I'd say B after an insane approximation, but....drawing it, I got:
Basically, the hypothenuse goes from (0,1.4) to (5,0). Using it as the base, since it's a diagonal respectively to the x axis, it's a little more than 5 units long.
The height would be from point (2, sqr of 10) to a 90 degree angle on the hypothenuse.
Again, using the fact that the hypothenuse is a diagonal, we can approximate to the height being a little more than 2.
Overall, Height*Base can be approximated to slightly more than 10, so /2 would be closest to 6. Definitely not 8 or 4.

There's probably a better way to solve this...

If the vertices of triangle are given,area of triangle is,

[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]/2

sanju09 wrote:In the x y-plane, which of the following is closest to the area of a triangle whose vertices are (√2, 0), (2, √10), and (5, 0)?
(A) 4
(B) 6
(C) 8
(D) 11
(E) 14

Any side of a triangle can be considered the base. Each base has a corresponding height. (So a triangle has three bases, and each base has a corresponding height.) To find the height that corresponds to a given base, draw a line from the vertex opposite the base so that it creates a right angle with the base.

In the problem above, one side of the triangle lies along the X axis. Let's call this side the base. The length of the base will then be the difference of the x values:

b = 5 - √2 = 5 - 1.4 = 3.6 approximately.

The corresponding height will be a line drawn from (2, √10) to the x axis so that the line and the x axis form a right angle. The length of this height will be √10.

h = √10 = 3.2 approximately.

So Area = 1/2(b)(h) = 1/2 * (3.6) * (3.2) = 5.8 approximately.

The correct answer is B.

@Guru.. nice explanation