△ABC has vertices A(−6,5), B(−2,5), and C(−6,0). What is the area of △ABC?

Question

20 units^2

14 units^2

9 units^2

10 units^2

Answer 1

10 square units

Plot the 3 points, connect them and they form a right triangle AB is 4 long AC is 5 long

call AB and AC the base and height. Multiply them, divide by 2: 4x5=20 20/2 = 10

AB's endpoints has the same y coordinate.

It's distance is the difference in the x coordinates, -2 minus -6 = 4

AC's endpoints each have the same x coordinate. It's length is the difference in the y coordinates, 5-0=5

BC is the hypotenuse. It's length is the square root of the sum of squares of AB and AC, the square root of 41. 4 squared plus 5 squared = 16+25=41

At that point you could have used Heron's formula, which would be useful if the triangle had not been a right triangle

Calculate half the perimeter: (9+sqr41)/2=d sqr41 = 6.4 approximately. Now calculate sqr[d(d-AB)(d-AC)(d-BC)]

=sqr[(9+sqr41)/2{(9+sqr41)/2 - 4}{(9+sqr41)/2 - 5}{(9+sqr41)//2 - sqr41}]

=sqr[7.7(3.7)((2.7)(1.3)] = sqr(99.9999) = 10.000