Find the area of a triangle whose vertices are D(-5,4), O(2,1), and G(6,5). Type only numbers in your answer

Answer 1

This is definitely a DOG of a question. a real DOG. actually real dogs are much nicer.

DO = sqr(7^2 + 3^2) = sqr58

DG = sqr(11^2+1^2) = sqr122

OG = sqr(4^2+4^2) = sqr32

sum them, divide by 2 which = p =[sqr58+sqr122+sqr32]/2

then use Heron's formula to calculate area, p is half the perimeter. then find the product of p times p minus each side. Then take the square root to get area.

A= sqr[(p)(p-sqr58)((p-sqr122)(p-sqr32)]

It should work. But there must be an easier way. Maybe, calculating angles.

Still calculate carefully, you can find the area with Heron's formula, without

calculating the angles.

or just leave it as is. They're all "numbers" as the problem asks. It's a number of a complicated DOG area.