What is the side length of the smallest square plate on which it 34-cm chopstick can fit along a diagonal without any overhang. round to the nearest tenth.

please i need help :((

24.04

Step-by-step explanation:

Let us set x as the side length of the square plate.

x² + x² = 34²

2x²= 1156

x² = 578

x ≈ 24.04

√578  = 24.0cm

Step-by-step explanation:

sin x = opp/hyp    sin (45)* 34 = 24.0416306 = 24.0cm to one tenth This is because diagonals of a square intersect (cross) in a 90 degree angle. This means that the diagonals of a square are perpendicular. The diagonals of a square are the same length (congruent).    You can check this with Pythagoras Theorem afterwards -  34^2 - 24.0416306^2 = 24.0416306^2  =  sqrt 1156 - sqrt 578 = sqrt 578

x intercept = (-1,0)   (4,0)

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.