Isosceles right triangle abc has a hypotenuse, ab, with a length of 18 feet. find the exact lengths of the other two sides, ac and bc. ac = a0√ a1 feet bc = a2√ a3 feet to the nearest square foot, find the area of abc . area of abc = a4 square feet

we know that

if ABC is an isosceles right triangle

then

side AC=side BC

angle A=angle B=45 degrees

cos B=adjacent side angle B/hypotenuse

adjacent side angle B=BC

hypotenuse=AB> 18 ft

angle B=45 degrees

cos 45°=(√2)/2

so

cos 45°=BC/AB> solve for BC

BC=AB*cos 45> BC=18*(√2)/2> BC=9√2 ft

AC=BC> AC=9√2 ft

the answer part 1) is

the exact lengths of the two sides, AC and BC is

AC=9√2 ft

BC=9√2 ft

Part b) Find the Area of triangle ABC

Area=b*h/2> AC*BC/2> (9√2)*(9√2)/2> 81 ft²

the answer part b) is

the area of triangle ABC is equal to 81 ft²

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.

b i think

step-by-step explanation:

not really sure