All related to triangles Part 1

The area of the rectangle is $48units^{2}$ .

The triangle with hypotenuse AB has area $3\times2\times\frac{1}{2}=3units^{2}$ .

The triangle with hypotenuse AC has area $5\times6\times\frac{1}{2}=15units^{2}$ .

The triangle with hypotenuse BC has area $4\times8\times\frac{1}{2}=16units^{2}$ .

Then by subtracting the other triangle areas we get $48-15-16-3=\boxed{14units^{2}}$ .

Using the Pick's formula $Area=12+6/2-1=14$

It can be solved by means of coordinate geometry.Along X axis and Y axis respectively A(3,6),B(0,4) and C(8,0).then using the formula for determining area in coordinate geometry which is: 0.5(x1y2+x2y3+x3y4-x2y1-x3y2-x4y3) we can determine the area of the inscribed triangle.

Start with the area of the rectangle, 6x8=48 then subtract the areas of the right angled triangles in the corners.

Interior points (Colored Orange = 12) -> i

Border points (Colored Green = 6) -> b

Area = i + 0.5b -1 = 12 + 3 -1 = 14

I used the Shoelace Formula:

Suppose the points are on a Cartesian plane. Then we can represent the points as $A = (3,6), B = (0,4), C = (8,0)$ . Then the Shoelace formula gives us:

$Area = \dfrac{1}{2}\left|(3*4-6*0)+(0*0-8*4)+(8*6-0*3)\right| = 14$

i simply don't have any idea... i just starred at the pic for a while and guessed area of triangle approximately occupies 30% of rectangular area. it gave me 14.4 but i answered 14... luckily that was ryt :D

Found the hypotenuse for each triangle on the outside which gave me the insides coordinates. From there i calculated the area as if it were a rectangle. Then simply divided it by 2. Mathematics brought to you by stoner chris