Mathematics, 04.08.2019 06:00 tsbrunson13
Arectangle has a perimeter of 42ft, it's length is 3ft more than twice it's width, great an equation for the perimeter of the rectangle
Answer from: Quest
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.
Answer from: Quest
Iagree that is the answer to the question
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