Question

The coordinates of midpoints of the sides of a triangle ABC are D(2, 1), E(5, 3), F(3, 7). Then sum of lengths of sides of the triangle is

• A. 2($\sqrt{20} + \sqrt{37} + \sqrt{13}$)
• B. 2($\sqrt{10} + \sqrt{35} + \sqrt{15}$)
• C. ($\sqrt{20} + \sqrt{37} + \sqrt{13}$)
• D. ($\sqrt{10} + \sqrt{35} + \sqrt{15}$)

Answer 1

Answer: (A)

2($\sqrt{20} + \sqrt{37} + \sqrt{13}$)

Answer 2

FE || BC ; FD || AC ; DE || AB

Also BC = 2FE, AC = 2FD; AB = 2DE

Now BC = 2. $\sqrt{(5 - 3)^{2} + (3–7)^{2}}$= 4$\sqrt{5}$

Similarly, AC = 2$\sqrt{37}$; AB = 2$\sqrt{13}$

AB + BC + CA = 2($\sqrt{20} + \sqrt{37} + \sqrt{13}$)