Area of a triangle (hard)

Let us consider OA as base.

then, length of base = $\sqrt { { 6 }^{ 2 }+{ 3 }^{ 2 } }$ = $\sqrt { 45 }$ .

slope of base = 3/6 = 1/2.

As we have to find the length of perpendicular onto the base,

slope of BD = -1/(slope of base) = -1/(1/2) = -2.

using slope-point form,

Equation of BD, 2x + y = -5.

Equation of base ( using slope-point form), x - 2y = 0.

Their intersection is (-2,-1).

thus distance BD = $\sqrt { { 6 }^{ 2 }+{ 3 }^{ 2 } }$ = $\sqrt { 45 }$ .

Thus area of the triangle is 45/2 = $\boxed{22.5}$

total area = 10 * 6 =>60

area(1)= $\frac{1}{2}$ *3 * 6 => 9

area(2)= $\frac{1}{2}$ *7 * 1 => 3.5

area(3)= $\frac{1}{2}$ *5 * 10 => 25

=>s=60 - 25 - 3.5 - 9 => 22.5

$\pm \frac{1}{2} \begin{bmatrix} 0 & 0 \\ 6 & 3 \\ 1 & -7 \\ 0 & 0 \end{bmatrix}=22.5$