## Trigonometry (11th Edition) Clone

$Area = 25$
Let $a$ be the length of the side from (0,0) to (3,4). We can find $a$: $a = \sqrt{3^2+4^2} = 5$ Let $b$ be the length of the side from (3,4) to (-8,6). We can find $b$: $b = \sqrt{[(-8)-3]^2+(6-4)^2} = \sqrt{125}$ Let $c$ be the length of the side from (-8,6) to (0,0). We can find $c$: $c = \sqrt{(-8)^2+6^2} = 10$ We can verify that this is a right angle triangle: $a^2+c^2 = 5^2+10^2 = 125 = b^2$ Thus, $b$ is the hypotenuse of this right angle triangle. Let angle B be the angle of $90^{\circ}$ We can find the area of the triangle: $Area = \frac{1}{2}ac~sin~B$ $Area = \frac{1}{2}(5)(10)~sin~90^{\circ}$ $Area = \frac{1}{2}(5)(10)(1)$ $Area = 25$