The length of one of the equal sides of an isosceles triangle is 4 cm. If its base is 2 cm, then what is its area?

A = 1/2bh

b = 2cm

h will be found using the side of the triangle with Pythagoras theorem

h² + 1² = 4²

h² = 15

h = √15 cm

then,

A = 2*√15

2

A = √15cm²

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.

these sign is used to tell weather the number is greater or less than the number   <   or >     but this sign means its equal =    

hope this you and

step-by-step explanation: