+{{(y}_2-y_1)}^2}$
So, length of side $AB$ is $\sqrt{{(4-3)}^2+{(8-5)}^2}=\sqrt{1+9}=\sqrt{10}$
Length of side $BC$ is $\sqrt{{(5-4)}^2+{(6-8)}^2}=\sqrt{1+4}=\sqrt{5}$
Length of side $AC$ is $\sqrt{{(5-3)}^2+{(6-5)}^2}=\sqrt{4+1}=\sqrt{5}$
Thus, perimeter of triangle $ABC=\sqrt{10}+\sqrt{5}+\sqrt{5}=2\sqrt{5}+\sqrt{10}=\sqrt{5}(2+\sqrt{2})$