The longest side of an acute isosceles triangle is 8 centimeters . rounded to the nearest tenth, what is the smallest possible length of one of the two congruent sides?

greater than 5.7 cm

Explanation:

1) The acute angle is restricted to an upper bound of to 90°, i.e. the greatest value of an acute angle must be less than 90°.

2) the smaller the acute angle the longer the two congruent sides of the isosceles triangle, the greater the acute angle the shorter the two congruent sides.

3) Find the lower bound of the two congruent sides by using the upper bound of the acute angle.

4) when the angle is 90°, you can use the Pytagorean theorem

c^2 = a^2 + b^2

c = 8 cm, a = b

c^2 = 2a^2

8^2 = 2a^2

2a^2 = 64

a^2 = 64/2

a^2 = 32

a = √32

a = 5.7 cm

5.7 cm is the lower bound of the length of the two small sides, therefore, the length of the two small sides has to be greater than 5.7 cm

Are you trying to figure out the volume?