(16.) GI = 23x - 33
KI = 6x + 22
Solve for x

The "x" value is 7

Step-by-step explanation:

If the shape in the image is a rhombus then we know that GI is consisted of two KI's. So now we can write the following equation

GI = (2)(KI)

23x - 33 = (2)(6x + 22)

23x - 33 = 12x + 44

23x = 12x + 77

11x = 77

x = 7

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.