Question

Area of traingle with verticals (-3,4) (1,-2), and (5,5) is in sq unit

Answer 1

Answer:

23.37 unit²

Step-by-step explanation:

a = $\sqrt{(1-5)^2+(-2-5)^2}$ = $\sqrt{65}$

b = $\sqrt{(-3-1)^2+(4+2)^2}$ = $\sqrt{52}$

c = $\sqrt{(-3-5)^2+(4-5)^2}$ = $\sqrt{65}$

ABC is isosceles triangle. "h" to "b" going through midpoint.

coordinates of midpoint are ($\frac{-3+1}{2}$, $\frac{4-2}{2}$)

h = $\sqrt{(-1-5)^2+(1-5)^2}$ = $\sqrt{42}$

Area of Δ = bh/2

√52×√42÷2 = $\sqrt{42 * 52}$ / 2 = 23.37 unit²