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.
If we assume the hypotenuse is of side length 5:
a^2+b^2=c^2
3^2+b^2=5^2
9+b^2=25
b^2=16
b=4
If the hypotenuse is the unknown:
a^2+b^2=c^2
3^2+5^2=c^2
9+25=c^2
c^2=34
c=(34)^(1/2)
I believe it equals -73 1/8.
