The longest side of an acute triangle measures 30 inches. The two remaining sides are congruent, but their length is unknown.

What is the smallest possible perimeter of the triangle, rounded to the nearest tenth?

A. 41.0 in.
B. 51.2 in.
C. 72.4 in.
D. 81.2 in.



Answer :

To determine the smallest possible perimeter of an acute triangle with the longest side measuring 30 inches and the two remaining sides being congruent, we follow these steps:

1. Let the length of each congruent side be denoted as [tex]\( x \)[/tex].

2. Since the triangle is acute, it must satisfy the property of acute triangles: the square of the longest side must be less than the sum of the squares of the other two sides. Mathematically, this is expressed as:
[tex]\[ (30)^2 < 2x^2 \][/tex]

3. Simplify the inequality:
[tex]\[ 900 < 2x^2 \][/tex]

4. Divide both sides of the inequality by 2:
[tex]\[ 450 < x^2 \][/tex]

5. Take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x > \sqrt{450} \][/tex]

6. Compute [tex]\( \sqrt{450} \)[/tex], which is approximately [tex]\( 21.213203435596427 \)[/tex].

7. The smallest possible value for the congruent sides is hence [tex]\( x \approx 21.213 \)[/tex].

8. The perimeter of the triangle is the sum of all its sides. Therefore, the perimeter is:
[tex]\[ 30 + 2x \][/tex]
Substituting the computed value of [tex]\( x \)[/tex] into the equation:
[tex]\[ 30 + 2 \times 21.213203435596427 \approx 30 + 42.42640687119285 \approx 72.42640687119285 \][/tex]

9. Finally, round the perimeter to the nearest tenth:
[tex]\[ 72.42640687119285 \approx 72.4 \][/tex]

Thus, the smallest possible perimeter of the triangle, rounded to the nearest tenth, is [tex]\( 72.4 \)[/tex] inches.

Therefore, the correct answer is [tex]\( 72.4 \)[/tex] inches.

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