A pentagon ABCDE has vertices located at A(-2, 6), B(-4, 3), C(-1, 1), D(2, 2), and E(2, 6). Max thinks that AE = DE. Which of the following is true? A. Max is incorrect. AE = 8 units, and DE = 4 units.
B. Max is incorrect. AE = 4 units, and DE = 8 units.
C. Max is correct. AE = 8 units, and DE = 8 units.
D. Max is correct. AE = 4 units, and DE = 4 units.

am busy Now I will do that later

5/6

step-by-step explanation:

if i remember correctly with perpendicular stuff you just take away the negative

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.