MATH SOLVE

7 months ago

Q:
# 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?

Accepted Solution

A:

Answer: greater than 5.7 cm

Explanation:

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

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 90Β°, 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 = β32

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

Explanation:

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

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 90Β°, 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 = β32

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